BUSINESS CALCULUS Notebook: Practicing Calculus, Statistics, and Technology [2 ed.]
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Table of contents :
About the author
Some words from the author
Juan José Prieto-Valdes
Preface
Wolfram Alpha and Calculus Course Assistant.
Microsoft Math. Add-in for Word Office.
Microsoft Excel
Desmos Graphing Calculator- DG
GeoGebra-GG
https://www.geogebra.org/
Symbolab-Sy
https://www.symbolab.com/
Math Solver (Matrix Calculator)-MS
https://www.math10.com/scripts/matrices/determinant-matrix-calculators.html
LaTeX Editor
CHAPTER-1. LIMITS
1.1. The Idea of a Limit
Examples of Limit Evaluation, an introductory view about the evaluation tactics:
1. Find the limit ( value) as approaches ϓ of the linear equation: =ϑ+ϕ
2. Find the limit ( value) as approaches ϐ of the quadratic function: =ϐ,-ϐ.+ϓ1+ϐ
3. Find the limit ( value) as approaches ϑ of the rational equation: =,ϐ-ϑ+.
4. Prove that ,-→ϒ. ,+ϑ-ϓ+,..=Ϗ
5. Find the limit as approaches Ϗ of the rational expression using numeric and graphic interpretation:
6. Find the limit as approaches ϒ of the rational expression:
7. Estimate the value of the limit as approaches ώ of the rational expression using numeric and graphic interpretation:
8. Estimate the value of the limit when the function exists in one side, using graphic interpretation:
9. Estimate the value of the limit when the function exists in one side, use graphic interpretation:
1.2. Formal Limit Definition
Steps for proving that exist:
Steps for proving that ,-→. ,.=∞ exist (Two-Side Infinite Limit):
Example 1. Finding ̹ for a given ̺:
Example 2. Finding ̹ for a given ̺:
Example 3. Prove that ,-→−Ϗ. ,,-ϐ.+ϑ.=ϒ.
1.3. Determining Limits and Algebraic Properties of Limits
Examples (accurate the limits evaluation applying some algebra simplification when possible:
Homework Topics 1.1, 1.2, and 1.3
1.4. Continuity, Intermediate Value Property, and One-Side Limits.
1.4.1. Continuity:
Some well-known facts related to continuity:
1.4.2. Intermediate Value Property of Continuous Function.
1.4.3. Bisection Algorithm for Approximating Roots
Example:
Method-1 (inaccurate approximation):
Method-2 (accurate approximation using bisection method):
1.4.4. Examples of Discontinuous Function and one Side Limit:
1.5. Limits at the Infinity (Analyzing also One Side Limit)
Consider the function ()=,ϐώ,-ϐ.++ϐ-,-ϐ.. for “large” values of x.
Examples
Homework Topic 1.4 and 1.5.
In 31-34 exercises find one-side limit, verify your answer with Mathematics or Wolfram Alpha.
In the next exercises decide if the given function is continuous at the given point.
In the next exercises find the limit at the infinity
1.6. Example of Limit Calculation Using Technology
PROJECT-1. Find the Limit Using technology.
You can print the Project from Professor Prieto-Valdes Web-page.
Example 1
Example 2.
Example 3.
Example 4:
Example 5:
Example 6:
PRACTICING QUIZ-1. Limits
CHAPTER-2. THE DERIVATIVE AND DIFFERENTIATION
2.1. Definition of Derivative
2.1.1. Linear Function and Slope of the Line:
2.1.2. Slope of the Secant Line
,=,∆-∆.=-,,+∆.−,.-∆..
2.1.3. Slope of the Tangent Line:
The next illustration serves to represent the derivative of the parabola function in three different positions.
Example 1: Find the derivative of ,,.=ϑ+ -ϒ.
Example 2: Compute the derivative of ,.=ϒ,-ϐ.−ϕ at =ϑ:
Example 3: Find the equation of the line tangent to the graph of ,.=,Ϗ-. at =ϐ
Example 4: Find the derivative of ,.=,.
Example 6: Graph the function ,.=,-ϑ.−ϒ , and its derivative ,-′.,.=ϑ,-ϐ.−ϒ.
Solve by yourself using the definition of the derivative.
Them, use Wolfram Alpha to interpreter the answer.
To succeed in this topic, you should understand
Homework Topic 2.1
PROJECT-2. The Derivative Using Limit Definition
Example 2. Compute the derivative of ,.=ϒ,-ϐ.−ϕ at =ϑ:
Example 3. Find the equation of the line tangent to the graph of ,.=,Ϗ-. at =ϐ
Complete the following exercises using limit definition
2.2. Techniques of differentiation.
2.3. Product and Quotient Rules
Example (product rule) 1
Example (quotient rule) 2:
Homework Topic 2.2 and 2.3.
2.4. The Chain Rule
Definition of the Chain Rule or differentiation of a composite function:
Example 1: Given: ,,.=-,,(-ϐ.−ϓ+ϓ)-ϑ.., find ,-′.,..
Example 2: =,+,-..
Example 3: =,,Ϗ−ϑ.-ϑ.,,ϐ,-ϐ.+ϑ.-ϓ.
Example 4: =,-ϐ.(,%'-(,-ϐ.)).
2.5. Implicit Differentiation.
2.6. High Order Derivatives (second, third, …):
Homework Topics 2.4, 2.5, and 2.6.
PRACTICING QUIZ-2. Derivatives
2.7. Real Case Examples (Marginal Analysis and Approximations)
To succeed in these topics, you should understand
Homework Topic 2.7.
2.8. Increasing and Decreasing Functions (critical points)
To succeed in these topics, you should understand
2.9. Rolle's Theorem and The Mean Value Theorem
Proof:
Example: 1
Example 2.
Homework Topic 2.8 and 2.9.
2.10. Concavity and Point of Inflexion
Sufficient condition of concavity (convexity) of a function:
Homework Topic 2.10.
PRACTICING QUIZ-3. Application of Derivatives
CHAPTER-3. EXPONENTIAL and LOGARITHMIC FUNCTIONS. OPTIMIZATIONS.
3.1. Remembering Exponential and Logarithmic Functions (an Algebra Topic):
3.1.2. Exponential Functions.
3.1.3. Logarithmic Functions.
Examples:
3.1.4. Properties of Logarithms:
3.1.5. Exponential and Logarithmic Equations
3.1.6. Compound Interest. Periodic and Continuous Compounding Interest:
Homework Topic 3.1
Exponential Functions.
Logarithmic Functions and Properties of Logarithms.
Exponential and Logarithmic Equations.
Compound Interest.
Additional Homework Topic 3.1.5-plus:
Exponential Growth and Decay exercises:
3.2. Differentiation of Exponential Functions
3.3. Differentiation of Logarithmic Functions
Homework Topic 3.2 and 3.3
Differentiate given expression
In exercises from 198 to 201 find an equation for the tangent line to =,.at the specified point:
3.4. L’ Hospital Rule / Revisiting Limits
How L’Hôpital’s Rule works?
Homework Topic 3.4.
3.5. Optimization and Business Applications Examples.
3.5.1. Application to Volume optimization
3.5.2. Application to Analytic Geometry:
3.5.3. Application to evacuation situation.
3.5.4. Application in Electrical Engineering:
3.5.5. More about Related Rate of Change in time.
Homework Topic 3.5.
PRACTICING QUIZ-4. Optimization, Exponential and Logarithmic Functions.
PROJECT-3. Quick Example of a Business Profit Optimization
I. General Information:
II. Specific Information/Data:
III. Analysis
IV. Conclusion
CHAPTER-4. INTEGRATION
4.1. Indefinite Integration and Simple Differential Equation
The General Power Rule for Integration (indefinite integral) and other rules for common functions:
4.2. Integration by substitution
Homework Topic 4.1 and 4.2.
Find the indicated integral. Apply substitution method.
4.3. Riemann Integration / Riemann Sum.
Homework Topic 4.3
4.4. Fundamental Theorem of Calculus and Definite Integral
Properties of Definite Integral
4.5. Application of Integrals (Average Function Value).
Homework Topic 4.4 and 4.5.
4.6. Application of Integrals (Area Under and Between Curves).
The definite integral can be used to find the area between a graph curve and the ‘e’ axis, between two given e values. This area is called the ‘area under the curve’ regardless of whether it is above or below the ‘e’ axis.
To succeed in this topic, you should understand
Homework Topic 4.6
PRACTICING QUIZ-5. Integrals and its applications
PROJECT – 4. Optimization of the Material Cost.
The Problem:
Analysis
First Conclusion:
Now let us work with the legs:
PROJECT – 5. Profit Optimization (Transport Marine Company)
PROJECT – 6. Optimization of fictitious city evacuation.
CHAPTER 5. INTRODUCTORY STATISTICS WITH CALCULUS.
5.1. Continuous Random Variable, Probability, and Histograms.
5.2. Probability Density Functions: Uniform, Exponential, and Normal.
5.3. Uniform Density Function
5.4. Exponential Density Functions
5.5. Normal Distribution
5.6. Mean and Median.
5.7. Variance and Standard Deviation
Now Let’s review some statistical properties of variance and deviation:
5.8. Variance in different types of distributions:
1. Case Study Uniform Distribution
2. Case Study Exponential Distribution
3. Case Study Normal Distribution
Homework Chapter 5
Homework and practicing exercises at Prieto-Valdes web-page
PROJECT – 7. Portfolio Optimization
Application of previously developed template:
REVIEW CHAPTER.
R-1. Some Elementary Algebra Concepts
R-1.1. Common Factors
R-1.2. Exponents
R-1.3. Roots; Products, Addition, and Subtraction Rules for Square Roots
R-1.4. Rationalizing the Denominators
R-1.5. Polynomials, Addition/Subtraction, Multiplication, Special Products, and Factoring
R-1.6. The Product of Two Binomials (Foil Method)
R-1.7. Special Product Rules
R-1.8. Greatest Common Factor (GCF) and Factoring by Grouping
R-1.9. Factoring Trinomials in the Form ,-ϐ.++ and ,-ϐ.++ where ≠Ϗ
R-1.10. Rational Expressions, Multiplication and Division:
R-1.11. Addition/subtraction. Finding the Low Common Denominator (LCD)
R-1.12. Compound Rational Expressions
R-2. Algebra Review
R-2.1. Lines, Slope, and Graphing Linear Functions.
R-2.2. Solving Linear Equations
R.2.3. Rational Equations
R-2.4. Complex Numbers.
Complex numbers are combination of real and imaginary number parts:
R-2.5. Quadratic Equation.
R-2.6. Radical Equations.
R-2.7. Other Types of Equations
R-2.8. Absolute Value Equations & Inequalities.
R.2.9. Factoring Before Calculus.
R-2.11. Functions and Their Graph.
R-2.12 Graphing Technics and Transformation of Functions.
R-2.13. General Characteristics of Functions
R-3.14. Combination and Composition of Functions.
R-2.15. Quadratic Functions.
The relation ,-.=−,-ϐ. also can be obtained using the middle point between zero solutions,-Ϗ,ϐ.:
Example: When no intercepts with axis X (left) and with interception (right)
R-2.16. Elementary Properties and Graph of Polynomial Functions.
End Behavior
Continuous and discontinuous function:
Multiplicity, an important attribute of the polynomial functions:
Number of solutions and turning points on the graph
Examples:
R-2.17. The Real Zeros of a Polynomial.
FACTOR THEOREM:
Proof:
Equivalencies:
Example:
Examples using ,-.=Ϗ. In this case is a possible factor of ,-ώ., then is a solution.
Examples:
Model Example: Finding the zeroes of a polynomial.
R-2.18. Complex Zeros and the Fundamental Theorem of Algebra
Steps for Finding the Real Zeros of a Polynomial Function. When graphing a Polynomial Function is convenient to follow the following steps:
R-2.19. Property and Graph of Rational Functions.
Descriptive examples:
R-2.21. Polynomial and Rational Inequalities
Tentative course content / schedule of regular course
Citation preview
Juan J. Prieto-Valdés
BUSINESS CALCULUS Notebook
2nd Edition
Practicing Calculus, Statistics, and Technology.
YOURS NAME: _____________________________________________________________
April 2021
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Business Calculus Notebook
About the author Juan Jose Prieto-Valdes (Juan Prieto); An accomplished educator, committed to student success; under whose guidance and supervision a large number of students have been prepared for careers, new programs, and a broad group of projects has been implemented. Prieto-Valdes holds a Doctor of Philosophy Degree in Physical & Mathematical Sciences. Over 20 years of educational experience teaching at undergraduate and graduate levels, advisor in MSc and PhD Thesis and Dissertations, author of more than 100 academic papers with internationally recognized results in materials research, environmental control technologies, math modeling, and business applications. Dr. Prieto-Valdés practices fine art and literature hobbies, he owns a valuable number of artistic copyrights, however, based in his own words: one of his preferred activities is to teach mathematics, mostly, using real application examples.
Some words from the author In this notebook is intended to improve business calculus learning conferring special attention to the today data driven society. This is not a regular textbook, it was firstly prepared to discuss new trends in teaching calculus at the international conference, and later was adopted to improve traditional business calculus course. In addition to traditional calculus topics, a group of new challenging computer-aided exercises are included; Microsoft Mathematics and Excel are introduced to solve calculus problems, as well as currently online available software and engines as referred at the preface pages. Elements of the big data analytics are also included and new calculus-statists basic overview for business application. A group of unpublished before exercises served to teach how to optimize the profit, the material consumption, the traffic light timeline intervals, the optimal route of evacuation, stocks, portfolio and other real case examples. I structured this book as a working notebook, leaving some free spaces to complete the lecture contents and to perform the homework practice. I have observed that when the students take their notes in the same place where the subject is exposed, they had better organized their questions and thoughts, notably improving learning process. I have to confess that I spend a lot of time organizing the material, creating new problems and formulating real live examples, all of them firstly published hereto. Truly yours;
Juan José Prieto-Valdes BUSINESS CALCULUS Notebook, 2nd Edition Practicing Calculus, Statistics, and Technology. ISBN: 9798729023608 All Rights Reserved. The Cataloging-in-Publication Data was filled at the US Library of Congress Included materials: Business Calculus. © 2014, Juan J. Prieto-Valdes, Library of Congress, Registration: TXu001954964 Pre-Calculus. © 2016, Juan J. Prieto-Valdes, Library of Congress, Registration TXu2014602 Trigonometry. © 2015, Juan J. Prieto-Valdes, Library of Congress, Registration TXu0019744949 College Algebra. © 2013, Juan J. Prieto-Valdes, Library of Congress, Registration TXu001880424 Screenshots to show software application with reference to the original online mechanism. Computer aided examples and projects developed by Prof. Prieto-Valdes.
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Juan J. Prieto-Valdés
TABLE OF CONTENTS About the author ................................................................................................................................................... 2 Some words from the author ................................................................................................................................ 2 Preface ................................................................................................................................................................... 5
CHAPTER-1. LIMITS ......................................................................................................................... 7 1.1. The Idea of a Limit ..................................................................................................................................... 7 1.2. Formal Limit Definition ............................................................................................................................ 12 1.3. Determining Limits and Algebraic Properties of Limits ........................................................................... 14 Homework Topics 1.1, 1.2, and 1.3 ..................................................................................................................... 18 1.4. Continuity, Intermediate Value Property, and One-Side Limits. ............................................................. 22 1.5. Limits at the Infinity (Analyzing also One Side Limit) .............................................................................. 28 Homework Topic 1.4 and 1.5............................................................................................................................... 30 1.6. Example of Limit Calculation Using Technology ...................................................................................... 33 PROJECT-1. Find the Limit Using technology. ..................................................................................................... 34 PRACTICING QUIZ-1. Limits................................................................................................................................. 41
CHAPTER-2. THE DERIVATIVE AND DIFFERENTIATION .................................................. 42 2.1. Definition of Derivative ................................................................................................................................ 42 Homework Topic 2.1............................................................................................................................................ 48 PROJECT-2. The Derivative Using Limit Definition .............................................................................................. 50 2.2. Techniques of differentiation. ................................................................................................................. 53 2.3. Product and Quotient Rules .................................................................................................................... 53 Homework Topic 2.2 and 2.3............................................................................................................................... 54 2.4. The Chain Rule ......................................................................................................................................... 60 2.5. Implicit Differentiation. ........................................................................................................................... 62 2.6. High Order Derivatives (second, third, …): .............................................................................................. 64 Homework Topics 2.4, 2.5, and 2.6. .................................................................................................................... 64 PRACTICING QUIZ-2. Derivatives ......................................................................................................................... 69 2.7. Real Case Examples (Marginal Analysis and Approximations) ................................................................ 70 Homework Topic 2.7............................................................................................................................................ 72 2.8. Increasing and Decreasing Functions (critical points) ............................................................................. 74 2.9. Rolle's Theorem and The Mean Value Theorem ..................................................................................... 76 Homework Topic 2.8 and 2.9............................................................................................................................... 78 2.10. Concavity and Point of Inflexion .......................................................................................................... 80 Homework Topic 2.10.......................................................................................................................................... 82 PRACTICING QUIZ-3. Application of Derivatives.................................................................................................. 87
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Business Calculus Notebook
CHAPTER-3. EXPONENTIAL and LOGARITHMIC FUNCTIONS. OPTIMIZATIONS. ....... 88 3.1. Remembering Exponential and Logarithmic Functions (an Algebra Topic): ............................................88 Homework Topic 3.1 ............................................................................................................................................94 3.2. Differentiation of Exponential Functions ...............................................................................................104 3.3. Differentiation of Logarithmic Functions ...............................................................................................106 Homework Topic 3.2 and 3.3 .............................................................................................................................107 3.4. L’ Hospital Rule / Revisiting Limits..........................................................................................................111 Homework Topic 3.4. .........................................................................................................................................113 3.5. Optimization and Business Applications Examples. ...............................................................................114 Homework Topic 3.5. .........................................................................................................................................120 PRACTICING QUIZ-4. Optimization, Exponential and Logarithmic Functions. ...................................................127 PROJECT-3. Quick Example of a Business Profit Optimization ...........................................................................128
CHAPTER-4. INTEGRATION .......................................................................................................132 4.1. Indefinite Integration and Simple Differential Equation ........................................................................132 4.2. Integration by substitution .....................................................................................................................132 Homework Topic 4.1 and 4.2. ............................................................................................................................134 4.3. Riemann Integration / Riemann Sum. ....................................................................................................138 Homework Topic 4.3 ..........................................................................................................................................140 4.4. Fundamental Theorem of Calculus and Definite Integral ......................................................................140 4.5. Application of Integrals (Average Function Value).................................................................................142 Homework Topic 4.4 and 4.5. ............................................................................................................................143 4.6. Application of Integrals (Area Under and Between Curves). .................................................................144 Homework Topic 4.6 ..........................................................................................................................................148 PRACTICING QUIZ-5. Integrals and its applications ............................................................................................150 PROJECT – 4. Optimization of the Material Cost. ...............................................................................................151 PROJECT – 5. Profit Optimization (Transport Marine Company) .......................................................................156 PROJECT – 6. Optimization of fictitious city evacuation. ...................................................................................157
CHAPTER 5. INTRODUCTORY STATISTICS WITH CALCULUS. ..........................................158 5.1. Continuous Random Variable, Probability, and Histograms. .................................................................158 5.2. Probability Density Functions: Uniform, Exponential, and Normal. ......................................................158 5.3. Uniform Density Function ......................................................................................................................162 5.4. Exponential Density Functions ...............................................................................................................163 5.5. Normal Distribution ................................................................................................................................166 5.6. Mean and Median. .................................................................................................................................170 5.7. Variance and Standard Deviation ...........................................................................................................174 5.8. Variance in different types of distributions: ..........................................................................................180 Homework Chapter 5 .........................................................................................................................................182 PROJECT – 7. Portfolio Optimization ..................................................................................................................183
REVIEW CHAPTER. ....................................................................................................................... 188 R-1. Some Elementary Algebra Concepts ......................................................................................... 188 R-2. Algebra Review .......................................................................................................................... 192
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Juan J. Prieto-Valdés
Preface Learning depends on the interest in gain knowledge that learners have. If you use technology to formalize a method to copy the answer of mathematical fitness, you will develop a brute skill to complete an assignment. However, using technology to learn different method of solutions and systematic procedures, checking graphical representations and physical meaning of the mathematical exercises will substantially help to develop your mathematical and analytical expertise. The following are the most popular technological recourses currently available for learning mathematics on Internet.
Wolfram Alpha and Calculus Course Assistant. https://www.wolframalpha.com/ and/or https://products.wolframalpha.com/courseassistants/calculus.html Using this application, you can evaluate any numeric expression or substitute a value for a variable. Plot basic, parametric, or polar plots of the function(s) of your choice. Determine the limit of a function as it approaches a specific value. Differentiate any function or implicit function. Find critical points and inflection points of a function; identify the local and absolute extrema of a function. Integrate a function, with or without limits, sum a function given a lower and upper bound. Find a closed form of a sequence or generate terms for a specific sequence; among other math applications.
Microsoft Math. Add-in for Word Office. https://www.microsoft.com/en-us/download/details.aspx?id=17786
Microsoft Mathematics Add-in for Microsoft Word makes it easy to plot graphs in 2D and 3D, solve equations or inequalities, and simplify algebraic expressions in your Word documents and OneNote. Using this add-in you can compute standard mathematical functions, such as roots and logarithms, trigonometric functions, find sums and products of series, matrix and complex numbers operations, solve equations and inequalities, factor polynomials or integers, simplify or expand algebraic expressions, and more advanced operation.
Microsoft Excel Microsoft Excel is a helpful and powerful program for data analysis and documentation. It is a spreadsheet program, which contains a number of columns and rows, where each intersection of a column and a row is a “cell.” Each cell contains one point of data or one piece of information. With Excel you can import, export, and convert data. Use editing formula to perform diverse mathematical calculations on your data, business application and analysis. You can use formatting to create your own style. In addition, you can create charts and functional graphics with your data.
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Business Calculus Notebook
Desmos Graphing Calculator- DG https://www.desmos.com/
Desmos is an advanced graphing calculator implemented as a web and mobile application. Users can create accounts and save the graphs and plots that they have created. Any algebraic function/expression can be graphically graphed and analyzed.
GeoGebra-GG https://www.geogebra.org/
GeoGebra is an interactive mathematics software suit for learning science, technology, engineering, and mathematics from primary school up to the university level. Constructions can be made with points, vectors, segments, lines, polygons, conic sections, inequalities, implicit polynomials and functions.
Symbolab-Sy https://www.symbolab.com/
Symbolab is an answer online engine/service that computes step-by-step solutions to mathematical problems in a range of subjects. Symbolab can interpret a user-entered equation or symbolic problem and find the solution if it exists and step-by-step solution.
Math Solver (Matrix Calculator)-MS https://www.math10.com/scripts/matrices/determinant-matrix-calculators.html
Math Solver offers solving fraction, metric conversions, power and radical problems. Area and volume of rectangles, circles, triangles, trapezoids, boxes, cylinders, cones, pyramids, spheres. You can simplify and evaluate expressions, and more advanced problems including multiplication, division, and matrix.
LaTeX Editor http://www.codecogs.com/latex/eqneditor.php http://www.numberempire.com/texequationeditor/equationeditor.php LaTeX is a document preparation system widely used in academia for the communication and publication of scientific documents in many fields, including mathematics, statistics, computer science, engineering, chemistry, physics, economics, linguistics, quantitative psychology, philosophy, and political science.
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Juan J. Prieto-Valdés
CHAPTER-1. LIMITS Read Chapter R (page 189) about basics on algebra before working with differentiation Nowadays graphs of functions are easy to produce using computers; however, calculus persists as the perfect tool for graphing, but most important, to explain and to predict the behavior of a function, extremely useful when solving real problems in different disciplines; business, economics, biology, engineering, environmental, physical and chemical sciences, among many other fields and applications. The two basic operations in calculus are derivatives and integrals, which give proper name to the two primary parts of the subject: differential and integral calculus. In order to understand these primaries calculus workings, is required to understand the concept of a limit. This concept is probably the most difficult to understand at the beginning, it took mathematicians 150 years to arrive at it; remaining today the most important and most useful to define continuity, derivatives, and integrals.
1.1. The Idea of a Limit In mathematics, a limit is the value that a function (𝑓(𝑥)) or sequence (𝑆(𝑥)) reach at some value of the variable 𝑥. Imagine a person walking up; let us consider that his horizontal position is measured by the value of 𝑥 (distance from the wall) and his vertical position (the high) by the coordinate 𝑦. As he gets closer and closer to 𝑐 horizontal position (𝑥 → 𝑐), he approaches L vertical position (the high). It means that his altitude gets nearer and nearer to L (except for a possible small error in accuracy) as + position approach 𝑐 distance. For example, suppose we set a particular accuracy goal for our traveler: He must get the high of 𝐿 = 120 inches. He reports that he can get 120 inches of L, when he is within 3-5 horizontal inches away from c. does not matter if he is going up or dawn, he notices that he is near to 120 inches in high when he is located around 𝑐 position. Previous informal statement can be formulated as follow: The limit of a function 𝑓(𝑥) as 𝑥 approaches 𝑐 is a number 𝐿 with the following property: given any target distance from 𝐿, there is a distance from 𝑐 within which the values of 𝑓(𝑥) remain within the target distance. This explicit statement is quite close to the formal definition of the limit we will study in the next topic.
Examples of Limit Evaluation, an introductory view about the evaluation tactics: 1. Find the limit (𝒚 value) as 𝒙 approaches 𝟓 of the linear equation: 𝒚 = 𝟑𝒙 + 𝟕 We have to write this equation using limit notation, and then substitute the 𝑥 approaching given value. We will follow the following rule: 𝑙𝑖𝑚[𝑓(𝑥) + 𝑔(𝑥)] = 𝑙𝑖𝑚𝑓(𝑥) + 𝑙𝑖𝑚𝑔(𝑥) = 𝑓(𝑐) + 𝑔(𝑐), hereto accepted as introductory 𝑥→c
𝑥→c
𝑥→c
method for solving given question (further, we will discuss the properties of the Limits operations) 𝑙𝑖𝑚 3𝑥 + 7 = 𝑙𝑖𝑚 3𝑥 + 𝑙𝑖𝑚 7 = 3(5) + 7 = 22 𝑥→5
𝑥→5
𝑥→5
2. Find the limit (𝒚 value) as 𝒙 approaches 𝟐 of the quadratic function: 𝒚 = 𝟐𝒙𝟐 + 𝟓𝐱 + 𝟐 We have to write this equation using limit notation, and then substitute the x (approaching) value. 𝑙𝑖𝑚 2𝑥 2 + 5x + 2 = 𝑙𝑖𝑚 2𝑥 2 + 𝑙𝑖𝑚 5𝑥 + 𝑙𝑖𝑚 2 = 2(22 ) + 5(2) + 2 = 20 𝑥→2
𝑥→2
𝑥→2
𝑥→2
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Business Calculus Notebook 𝟐
3. Find the limit (𝒚 value) as 𝒙 approaches 𝟑 of the rational equation: 𝒚 = 𝟑+𝒙 We have to write this equation using limit notation, and then substitute the x (approaching) value. 𝑙𝑖𝑚𝑓(𝑥)
𝑓(𝑥)
𝑓(𝑐)
𝑥→c In this case, we will consider the following rule: 𝑙𝑖𝑚 𝑔(𝑥) = 𝑙𝑖𝑚 = 𝑔(𝑐) ; 𝑖𝑓 𝑙𝑖𝑚𝑔(𝑥) ≠ 0 𝑔(𝑥)
𝑥→c
𝑙𝑖𝑚
𝑥→3
4. Prove that 𝒍𝒊𝒎
𝒙+𝟑
𝒙→𝟒 𝟓+√𝒙
𝑥→c
𝑥→c
𝑙𝑖𝑚 2 𝑙𝑖𝑚 2 2 2 1 𝑥→3 𝑥→3 = = = = 3 + 𝑥 𝑙𝑖𝑚 (3 + 𝑥) 𝑙𝑖𝑚 3 + 𝑙𝑖𝑚 𝑥 3 + 3 3 𝑥→3
𝑥→3
𝑥→3
=𝟏
Formally, we have to decompose the limit question into a group of single elements; however, direct substitution
will produce the same answer: (𝑙𝑖𝑚 𝑥 + 𝑙𝑖𝑚 3)/(𝑙𝑖𝑚 5 + 𝑙𝑖𝑚 √𝑥) = 1 𝑥→4
𝑥→4
𝑥→4
𝑥→4
At this moment we know that the direct substitution of 𝑥 = 4 will produce same result. In this case
4+3 5+2
= 1. This
“direct” substitution works because there is not undefinition at the requested 𝑥 value. Further, we will see different situation when having undefined position, as it is observed in the next example.
5. Find the limit as 𝒙 approaches 𝟏 of the rational expression using numeric and graphic interpretation: Notice that we cannot solve by substituting 𝑥 = 1 because 0⁄ is undefined, and therefore it is not continuity: However, 0 we can solve the question by conducting a numeric investigation, approaching 𝑥 = 1 from the left and from the right using a very small deviation (see the table below):
√𝑥 − 1 𝑥→1 𝑥 − 1
𝑦 = 𝑙𝑖𝑚
√1 − 1 0 = = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑. 𝑥→1 1 − 1 0 𝑙𝑖𝑚
0.8
(𝑥) value √(𝑥) − 1 𝑥→1 (𝑥) − 1
0.9
0.999
1.0
0.527864... 0.513167.. 0.500125...
𝑙𝑖𝑚
1.001
Und.
1.1
0.499875... 0.488088...
1.2 0.477226...
𝑥−1
Resulting that: 𝑙𝑖𝑚 √𝑥−1 = 0.5, as from both sides it towards to 0.5. 𝑥→1
We can graph it and analyze its value ( 𝑦 ) when 𝑥 → 1. When 𝑥 = 1, the value will be undefined (there is a hole in the function), however, from the left side and from the right side around 𝑥 = 1, the function gets 𝑦 = 0.5.
6. Find the limit as 𝒙 approaches 𝟒 of the rational expression: 3
2
𝑥 + 𝑥 − 16 𝑥 − 16 =8 𝑥→4 𝑥 2 − 3𝑥 − 4 𝑙𝑖𝑚
8
We cannot simplify by substituting 𝑥 = 4 because the expression is undefined at this value of 𝑥 , and therefore the function is not continuous. We can get the result approaching 𝑥 = 4 from the left and from the right using a 0.01, 0,001, and 0.001 deviation (see the table):
𝑥 value
3.9
3.99
3.999
4
4.001
4.01
4.1
𝑙𝑖𝑚
7.9
7.99
7.999
___
8.001
8.01
8.1
Juan J. Prieto-Valdés
9
Business Calculus Notebook
7. Estimate the value of the limit as 𝒙 approaches 𝟎 of the rational expression using numeric and graphic interpretation: √𝑥 2 + 9 − 3 𝑥→0 𝑥2 𝑙𝑖𝑚
√9 − 3 0 = = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑. 𝑥→1 0 0 𝑙𝑖𝑚
Notice that we cannot solve by substituting 𝑥 = 0 because 0⁄ is undefined, and therefore it is not continuity: However, 0 we can solve the question by conducting a numeric investigation, approaching 𝑥 = 0 from the left and from the right using a very small deviation (see the table):
(𝑥) value
-0.5
-0.1
-0.05
0.001
0.05
0.1
0.5
√𝑥 2 + 9 − 3 𝑥→1 𝑥2
0.165525…
0.166620…
0.166655…
0.16666…
0.166655…
0.166620…
0.165525…
𝑙𝑖𝑚
√𝑥 2 +9−3 𝑥2 1 =6
Resulting that: 𝑙𝑖𝑚
𝑥→0
0.166666 …
=
We can graph it and analyze its value when 𝑥 → 0. There is a oscillating process around 𝑥 = 0 as it can be observed in the right graph.
8. Estimate the value of the limit when the function exists in one side, using graphic interpretation: 𝑙𝑖𝑚 √𝑥 2 − 1 = 0
𝑥→1+
Notice that we 0 ≤ 𝑥 ≤ 1 the solution will be imaginary. Therefore, the limit only will exist when approaching 𝑥 = 1 from the positive side. Further, we will discuss such type of example; Now it is included for illustration about the one side limit.
9. Estimate the value of the limit when the function exists in one side, use graphic interpretation: ln(𝑥) = −∞ 𝑥→0 𝑥 Notice that we 𝑥 approach 0 from the negative side, the function does not exist. However, when approaching 0 from the positive side the function towards to negative infinite. 𝑙𝑖𝑚
10
Juan J. Prieto-Valdés
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Business Calculus Notebook
1.2. Formal Limit Definition The limit of a function 𝑓(𝑥) when 𝑥 towards 𝑐 is equal to 𝐿. The statement 𝑙𝑖𝑚 𝑓(𝑥) = 𝐿 has the following precise 𝑥→𝐶
definition: Given any real number 𝜉 > 0, there exists another real number δ > 0, so that if 0 < |𝑥 − 𝑐| < 𝛿 , then |f(x) − L| < ξ .
In general, the value of 𝛿 will depend on the value of ξ. That is, we will always begin with ξ > 0 and then determine an appropriate corresponding value for 𝛿 > 0. There are many values of 𝛿 which work. Once you find a value that works, all smaller values of 𝛿 also work. Sometimes, finding the limiting value of an expression means simply substituting a number.
Steps for proving that 𝒍𝒊𝒎𝒊𝒕 exist: On the graph located on the left side we can observed that 𝜀 = |𝑓(𝑥) − 3| < 1, while 𝛿 = |𝑥 − 2| < 2 . We can reduce the intervals until approaching the smallest value of 𝛿 we can (of course, suitable for the calculation), as it is shown on the right side. The limit of 𝑓(𝑥) exist if for any number 𝜀 > 0 there is a corresponding number 𝛿 > 0, such that: |𝑓(𝑥) − 3| < 𝜀, whenever |𝑥 − 2| < 𝛿.
Steps for proving that 𝒍𝒊𝒎 𝒇(𝒙) = ∞ exist (Two-Side Infinite Limit): 𝒙→𝐚
FIRST: Find 𝛿; let 𝑁 be an arbitrary positive number. Use the statement that 𝑓(𝑥) > 𝑁 to find an inequality of the form |𝑥 − 𝑎|
0, assume 0 < |𝑥 − 𝑎| < 𝛿
and use the relationship between N and 𝛿 found in step FIRST to prove that 𝑓(𝑥) > 𝑁. Example 1. Finding 𝜹 for a given 𝜺: Given the limit 𝑙𝑖𝑚 (2𝑥 − 5) = 1, find 𝛿 such that |(2𝑥 − 5) − 1| < 0.01. Whenever 0 < |𝑥 − 3| < 𝛿. 𝑥→3
Solution: First; in this problem we are working with a given value of ε = 0.01. Secondly; to find an appropriated 𝛿, we have to establish a connection between the absolute values |(2𝑥 − 5) − 1| 𝑎𝑛𝑑 |𝑥 − 3| Because |(2𝑥 − 5) − 1| < 0.01 equivalently to 2|𝑥 − 3| < 0.01 Where |(2𝑥 − 5) − 1| = |2𝑥 − 6| = 2|𝑥 − 3| 1 2
We can chose: 𝛿 = (0.001) = 0.005 Which implies that: |(2𝑥 − 5) − 1| = 2|𝑥 − 3| < 2(0.005) = 0.01 12
Juan J. Prieto-Valdés
Example 2. Finding 𝜹 for a given 𝜺: Given the limit 𝑙𝑖𝑚 𝑥 2 = 4. 𝑥→2
find 𝛿 such that |𝑥 2 − 4| < 𝜀. Whenever 0 < |𝑥 − 2| < 𝛿. Where 𝑐 = 𝑥. To find 𝛿 we write |𝑥 − 2||𝑥 + 2| = |𝑥 2 − 4| for all 𝑥 in the interval (1, 3), |𝑥 + 2| < 5, 𝜀 |𝑥 2 − 4| = |𝑥 − 2||𝑥 + 2| < ( ) 5 = 𝜀 5
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Business Calculus Notebook
Example 3. Prove that 𝒍𝒊𝒎 (𝒙𝟐 + 𝟑) = 𝟒. 𝒙→−𝟏
Consider 𝜀 > 0 and find 𝛿 > 0 which depends on 𝜀, so that 0 < |𝑥 − (−1)| < 4, then |𝑓(𝑥) − 4| < 𝜀 |𝑓(𝑥) − 4| = |(𝑥 2 + 3) − 4| = |𝑥 2 − 1| < 𝜀 So |𝑥 − 1||𝑥 + 1| < 𝜀 If replacing the term |𝑥 − 1| with an appropriate constant and keep the term |𝑥 + 1| (since this is the term we want to solve for). Now, assuming that 𝛿 ≤ 1, we can write that |𝑥 + 1| < 𝛿 ≤ 1, which implies that −1 < 𝑥 + 1 < 1, equivalently −2 < 𝑥 < 0 and 1 < |𝑥 − 1| < 3. So |𝑥 − 1||𝑥 + 1| < (3)|𝑥 + 1| < 𝜀 𝜀
resulting that |𝑥 + 1| = 3 𝜀
Now we can choose 𝛿 = min {1, 3}. This guarantees that both assumptions made about 𝛿 in the course of this proof are taken into a consideration simultaneously. Thus: 0 < |𝑥 − (−1)| = |𝑥 + 1| < 𝛿 . It follows that |𝑓(𝑥) − 4| < 𝜀 and this completes the proof.
1.3. Determining Limits and Algebraic Properties of Limits 1. Addition
𝑙𝑖𝑚[𝑓(𝑥) + 𝑔(𝑥)] = 𝑙𝑖𝑚𝑓(𝑥) + 𝑙𝑖𝑚𝑔(𝑥) = 𝑓(𝑐) + 𝑔(𝑐)
2. Subtraction
𝑙𝑖𝑚[𝑓(𝑥) − 𝑔(𝑥)] = 𝑙𝑖𝑚𝑓(𝑥) − 𝑙𝑖𝑚𝑔(𝑥) = 𝑓(𝑐) − 𝑔(𝑐)
3. Product
𝑙𝑖𝑚[𝑓(𝑥) × 𝑔(𝑥)] = 𝑙𝑖𝑚𝑓(𝑥) × 𝑙𝑖𝑚𝑔(𝑥) = 𝑓(𝑐) × 𝑔(𝑐)
𝑥→c
𝑥→c
𝑥→c
𝑥→c
𝑥→c
4. Division
𝑥→c 𝑥→c
𝑥→c
𝑥→c
𝑓(𝑥) 𝑓(𝑐) 𝑓(𝑥) 𝑙𝑖𝑚 = 𝑥→c = ; 𝑖𝑓 𝑙𝑖𝑚𝑔(𝑥) ≠ 0 𝑥→c 𝑔(𝑥) 𝑥→c 𝑙𝑖𝑚𝑔(𝑥) 𝑔(𝑐)
𝑙𝑖𝑚
𝑥→c
5. Constant
𝑙𝑖𝑚[𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡] = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑥→c
Examples (accurate the limits evaluation applying some algebra simplification when possible: = lim(𝑥 3 ) + lim(− 2𝑥 2 ) + lim(6𝑥) + lim(− 4) = 𝑥→2
1. Polynomial function: 3
𝑥→2
lim(𝑥 − 2𝑥 + 6𝑥 − 4) = 8
𝑥→2
𝑥→2
𝑥→2
𝑥→2
= [lim(𝑥)]3 − 2 [lim(𝑥)]2 + 6 [lim(𝑥) − lim(4) = =
𝑥→2 (2)3
3
lim(𝑥 − 2)(𝑥 − 5) = −50
𝑥→2
𝑥→2
𝑥→2
2
− 2(2) + 6(2) − 4 = 8 − 8 + 12 − 4 = 8 = [lim(𝑥 3 − 2)][lim(𝑥 − 5)] =
2. Product of two binomials: 𝑥→3
𝑥→2 2
= lim(𝑥 ) − lim(2𝑥 ) + lim(6𝑥) − lim(4) =
2
𝑥→2
𝑥→2 3
𝑥→3
𝑥→3
{[lim(𝑥)]3 − [lim(2)]}{[lim(𝑥)] − [lim(5)]} = 𝑥→3
𝑥→3
𝑥→3
𝑥→3
= (27 − 2)(3 − 5) = −50 14
Juan J. Prieto-Valdés
15
Business Calculus Notebook
3. Indeterminate at 𝒙 = 𝟓; simplify by factoring (dividing out technique):
(𝑥 + 5)(𝑥 − 5) = 𝑙𝑖𝑚(𝑥 + 5) = 10 𝑥→5 𝑥→5 (𝑥 − 5)
= 𝑙𝑖𝑚
𝑥 2 − 25 = 10 𝑥→5 𝑥 − 5 𝑙𝑖𝑚
4. Indeterminate at 𝒙 = 𝟗; solve multiplying by the conjugate of the numerator (rationalizing the numerator):
𝑥−9 √𝑥 − 3 √𝑥 − 3 √𝑥 + 3 = 𝑙𝑖𝑚 × = 𝑙𝑖𝑚 𝑥→9 𝑥 − 9 𝑥→9 𝑥 − 9 √𝑥 + 3 𝑥→9 (𝑥 − 9)(√𝑥 + 3) = 1 1 = 𝑙𝑖𝑚 = 𝑥→9 (√𝑥 + 3) 6 𝑙𝑖𝑚
√𝑥 − 3 1 = 𝑥→9 𝑥 − 9 6 𝑙𝑖𝑚
5. Indeterminate at 𝒙 = −𝟑; simplify by factoring (dividing out technique): 𝑥2 + 𝑥 − 6 = −5 𝑥→−3 𝑥+3
(𝑥 + 3)(𝑥 − 2) 𝑥2 + 𝑥 − 6 = 𝑙𝑖𝑚 = 𝑙𝑖𝑚 (𝑥 − 2) = −5 𝑥→−3 𝑥→−3 𝑥→−3 𝑥+3 𝑥+3 𝑙𝑖𝑚
𝑙𝑖𝑚
6. Function is indeterminate at 𝒙 = 𝟒; solve by rationalizing denominator and simplifying: 𝑙𝑖𝑚
𝑥→4
𝑥 2 − 16 √𝑥 − 4
𝑙𝑖𝑚
𝑥→4
𝑥 2 − 16 √𝑥 − 4
= 𝑙𝑖𝑚
=0
𝑥 2 − 16
×
√𝑥 − 4
√𝑥 − 4 √𝑥 − 4 (𝑥 + 4)(𝑥 − 4)√𝑥 − 4 = 𝑙𝑖𝑚 𝑥→4 (𝑥 − 4) = 𝑙𝑖𝑚(𝑥 + 4)√𝑥 − 4 = 0
𝑥→4
𝑥→4
7. Function is undefined (infinite) but not indeterminate; the solution can be graphically observed; at 𝒙 = 𝟐 the function is discontinue; no limit. 𝑥2 − 𝑥 + 5 lim = 𝑢𝑛𝑑 𝑥→2 𝑥−2
𝑥2 − 𝑥 + 5 = −∞ 𝑥→2 𝑥−2 𝑥2 − 𝑥 + 5 𝑙𝑖𝑚+ =∞ 𝑥→2 𝑥−2 𝑥2 − 𝑥 + 5 𝑙𝑖𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑. 𝑥−2 𝑥→2 𝑙𝑖𝑚−
The function can be analyzed numerically, approaching 𝑥 = 2 from both sides: 𝑥 Value 𝑦= 16
𝑥2 − 𝑥 + 5 𝑥−2
1.6
1.9
1.99
1.999
2
-14.9
-67.1
-697.01
-6997
DIV/0
2.01
2.1
703.01 73.1
2.3
2.6
26.633
15.266
Juan J. Prieto-Valdés
Requirement to succeed in topic 1.1 – 1.3. (1) Develop an intuitive understanding of the limiting process of a function. (2) Be able to calculate the limit using numeric evaluation and tables of values. (3) Be able to calculate the limit using analytic solution (algebra). (4) Be able to estimate the limit from graphs.
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Business Calculus Notebook
Homework Topics 1.1, 1.2, and 1.3 Solve using limit properties -8
1) lim( 1 − 𝑥 2 ) 𝑥→−3
2) lim(2𝑥 2 + 4)
4
𝑥→0
3) lim(8)
8
𝑥→2
4) lim(𝑥 3 − 1)
26
𝑥→3
5) lim(𝑥 3 − 2𝑥 2 + 3𝑥 − 6)
0
𝑥→2
6)
𝑥 2 + 7𝑥 + 10 𝑙𝑖𝑚 ( ) 𝑥→1 𝑥+2
7)
𝑥 2 + 7𝑥 + 10 𝑙𝑖𝑚 ( ) 𝑥→−2 𝑥+2
8)
𝑥 3 + 4 𝑥 2 − 11 𝑥 − 30 𝑥→−2 𝑥2 + 𝑥 − 2 𝑙𝑖𝑚
2
36
2
9
Factor numerator and denominator
(X − 3) (x + 2) (x + 5) and (𝑥 + 2)(𝑥 − 1) Cancel (x + 2) term, and substitute 𝑥 = −2. Answer: 5
18
Juan J. Prieto-Valdés
9) Find the limit numerically (A), graphically (B), and try to rationalize the numerator(C). Compare with the example No. 5 on page No 6: √𝑥 − 1 𝑙𝑖𝑚 𝑥→1 𝑥 − 1 Rationalize the numerator and simplify. Answer: 10) lim[(2𝑥 2 + 3𝑥)√𝑥 + 2]
1 2
28
𝑥→2
11)
𝑥+2 2 lim ( ) 𝑥→4 𝑥
9 4
12) lim 𝑥 3 − 5𝑥 + 10
8
𝑥→2
13) lim 𝑥 3 − (𝑥 + 2)2
−2
14)
−3
𝑥→−1
lim
𝑥→−2
15)
𝑥 2 + 7𝑥 + 10 𝑥 2 + 3𝑥 + 2
√9 − 𝑥 𝑥→2 𝑥 − 2 lim
𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 (𝑙𝑖𝑚𝑖𝑡 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡)
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Business Calculus Notebook
16)
𝑥−2 𝑎𝑛𝑑 𝑥→∞ 𝑥 2 − 16 lim
𝑥−2 𝑥→−∞ 𝑥 2 − 16
0
lim
Evaluate the limit numerically 17)
𝑥+2 2 lim ( ) 𝑥→0 𝑥 − 2 -0.1 𝑥
1 -0.01
-0.001
0
0.001
0.01
0.1
𝑓(𝑥) 18)
√2𝑥 − 2 𝑥→2 𝑥 − 2 1.9 𝑥
0.5
lim
1.99
1.999
𝑥 3 − 6𝑥 + 8 lim 𝑥→0 𝑥−2 -.1 -0.01 𝑥
-0.001
2.1
0
0.001
0.01
0.1
___
𝑥3 + 1 lim 𝑥→2 𝑥 − 2 1.9 𝑥
𝐼𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒 1.99
1.999
2
2.001
2.01
2.1
____
𝑓(𝑥) 21)
2.01
-4
𝑓(𝑥) 20)
2.001
____
𝑓(𝑥) 19)
2
𝑥3 + 1 lim 𝑥→2− 𝑥 − 2
𝑥 3 +1 𝑥→2− 𝑥−2
lim
= −∞ and
𝑥 3 +1 𝑥→2+ 𝑥−2
= +∞
𝑦=
𝑥3 + 1 𝑥−2
lim
22)
23)
𝑥3 + 1 𝑥→2+ 𝑥 − 2 lim
lim
3−𝑥
1
𝑥→3 √9 −
𝑥 𝑓(𝑥) 20
𝑥 2.9
2.99
2.999
3 __
3.001
3.01
3.1
Juan J. Prieto-Valdés
24) Solve algebraically: √2𝑥 − 2 lim 𝑥→2 𝑥 − 2 0.5 25)
−
8 3
26) Use the graph of 𝑦 = 𝑥 2 − 2 to find a 𝛿 > 0 such that for all 𝑥, 0 < < 𝛿 ⇒ < 𝜀 𝑥0 = 2, 𝐿 = 2, 𝜀 = 0.1
27) Given 𝑓(𝑥) = 5𝑥 + 2, 𝐿 = 22, 𝑥0 = 4, and 𝜀 = .01, find the greatest value for 𝛿 > 0 such that 0 < |𝑥 − 𝑥0 | < 𝛿 the inequality holds.
28) Given 𝑓(𝑥) = −9𝑥 − 2, 𝐿 = −20, 𝑥0 = 2, and 𝜀 = 0.01, find the greatest value for 𝛿 > 0 such that 0 < |𝑥 − 𝑥0 | < 𝛿 the inequality holds.
0.002
0.0012
29) The cross-sectional area of a cylinder is 𝐴 = 𝜋𝑟 2 , where 𝑟 = 𝑟𝑎𝑑𝑖𝑢𝑠. Find the tolerance 𝛿 of the radius such that |𝐴 − 20| < 0.01, as long as 𝑟 = 𝑟0 ± 𝛿
𝑟 = 𝑐 ± 2.52
30) The current in an electrical circuit is given by 𝐼 = 𝑉/𝑅, in ampers (where V is the voltage in volts and R the resistance in ohms. At 𝑉 = 21 𝑣𝑜𝑙𝑡𝑠, what is the resistance R tolerance to keep the current to be within 1 ± 0.02 𝑎𝑚𝑝𝑒𝑟
2%
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Business Calculus Notebook
1.4. Continuity, Intermediate Value Property, and One-Side Limits. 1.4.1. Continuity: Differentiation and integration operations require that functions be continuous on intervals of real numbers. Many other operations in calculus also require continuity of the function. The following definition is the most accepted: A function 𝑓 is continuous at a point (𝑐, 𝑓(𝑐)) [equivalent to (𝑥, 𝑦)] if the next conditions are satisfied: An output of c exists, so 𝑓(𝑐) 𝑒𝑥𝑖𝑠𝑡, 𝑓𝑜𝑟 𝑒𝑥𝑎𝑚𝑝𝑙𝑒 𝑓(𝑥) = 𝐿 and also the limit exists for 𝑥 = 𝑐 , so 𝑙𝑖𝑚 𝑓(𝑥) = 𝐿 and for 𝑙𝑖𝑚 𝑓(𝑥) = 𝐿 𝑥→𝑐+
𝑥→𝑐−
This definition, basically means that there is no missing point, gap, or split for 𝑓(𝑥) when 𝑥 = 𝑐. In other words, you can move draw the graph without lifting your pencil from the paper. To see if the three conditions of the definition are satisfied is a simple process: Plug in the value assigned to 𝑐 into the function and see if 𝑓(𝑐) exists. Use the limit definition to see if the limit exists as 𝑥 approaches 𝑐. 𝑙𝑖𝑚 𝑓(𝑥) = 𝑙𝑖𝑚 𝑓(𝑥) = 𝑙𝑖𝑚𝑓(𝑥)
𝑥→𝑐+
𝑥→𝑐−
𝑥→𝑐
The limit is the same coming from the left and from the right of 𝑓(𝑐), so the limit exists; and if the limit is the same as 𝑓(𝑐), in this case the function is continuous. 𝑙𝑖𝑚𝑓(𝑥) = 𝑓(𝑐) 𝑥→𝑐
Some well-known facts related to continuity: Function f is said to be continuous on a given interval if f is continuous at each point 𝑥 in that interval. Here is a list of some well-known facts related to continuity: 1. The SUM of continuous functions is continuous. 2. The DIFFERENCE of continuous functions is continuous. 3. The PRODUCT of continuous functions is continuous. 4. The QUOTIENT of continuous functions is continuous at all points 𝑥 where the denominator is nor zero. 5. The FUNCTIONAL COMPOSITION of continuous functions is continuous at all points x where the composition is properly defined. 6. Any polynomial is continuous for all values of 𝑥. 7. Function 𝑒 𝑥 and trigonometry functions sin(𝑥) and 𝑐𝑜𝑠(𝑥) are continuous for all values of 𝑥. When solving limits problems, you can avoid common mistakes by using the above step-by-step definition of continuity at a point and by considering the indeterminate form
0 0
during the computation of limits. Knowledge of
one-sided limits will be required as illustrated in 1.4.4, below.
1.4.2.
Intermediate Value Property of Continuous Function.
Since the graph of a continuous function is connected and does not have any holes or breaks in it, the values of the function cannot "skip" or "jump over" a horizontal line. If one value of the continuous function is below the line and another value of the function is above the line, then somewhere the graph will cross the line. The next theorem makes this statement more precise. If 𝑓 is continuous on the interval [𝑎, 𝑏] and 𝑉 is any value between 𝑓(𝑎) and 𝑓(𝑏), then there is a number c between a and b so that 𝑓(𝑐) = 𝑉. Yhat is 𝑓 takes each intermediate value between 𝑓(𝑎) 𝑎nd 𝑓(𝑏). The application of Intermediate Value Property for approximating roots (to get a solution of an equation) can be perform using Bisection Algorithm. 22
Juan J. Prieto-Valdés
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Business Calculus Notebook
1.4.3.
Bisection Algorithm for Approximating Roots
The Intermediate Value Theorem is an example of an "existence theorem” because it concludes that something exists, it can be performing follow the next step: Find two values of 𝑥 , say 𝑎 and 𝑏, so that 𝑓(𝑎) 𝑎𝑛𝑑 𝑓(𝑏) have opposite signs, Calculate the midpoint (bisection point) of the interval [𝑎, 𝑏], 𝑚 = (𝑎 + 𝑏)/2, and evaluate 𝑓(𝑚). (a) If 𝑓(𝑚) = 0, then 𝑚 is a root of 𝑓, and we are done. (b) If 𝑓(𝑚) ≠ 0, then 𝑓(𝑚) has the sign opposite one of 𝑓(𝑎) 𝑜𝑟 𝑓(𝑏): then 𝑓(𝑥) has a root between 𝑎 and 𝑏, a root in the interval [𝑎, 𝑏]. If 𝑓(𝑎) 𝑎𝑛𝑑 𝑓(𝑚) have opposite signs, then 𝑓 has a root in [𝑎, 𝑚] so put 𝑏 = 𝑚 if 𝑓(𝑏) 𝑎𝑛𝑑 𝑓(𝑚) have opposite signs, then 𝑓 has a root in [𝑚, 𝑏] so put 𝑎 = 𝑚 Repeat steps (1) - (3) until a root is found exactly or is approximated closely enough.
Example: Use the Intermediate Value Theorem to verify that given function has a root in the given interval. Then use the Bisection Algorithm to narrow the location of that root to an interval of 𝑦 = 𝑥 3 − 3𝑥 2 + 3 on [–1, 0]. After calculating, represent your answer graphically.
Method-1 (inaccurate approximation): 𝑦(−1) = 𝑥 3 − 3𝑥 2 + 3 = −1,
𝑦(0) = 𝑥 3 − 3𝑥 2 + 3 = 3
𝑅𝑒𝑠𝑢𝑙𝑡𝑖𝑛𝑔 2 𝑝𝑜𝑖𝑛𝑡𝑠: (−1, −1) 𝑎𝑛𝑑 (0, 3) The equation of the line passing through these points has de slope: 𝑚=
(3) − (−1) = 4 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑌𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑖𝑜𝑛 = 3 (0) − (−1)
The line between these two points has the equation: 𝑦 = 𝑚𝑥 + 𝑏 𝑜𝑟 𝑦 = 4𝑥 + 3, 3 4
𝑤ℎ𝑖𝑐ℎ 𝑐𝑢𝑡 𝑡ℎ𝑒 𝑎𝑥𝑖𝑠 𝑥 (𝑤ℎ𝑒𝑛 𝑦 = 0) 𝑎𝑡 𝑥 = − = −0.75. This inaccurate approximation is illustrated in the graph on the right above.
Method-2 (accurate approximation using bisection method): You can calculate manually or using Microsoft Excel (for example) in the next table, for 𝑓(𝑎), 𝑓(𝑏), 𝑎𝑛𝑑 𝑓(𝑚) successively, until getting a good approximation.
𝑓(−0.875) = 0.03 and 𝑓(−0.879385) = 0.000001, Which is a good approximation. Note that 𝑓(−0.879385241571817) = 0.000000000000001762. Use excel or Wolfram Alpha to verify your calculation. 24
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1.4.4.
Examples of Discontinuous Function and one Side Limit: The left-hand limit:
Example 1. Analize continuity at 𝒙 = −𝟐.
At 𝑥 = −2, (−2)2 + 2(−2) = 0 The right-hand limit 𝑥 = −2+ : At 𝑥 = −2, (−2)3 + 6(−2) = 4 Since the left and right-hand limits are different, the limit at 𝑥 = −2 do not exist.
Example 2. Analyze the continuity at 𝒙 = 𝟏 𝒂𝒏𝒅 𝟐 𝒙, 𝒚=
𝟐,
𝒇𝒐𝒓 𝒙 < 𝟏 𝒇𝒐𝒓 𝟏 ≤ 𝒙 ≤ 𝟐
𝒙𝟐 + 𝟓,
𝒇𝒐𝒓 𝒙 > 𝟐
Example 3. For what values of 𝒙 the following function is continuos?
In the first case, 𝑦 = 𝑥 exist before 𝑥 = 1 When 𝑦 = 2, it exists on the closed interval [1,2] While 𝑦 = 𝑥 2 + 5 exist only when 𝑥 > 2 In all cases Limit exist in one side The numerator and denominator functions are continuous because they are polynomials. The quotient of these functions 𝑓(𝑥) is discontinuous at 𝑥 = −4 𝑎𝑛𝑑 1 𝑥 2 + 3𝑥 − 4 = (𝑥 + 4)(𝑥 − 1) = 0. Because division by zero is not allowed.
Example 4. Finding Vertical azymptote at 𝒙 = ± 𝟏. 𝒙𝟐 + 𝟏 𝒚= 𝟐 𝒙 −𝟏
Example 5. For what values of 𝒙 the following function is continuous 𝒇(𝒙) = √𝒙𝟐 − 𝟐𝒙
Example 6. (for illustration purposes) Analize the continuity at 𝒙 = 𝟎 𝟏 𝒇(𝒙) = 𝒙 𝐜𝐨𝐬 ( ) 𝒙 26
lim+
𝑥2 + 1 = +∞ 𝑥2 − 1
lim−
𝑥2 + 1 = −∞ 𝑥2 − 1
𝑥→1
𝑥→1
Because the polynomial 𝑥 2 − 2𝑥 has to be positive or zero (otherwise imaginary solution because inside the 2nd degree radical) 𝑥 2 − 2𝑥 = 𝑥(𝑥 − 2) ≥ 0 It happens at 𝑥 ≤ 0 𝑎𝑛𝑑 𝑥 ≥ 2 We can solve this example applying Squeeze Theorem, but this topic is not covered in business calculus, this is only for illustration purposes. The limit goes to zero (on the right graph) while the function is undefined (left equation) at 𝑥 = 0, resulting that the function 𝑓(𝑥) is discontinuous, but the limits exist and it is equal to zero
Juan J. Prieto-Valdés
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Business Calculus Notebook
1.5. Limits at the Infinity (Analyzing also One Side Limit) Consider the function 𝒇(𝒙) =
𝟐𝟎𝒙𝟐 +𝒙+𝟐 𝒙𝟐
for “large” values of x.
Definition (on the right): Let f be a function defined on the interval (𝑎, ∞), where a is a real number. The statement lim 𝑓(𝑥) = 𝐿 means as 𝑥 approaches (positive) infinity (“as 𝑥 gets 𝑥→∞
larger and larger positively”), 𝑓(𝑥) approaches 𝐿. (Use Microsoft Mathematics or Wolfram Alpha to see the result graphically) Definition (on the left): Similar to the previous, let 𝑓 be a function defined on the interval (∞, 𝑏), where b is a real number. The statement lim 𝑓(𝑥) = 𝐿 means as 𝑥 approaches 𝑥→−∞
negative infinity (“as 𝑥 gets larger and larger negatively”), 𝑓(𝑥) approaches 𝐿. (Use Microsoft Mathematics or Wolfram Alpha to see the result). For large x, 𝑦 = 20
Examples Example 1: The limit results in indeterminate form
∞ ∞
. To solve this problem, we can divide both the numerator
2
and denominator by 𝑥 , which is the higher degree of 𝑥 on denominator position. 20𝑥 2 𝑥 2 20𝑥 2 + 𝑥 + 2 20𝑥 2 + 𝑥 + 2 20 + 0 + 0 2 + 𝑥2 + 𝑥2 𝑥 Lim− = lim+ = lim == lim = 20 2 2 2 𝑥→∞ 𝑥→∞ 𝑥→∞ 𝑥→∞ 𝑥 𝑥 𝑥 1 𝑥2 Example 2: Next limit also results in indeterminate form
∞ . ∞
To solve this problem, we can divide both the numerator
and denominator by 𝑥 2𝑥 − 1 1 2−𝑥 2𝑥 − 1 2𝑥 − 1 𝑥 Lim = Lim = = Lim = Lim = =2 𝑥+1 1 𝑥→∞ 𝑥 + 1 𝑥→∞ 𝑥→∞ 𝑥 + 1 𝑥→∞ 1+𝑥 𝑥 Example 3: Find the horizontal asymptote:
𝑦𝐻𝐴
2 1 4−𝑥− 3 𝑥 = Lim = 4 Lim 3 2 4𝑥 2𝑥 1 4 1 𝑥→∞ 2 − 2 + 3 𝑥→∞ 4𝑥 3 − 2𝑥 2 − 1 3 − 𝑥3 − 𝑥3 𝑥 𝑥 𝑥 = Lim = Lim = = =2 𝑥→∞ 2𝑥 3 − 4𝑥 + 1 𝑥→∞ 2𝑥 3 4𝑥 1 2 − + 𝑥3 𝑥3 𝑥3
Remembering an Algebra Rule to locate the Horizontal Asymptote of the rational function: 𝑎
Horizontal Asymptote can be found follow the next rule:
If 𝑛 = 𝑚
the Horizontal Asymptote is located at 𝑦ℎ.𝑎. = 𝑏
𝑎𝑥 𝑛 + ⋯ 𝑓(𝑥) = 𝑚 𝑏𝑥 + ⋯
If 𝑛 > 𝑚
there is NO Horizontal Asymptote (will be slant asymptote)
If 𝑛 < 𝑚
the Horizontal Asymptote is located at 𝑦ℎ.𝑎. = 0
Example 4: Find the limit
28
√5𝑥 2 +2𝑥 Lim 𝑥 𝑥→∞
= Lim
𝑥→∞
√5𝑥2 +2𝑥 √𝑥2 𝑥 𝑥
2
= Lim √5 + 𝑥 = √5 𝑥→∞
Juan J. Prieto-Valdés √𝑥 2 +𝑥+𝑥
Example 5: Find the limit Lim √𝑥 2 + 𝑥 − 𝑥 = Lim (√𝑥 2 + 𝑥 − 𝑥) ( 𝑥→∞
= Lim
(𝑥 2 +𝑥)−𝑥 2
𝑥→∞ √𝑥 2 +𝑥+𝑥
= Lim
√𝑥 2 +𝑥+𝑥
𝑥→∞
𝑥
=Lim
𝑥 𝑥
𝑥→∞ √𝑥 2 +𝑥+𝑥 𝑥→∞ √𝑥2 𝑥 𝑥 + + 𝑥2 𝑥2 𝑥
=
1 1+1
=
)=
1 2
1 𝑥 𝑥
Example 4: Evaluate numerically and graphically the following limit lim (1 + ) 𝑥→∞
𝑥 = 1,
1 1 (1 + 1)
=2;
𝑥 = 10, (1 + 100
1
𝑥 = 100, (1 + 100) 1
𝑥 = 10000000, (1 +
= 2.594
= 2.705
1000
𝑥 = 1000, (1 + 1000)
1 10 ) 10
= 2.717
1000000 1 ) 10000000
= 2.718282 …
When 𝑥 towards to infinity, the 𝑓(𝑥) towards 𝑒 = 2.718282 … In Chapter III we will study more advanced methods to evaluate such type of limits.
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Business Calculus Notebook
Homework Topic 1.4 and 1.5. In 31-34 exercises find one-side limit, verify your answer with Mathematics or Wolfram Alpha. 31)
lim 𝑥 3 − 5𝑥 and lim+ 𝑥 3 − 5𝑥
𝑥→2−
𝑥→2
Answer: -2
32)
lim √5𝑥 − 4 and lim+ √5𝑥 − 4
𝑥→4−
𝑥→4
Answer: 4
33)
lim−
𝑥→3
√5𝑥 − 4 𝑎𝑛𝑑 𝑥−3
lim+
𝑥→3
√5𝑥 − 4 𝑥−3 Answer: +∞ 𝑎𝑛𝑑 − ∞
34)
𝑥2 − 4 𝑥2 − 4 lim 𝑎𝑛𝑑 lim− 𝑥→2 𝑥 − 2 𝑥→2+ 𝑥 − 2 𝑥 ≠ 2, 𝑏𝑢𝑡 𝑙𝑖𝑚𝑖𝑡 = 4
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Juan J. Prieto-Valdés
In the next exercises decide if the given function is continuous at the given point. 35)
36)
𝑓(𝑥) =
𝑥3 + 1 𝑥−2
𝑎𝑡 𝑥 = 2
𝑓(𝑥) =
4 𝑥+3
𝑎𝑡 𝑥 = −3 Discontinue Lim
4
𝑥→−3 𝑥+3
37)
𝑓(𝑥) =
12 2 𝑥 +5𝑥+6
𝑎𝑡 𝑥 = 12 𝑎𝑛𝑑 𝑥 = −3
= 𝐼𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒 12
Continuous, lim
=
𝑥→12 𝑥 2 +5 𝑥+6 12 2
122 +5(12)+6
=−
38
12
Discontinuous, lim
𝑥→−3 𝑥 2 +5 𝑥+6
=
𝐼𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒
38)
𝑓(𝑥) =
3−𝑥 𝑥
𝑎𝑡 𝑥 = 0 Discontinuous, lim
3−𝑥
𝑥→0 𝑥
=
𝐼𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒
39)
𝑓(𝑥) =
3−𝑥 𝑥+2
𝑎𝑡 𝑥 = 2 Continuous, lim
3−𝑥
𝑥→2 𝑥+2
40)
𝑓(𝑥) =
√𝑥 − 2 𝑥−4
=
1 4
𝑎𝑡 𝑥 = 4
√𝑥−2 𝑥→4 𝑥−4
Discontinuous, lim
×
√𝑥+2 √𝑥+2
=
1 4
41) Find the limit. Provide approximated values if exist: lim 𝑓(𝑥) 𝑐 𝑥→𝑐
−1 0 1+ 1− 1 2+ 2 3 31
Business Calculus Notebook
42) Find the limit. Provide approximated values if exist: lim 𝑓(𝑥)
𝑐
𝑥→𝑐
−1 0 1+ 1− 1 2+ 2 3
In the next exercises find the limit at the infinity 43)
lim
𝑥→−∞
8 6 8 − ( 2) 𝑥 1
44)
6𝑥 3 + 2𝑥 2 𝑥→−∞ 𝑥 − 3𝑥 2 lim
∞ 45)
100𝑥 2 lim √ 𝑥→∞ 8 + 49𝑥 2 10 7
46)
lim (𝑥 − √𝑥 2 − 2𝑥 + 3)
𝑥→−∞
1 47)
𝑥3 + 1 lim 𝑥→2− 𝑥 − 2
𝑥 3 +1 𝑥→2− 𝑥−2
lim
= −∞ and
𝑥 3 +1 𝑥→2+ 𝑥−2
= +∞
𝑦=
𝑥3 + 1 𝑥−2
lim
48)
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𝑥3 + 1 𝑥→2+ 𝑥 − 2 lim
Juan J. Prieto-Valdés
1.6. Example of Limit Calculation Using Technology 𝑥 2 + 4𝑥 − 12 𝑥→2 𝑥 2 − 2𝑥 We can apply three different methods to solve this limit: lim
1.6.1.
Algebraic by factoring. (𝑥 − 2)(𝑥 + 6) (𝑥 + 6) (𝑥 + 6) 𝑥 2 + 4𝑥 − 12 = lim = lim , 𝑎𝑡 𝑥 = 2, =4 2 𝑥→2 𝑥→2 𝑥→2 𝑥 − 2𝑥 𝑥(𝑥 − 2) 𝑥 𝑥 lim
1.6.2.
Numeric. Approaching 2 ( 𝒙 → 𝟐 )
. We will perform these operations automatically using Microsoft Excel Sheet as displayed in the next Table. We will consider the limit from the left and right side (use Microsoft Help if required): 𝑓(𝑥) =
1.6.3.
𝑥 2 + 4𝑥 − 12 𝐶1 ∗ 𝐶1 ∗ +4 ∗ 𝐶1 − 12 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑒𝑑 𝑎𝑠: 𝑓𝑥 = 2 𝑥 − 2𝑥 𝐶1 ∗ (𝐶1 − 2)
Graphic.
Use Microsoft Mathematic or Wolfram Alpha to graph the function and analyze the Limit value when x approaches 2. y=
𝑥 2 + 4𝑥 − 12 𝑥 2 − 2𝑥 or 𝑦=
(𝑥 + 6) 𝑥
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Business Calculus Notebook
PROJECT-1. Find the Limit Using technology. You can print the Project from Professor Prieto-Valdes Web-page. Step-1: Follow the 6 examples on pages 35-40. Step-2: After completing Step-1, solve by yourself the next six examples to test your abilities. Solve analytically when possible and complete your answer using Wolfram Alpha, Wolfram Alpha App, or Microsoft Math Add-in, and/or Excel. 𝑥3 + 1 𝑥→1 (𝑥 − 1)2 lim
4𝑥 2 − 4𝑥 − 8 𝑥→2 𝑥 2 + 𝑥 − 6 lim
𝑥3 + 1 𝑥→3 𝑥 − 3 lim
2𝑥 3 + 8𝑥 2 + 6𝑥 𝑥→−3 𝑥 2 + 5 𝑥 + 6 lim
√𝑥 − 4 𝑥→16 𝑥 − 16 lim
lim
𝑥→3
𝑥(𝑥 − 3) √9 − 𝑥
Your answer has to satisfy the following criteria. (1). Calculate the Limit algebraically when possible. You can use Microsoft or Wolfram programs for factoring and/or to simplify rational and radical functions (50%) (2). Write the Microsoft Excel Function and evaluate numerically the limit. You can use another software or any calculator to complete your answer/calculation (25%) (3). Graph and analyze using Microsoft Mathematics, Wolfram Alpha, or Wolfram Alpha APP (one of them) (25%) Use the following format to provide your answer:
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Juan J. Prieto-Valdés
Example 1 𝑥 2 + 4𝑥 − 12 𝑥→2 𝑥 2 − 2𝑥 lim
We can apply three different methods to solve this limit: 1. Algebraic by factoring. (𝑥 − 2)(𝑥 + 6) (𝑥 + 6) (𝑥 + 6) 𝑥 2 + 4𝑥 − 12 lim = lim = lim , 𝑎𝑡 𝑥 = 2, =4 2 𝑥→2 𝑥→2 𝑥→2 𝑥 − 2𝑥 𝑥(𝑥 − 2) 𝑥 𝑥 2. Numerically; approaching 2 ( 𝒙 → 𝟐 ) from the left and right side. We will perform these operations automatically using Microsoft Excel Sheet.
3. Graphically. We can use Microsoft Mathematics or Wolfram Alpha. Microsoft Mathematics Add-in
Wolfram Alpha (on Line)
Wolfram Alpha APP
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Example 2. 𝑥 − √2𝑥 + 3 𝑥→3 3−𝑥 1. Algebraically. Rationalizing the numerator: lim
𝑥 − √2𝑥 + 3 𝑥 − √2𝑥 + 3 𝑥 + √2𝑥 + 3 𝑥 2 − (2𝑥 + 3) = lim × = lim = 𝑥→3 𝑥→3 3−𝑥 3−𝑥 𝑥 + √2𝑥 + 3 𝑥→3 (3 − 𝑥)(𝑥 + √2𝑥 + 3) lim
= lim
𝑥→3 (3
𝑥 2 − 2𝑥 − 3 − 𝑥)(𝑥 + √2𝑥 + 3)
= lim
(𝑥 − 3)(𝑥 + 1)
𝑥→3 (3 −
𝑥)(𝑥 + √2𝑥 + 3)
4 = − = 0.666 … 𝑥→3 (𝑥 + √2𝑥 + 3) 6
= lim
−(𝑥 + 1)
2. Numerically. Table of values approaching 𝒙 = 𝟑
3. Graphically (Microsoft Mathematics & Wolfram Alpha):
𝑦=
𝑥 − √2𝑥 + 3 3−𝑥
4. Graph and provide your solution using Wolfram Alpha APP (handwritten reproduction is accepted)
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Juan J. Prieto-Valdés
Example 3. 2(−3 + 𝑥)2 − 18 𝑥→0 𝑥 lim
1. Algebraically: 2(−3+𝑥)2 −18 𝑥 𝑥→0
Lim
18−12𝑥+2𝑥 2 −18 𝑥 𝑥→0
= lim
−12𝑥+2𝑥 2 𝑥 𝑥→0
= lim
2𝑥(−6+𝑥) 𝑥 𝑥→0
= lim
= −12
2. Numerically:
3. Graphically (Microsoft Mathematics & Wolfram Alpha):
𝑦=
2(−3 + 𝑥)2 − 18 𝑥
4. Graph and provide your solution using Wolfram Alpha APP (handwritten reproduction is accepted)
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Business Calculus Notebook
Example 4: 𝑥 − √3𝑥 + 4 𝑥→4 4−𝑥 1. Algebraically. Rationalizing the numerator: lim
𝑥 − √3𝑥 + 4 𝑥 2 − (3𝑥 + 4) 𝑥 2 − 3𝑥 + 4 lim = lim = lim = 𝑥→4 𝑥→4 (4 − 𝑥)(𝑥 + √3𝑥 + 4) 𝑥→4 (4 − 𝑥)(𝑥 + √3𝑥 + 4) 4−𝑥 lim
𝑥→4 (4
(𝑥 − 4)(𝑥 + 1) − 𝑥)(𝑥 + √3𝑥 + 4)
= lim
𝑥→4 −(𝑥
(𝑥 + 1) + √3𝑥 + 4)
=−
5 8
2. Numerically:
3. Graphically (Microsoft Mathematics & Wolfram Alpha):
𝑦=
𝑥 − √3𝑥 + 4 4−𝑥
4. Graph and provide your solution using Wolfram Alpha APP (handwritten reproduction is accepted)
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Juan J. Prieto-Valdés
Example 5: 3 𝑥→2 (𝑥 − 2)2 lim
1. There is not required specific algebraic operation, however is required analyze two different situations: 3 3 lim+ → +∞ 𝑎𝑛𝑑 lim− → +∞ 2 𝑥→2 (𝑥 − 2)2 𝑥→2 (𝑥 − 2) 3 so: lim 𝑒𝑥𝑖𝑠𝑡 𝑎𝑡 𝑡ℎ𝑒 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑦 𝑥→2 (𝑥 − 2)2 2. Numerically:
3. Graphically (Microsoft Mathematics & Wolfram Alpha):
𝑦=
3 (𝑥 − 2)2
4. Graph and provide your solution using Wolfram Alpha APP (handwritten reproduction is accepted)
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Business Calculus Notebook
Example 6: 3 𝑥→4 (4 − 𝑥)3 lim
1. There is not required specific algebraic operation, however is required to analyze two different situations: 3 3 lim+ → −∞ 𝑎𝑛𝑑 lim− → +∞ 3 𝑥→4 (4 − 𝑥)3 𝑥→4 (4 − 𝑥) 3 so: lim 𝑑𝑜 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡 𝑎𝑠 ∆𝜀 → ∞ 𝑤ℎ𝑒𝑛 ∆𝛿 → 0 (𝑏𝑦 𝑙𝑖𝑚𝑖𝑡 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛) 𝑥→4 (4 − 𝑥)3 2. Numerically:
3. Graphically (Microsoft Mathematics & Wolfram Alpha):
𝑦=
3 (4 − 𝑥)3
4. Graph and provide your solution using Wolfram Alpha APP (handwritten (handwritten reproduction is accepted) is accepted)
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Juan J. Prieto-Valdés
PRACTICING QUIZ-1. Limits 1. What is the slope of the line through (–1, –2) and (𝑥, 𝑦) for 𝑦 = 𝑥 2 + 𝑥 – 2 when 𝑎) 𝑥 = – 0.98, 𝑎𝑛𝑑 𝑏. ) 𝑥 = – 1.03. 2. Sketch the graph of 𝑦 = 𝑥 2 + 𝑥 – 2 and the lines passing through 𝑥 = – 0.98 and 𝑥 = – 1.03 which slope was calculated before in 1.a and 1.b above (use large graph on the interval [-2, 0] to make possible your graphic representation. 3. Evaluate the limit as x approaches 4 of the following rational expression: 𝑥 3 + 𝑥 2 − 16 𝑥 − 16 𝑙𝑖𝑚 = 𝑥→4 𝑥 2 − 3𝑥 − 4 Use table of values to find the limit; write only the values, not the arithmetic operations: 3.9 3.99 3.999 4 4.001 4.01 4.1 𝑥 value 𝑙𝑖𝑚 4. For what values of 𝒙 the following function is discontinuous (explain your answer, you can use the graph of the function) ? 𝑥 2 + 3𝑥 + 5 𝑓(𝑥) = 2 𝑥 + 3𝑥 − 4 5. Evaluate √𝑥 − 1 − 2 lim = 𝑥→5 𝑥−5
6. Evaluate 2𝑥 2 − 𝑥 − 1 lim = 𝑥→1 𝑥−1
7. Evaluate lim[(2𝑥 2 + 3𝑥)√𝑥 + 2] 𝑥→2
8. Find the vertical asymptotes 𝑦=
𝑥2 + 1 𝑥2 − 1
9. (10 Points) Given the limit 𝑙𝑖𝑚 (2𝑥 − 5) = 1, find 𝛿 such that |(2𝑥 − 5) − 1| < 𝜀 = 0.01. Whenever 𝑥→3
0 < |𝑥 − 3| < 𝛿. Use Formal definition of Limits. 10. (10 Points) The cross-sectional area of a cylinder is 𝐴 = 𝜋𝑟 2 , where 𝑟 = 𝑟𝑎𝑑𝑖𝑢𝑠. Find the tolerance of the radius such that |𝐴 − 20| < 0.01, as long as 𝑟𝑚𝑖𝑛 < 𝑟 < 𝑟𝑚𝑎𝑥
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Business Calculus Notebook
CHAPTER-2. THE DERIVATIVE AND DIFFERENTIATION Derivatives enable us to talk about the instantaneous rate of change of a function, which, leads to concepts such as velocity and acceleration, population growth rates, marginal cost, flow rates, and other applications notable more sophisticated, which will be reviewed later in the next topics.
2.1. Definition of Derivative The derivative of a function of a real variable measures the sensitivity to change its quantity. Derivatives are a fundamental tool of calculus. The derivative of a function of a single variable at a chosen input value of the variable 𝑥, is the slope of the tangent line to the graph of the function at that point. The tangent line is the best linear approximation of the function near to that input value. For this reason, the derivative is described as the "instantaneous rate of change". Differentiation is the action of computing a derivative. The derivative of a function f(x) is a measure of the rate at which the value of the function changes with respect to the change of the variable 𝑥. It is called the derivative of 𝑓(𝑥) with respect to 𝑥 . Let’s see some preliminary concepts required to understand and to define concept of the derivative:
2.1.1. Linear Function and Slope of the Line: If 𝑦 is a linear function 𝑓(𝑥) of x, means that the value of y divided by x is a linear function. In this case, 𝑦 = 𝑚 𝑥 + 𝑏, where the slope m is given by 𝒎=
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑦 ∆𝑦 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥 ∆𝑥
The symbol Δ (Delta) is an abbreviation for the piece of axis 𝑥 and/or 𝑦 representing “change in”. So, for any value of 𝑥 , the function 𝑦 will be equal to the product of 𝑚 𝑥 plus the constant value 𝑏 .
2.1.2. Slope of the Secant Line A secant line is a straight line joining two points on a function. (See below.) It is also equivalent to the average rate of change, or simply the slope of the line connecting (between) two points. The slope of the secant line passing through the points 𝑃1 𝑎𝑛𝑑 𝑃2 can be written as follow:
𝒎=
∆𝒚 𝒇(𝒙 + ∆𝒙) − 𝒇(𝒙) = ∆𝒙 ∆𝒙
2.1.3. Slope of the Tangent Line:
Figure 1. Secant line between 𝑃1 𝑎𝑛𝑑 P2
The tangent line (or simply tangent) to a curve at a given point is the straight line that "just touches" the curve at that point. To calculate the slope of the tangent line to the function 𝑓(𝑥) at 𝑥 we have to vary the position of the point 𝑃2 (in Fig. 1) as illustrated in the next Figure. Note that: as 𝑃1 approaches 𝑃2 the space ∆𝑥 between 𝑃1 𝑎𝑛𝑑 𝑃2 is reduced towards zero, resulting that the equation (2) becomes a limit exercise as the only way to calculate the slope of the tangent line, precisely when 𝛥𝑥 towards zero ( 𝛥𝑥 → 0). 𝑑𝑦 𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥) = lim 𝑑𝑥 ∆𝑥→0 ∆𝑥 Previous solution was created by Isaac Newton in the mid-17th century. 42
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Business Calculus Notebook
This solution was named the Derivative of a function, while the Differentiation is the algebraic method of finding the Derivative for a function at any point having the following 2 interpretations: Geometrical (as the slope of a tangent to the function/curve at the given 𝑥 point), Figure 2. The secant line becomes a tangent line and Physical (as a rate of change). Either way, both the slope and the instantaneous rate of change are equivalent, and the function to find both of these at any point is called the derivative. Using geometrical concept, we can define the derivative as the slope of a tangent line on a graph at that point where we are calculating the derivative. Equivalently:
𝐝𝐲 𝐝𝐱
= tan ∝, where ∝ is the angle of the tangent line with the horizontal.
The next illustration serves to represent the derivative of the parabola function in three different 𝒙 positions.
Example 1: Find the derivative of 𝒇(𝒙) = 𝟑𝒙 + 𝟒 𝑑𝑦 𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥) [3(𝑥 + ∆𝑥) + 4] − [3𝑥 + 4)] 3∆𝑥 = lim = lim = =3 ∆𝑥→0 𝑑𝑥 ∆𝑥→0 ∆𝑥 ∆𝑥 ∆𝑥
Example 2: Compute the derivative of 𝒇(𝒙) = 𝟒𝒙𝟐 − 𝟕𝒙 at 𝒙 = 𝟑: [4(𝑥 + ∆𝑥)2 − 7(𝑥 + ∆𝑥)] − [4𝑥 2 − 7𝑥] 𝑑𝑦 𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥) ∆𝑥(8𝑥 + 4∆𝑥 − 7) = lim = lim = lim = 8𝑥 − 7 ∆𝑥→0 ∆𝑥→0 𝑑𝑥 ∆𝑥→0 ∆𝑥 ∆𝑥 ∆𝑥 𝑑𝑦
At 𝑥 = 3, 𝑑𝑥 = 17 Also you can write:
𝑑𝑦 𝑑𝑥
= lim
𝑓(3+∆𝑥)−𝑓(3) ∆𝑥
∆𝑥→0
= lim
∆𝑥→0
[4(3+∆𝑥)2 −7(3+∆𝑥)]−[4(32 )−7𝑥] ∆𝑥
Resulting: ∆𝑥(17 + 4∆𝑥) = lim 17 + 4∆𝑥 = 17 ∆𝑥→0 ∆𝑥→0 ∆𝑥 lim
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Juan J. Prieto-Valdés
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Business Calculus Notebook 𝟏
Example 3: Find the equation of the line tangent to the graph of 𝒇(𝒙) = 𝒙 at 𝒙 = 𝟐 1
We have to write the equation of the line 𝑦 = 𝑚𝑥 + 𝑏, tangent to 𝑦 = 𝑥 at 𝑥 = 2. The numeric value of the slope 𝑚 1 𝑥
of the requested line is equal to the value of the derivative of the function at 𝑥 = 2. 1 1 − 𝑑𝑦 𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥) 𝑥 + ∆𝑥 𝑥] = = lim = lim [ ∆𝑥→0 𝑑𝑥 ∆𝑥→0 ∆𝑥 ∆𝑥 1𝑥 1(𝑥 + ∆𝑥) −∆𝑥 − 2 + ∆𝑥 1 (𝑥 + ∆𝑥)𝑥 𝑥(𝑥 + ∆𝑥) 𝑥 = lim [ ] = = lim [ ]=− ∆𝑥→0 ∆𝑥→0 ∆𝑥 ∆𝑥 4
1
The point of tangency is located at (𝑥, 𝑦) = (2, 2). After replacing on the equation of the line the values of 1 2
(𝑥, 𝑦) and the slope 𝑚 we get:
1
= (− 4) (2) + 𝑏, resulting that 𝑏 = 1. Finaly, we can write the equation of the
1 4
requested line as follow: 𝑦 = − 𝑥 + 1
Example 4: Find the derivative of 𝒇(𝒙) = √𝒙 𝑑𝑓(𝑥) √(𝑥 + ∆𝑥) − √𝑥 √(𝑥 + ∆𝑥) − √𝑥 √(𝑥 + ∆𝑥) + √𝑥 = lim = lim ( ) ∆𝑥→0 ∆𝑥→0 𝑑𝑥 ∆𝑥 ∆𝑥 √(𝑥 + ∆𝑥) + √𝑥 = lim
∆𝑥→0
√𝑥
𝑥 + ∆𝑥 − 𝑥 ∆𝑥 (√(𝑥 + ∆𝑥) + √𝑥)
1
lim
∆𝑥→0 √(𝑥
+ ∆𝑥) + √𝑥
=
1 2√𝑥
Example 6: Graph the function 𝒇(𝒙) = 𝒙𝟑 − 𝟒𝒙 , and its derivative 𝒇′ (𝒙) = 𝟑𝒙𝟐 − 𝟒. Solve by yourself using the definition of the derivative.
𝑓(𝑥) = 𝑥 3 − 4𝑥
𝑑𝑓(𝑥) 𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥) = lim ∆𝑥→0 𝑑𝑥 ∆𝑥
Them, use Wolfram Alpha to interpreter the answer.
To succeed in this topic, you should understand The definitions of a derivate as a limit, as two points of a function get infinitesimally close. The relationship between differentiability and continuity. How derivatives are represented graphically, and numerically-analytically, and how they are conceived as an instantaneous rate of change. Also, you have to perform abundant practice to get required mastery 46
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Homework Topic 2.1 Differentiate using limit definition 49)
𝑑𝑦 𝑑𝑥
= lim
∆𝑥→0
𝑓(𝑥+∆𝑥)−𝑓(𝑥) ∆𝑥
𝑥+2 2 y =( ) 4 𝑥 1 + 8 4
50) 𝑓(𝑥) = 𝑥 2 − 5𝑥 + 10
2𝑥−5
51) 𝑓(𝑥) = 3𝑥 + 4
52)
3
Graph 𝑓(𝑥) = 3𝑥 + 4 and compare your result with previous exercise; explain:
53) 𝑦 = 6 + (𝑥 + 1)2
54) 𝑦 = 6𝑥 + (𝑥 + 1)3
55) 𝑓(𝑥) =
𝑥 𝑥−3
56)
3𝑥 2 − 𝑥 3 + 𝑥 4
𝑓(𝑥) =
2 (𝑥 + 1)
3 (𝑥 + 1)2 + 6
−
3 𝑥2 − 6 𝑥 + 9
3 48
Juan J. Prieto-Valdés
57)
58)
𝑓(𝑥) =
2𝑥 2 + 5𝑥 5𝑥
𝑓(𝑥) =
(𝑥 2 − 16𝑥) 𝑥
2 5
1
Graph the tangent line to the given graph at the indicated point. Estimate the slope graphically and evaluate it using limit definition of derivative. 59)
𝑦 2 + (x − 2)2 = 25; , at 𝑥 = 2
𝑚=0 60)
𝑦 = (𝑥 + 2)2 + 3; 𝑎𝑡 𝑥 = −2
𝑚=0 61)
𝑦 = 3𝑥 + 2, at 𝑥 = 1
𝑚=3 62) 𝑦 = √𝑥 − 2; 𝑎𝑡 𝑥 = 5
𝑚=
1 2√3
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Business Calculus Notebook
PROJECT-2. The Derivative Using Limit Definition Complete this Project using Graphic Calculator T1-83, Ti-84, or any other with similar features. The project also will be accepted if you use Microsoft Mathematics, Wolfram Alpha, or any other software or App with similar capability. When completing this project, pay attention to the following: (1)- The definitions of a derivate as a limit operation, as two points of a function get infinitesimally close. (2)- How derivatives are represented graphically and how they are conceived as an instantaneous rate of change. (3)- How you evaluate the derivative, numerically, graphically, and using digital technology. Examples 1, 2, and 3 are training examples. Solve by yourself to get training. Exercises 1-6 are project questions. Your answer has to satisfy the following criteria. (1). Calculate the derivative algebraically using the definition (Formula on topic 2.1.2, page 42. (50%) (2). Show the value or obtained function from T-83/84 or Wolfram Alpha. (25%) (3). Graph your answer. (25%)
Example 1: Find the derivative of 𝒇(𝒙) = 𝟑𝒙 + 𝟒 𝑑𝑦 𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥) [3(𝑥 + ∆𝑥) + 4] − [3𝑥 + 4)] = lim = lim = ∆𝑥→0 𝑑𝑥 ∆𝑥→0 ∆𝑥 ∆𝑥 [3𝑥 + 3𝛥𝑥 + 4 − 3𝑥 − 4] 3∆𝑥 = lim = = lim =3 ∆𝑥→0 ∆𝑥→0 ∆𝑥 ∆𝑥 Using Graphic Calculator 1. Entering the information
2. Graph the function
Using Wolfram Alpha Use ℎ instead ∆𝑥 to facilitate the calculation
𝒅𝒚
3. Input 𝒅𝒙 command
4. Go to GRAPF and press ENTER
Example 2. Compute the derivative of 𝒇(𝒙) = 𝟒𝒙𝟐 − 𝟕𝒙 at 𝒙 = 𝟑: [4(𝑥 + ∆𝑥)2 − 7(𝑥 + ∆𝑥)] − [4𝑥 2 − 7𝑥] 𝑑𝑦 𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥) = lim = lim = ∆𝑥→0 𝑑𝑥 ∆𝑥→0 ∆𝑥 ∆𝑥
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Juan J. Prieto-Valdés
For 𝑥 = 3 :
𝑑𝑦 𝑑𝑥
= lim
[4(3+∆𝑥)2 −7(3+∆𝑥)]−[4(32 )−7𝑥] ∆𝑥
∆𝑥→0
= lim
∆𝑥→0
∆𝑥(17+4∆𝑥) ∆𝑥
Using Graphic Calculator Entering the information
= lim 17 + 4∆𝑥 = 17 ∆𝑥→0
Using Wolfram Alpha
Graph the function Use 𝒉 instead ∆𝒙 to facilitate the calculation
Input
𝒅𝒚 𝒅𝒙
command
Use TANGENT command
Go to GRAPF, ENTER 𝒙 = 𝟑
Go to GRAPF, ENTER 𝒙 = 𝟑
𝑎𝑡 𝑥 = 3 8(3) − 7 = 17 Note, when using graphic calculator you can get directly the equation of the line tangent to the given parabola at 𝑥 = 3 𝒚 = 𝟏𝟕𝒙 − 𝟑𝟔
𝟏
Example 3. Find the equation of the line tangent to the graph of 𝒇(𝒙) = at 𝒙 = 𝟐 𝒙
1
We have to write the equation of the line 𝑦 = 𝑚𝑥 + 𝑏, tangent to 𝑦 = 𝑥 at 𝑥 = 2. The numeric value of the slope 𝑚 1
of the requested line is equal to the value of the derivative of the function 𝑓(𝑥) = 𝑥 at 𝑥 = 2. 1 1 −𝑥 𝑑𝑦 𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥) 𝑥 + ∆𝑥 = lim = lim [ ]= ∆𝑥→0 𝑑𝑥 ∆𝑥→0 ∆𝑥 ∆𝑥 1𝑥 1(𝑥 + ∆𝑥) −∆𝑥 − 2 + ∆𝑥 1 (𝑥 + ∆𝑥)𝑥 𝑥(𝑥 + ∆𝑥) 𝑥 = lim [ ] = lim [ ]=− ∆𝑥→0 ∆𝑥→0 ∆𝑥 ∆𝑥 4 1
The point of tangency is located at (𝑥, 𝑦) = (2, 2). After replacing on the equation of the line the values of (𝑥, 𝑦) 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒 𝑚 we get:
1 2
1
1
= (− 4) (2) + 𝑏, resulting that 𝑏 = 1 and 𝑦 = − 4 𝑥 + 1 (the line). 51
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Graphic Calculator Entering the information
𝒅𝒚
Input 𝒅𝒙 command
Wolfram Alpha
Graph the function
Go to GRAPF, ENTER 𝒙 = 𝟐 𝑎𝑡 𝑥 = 2 1 1 − 2 = − = −0.25 𝑥 4 Replacing (𝑥, 𝑦) = (2. 0.5)
Use TANGENT command
Go to GRAPF, ENTER 𝒙 = 𝟑
in the equation of the line 𝑦 = −0.25𝑥 + 𝑏 We can find 𝑏 = 0.5 + 0.25(2) = 1 Having 𝑚 𝑎𝑛𝑑 𝑏 We can write the equation of the tangent line 𝑦 = −0.25𝑥 + 1
Complete the following exercises using limit definition 1.
𝑥+2 2 y =( ) 𝑎𝑡: 𝑥 = 2 4
2.
𝑓(𝑥) = 𝑥 2 − 5𝑥 + 10 at 𝑥 = 3
3.
𝑦 = 6 + (𝑥 + 1)2 at 𝑥 = 1
4.
𝑦 = 6𝑥 + (𝑥 + 1)3
5.
𝑓(𝑥) =
6.
𝑦 = √𝑥 − 2; 𝑎𝑡 𝑥 = 5
𝑥 𝑥−3
𝑥 1 1 + = 8 4 2
2𝑥−5=1 2(𝑥 + 1) = 4 3 (𝑥 + 1)2 + 6 −
3 𝑥2 − 6 𝑥 + 9 1 2√3
REMEMBER: To succeed in this topic you should understand the definitions of a derivate as a limit, as two points of a function get infinitesimally close. The relationship between differentiability and continuity. How derivatives are represented graphically, and numerically-analytically, and how they are conceived as an instantaneous rate of change. Also, you have to perform abundant practice to get required mastery. 52
Juan J. Prieto-Valdés
2.2. Techniques of differentiation. 𝒅
The Power Rule: 𝒅𝒙 𝒙𝒏 = 𝒏𝒙𝒏−𝟏
Example using all four properties
𝑐𝑜𝑛𝑠𝑡. = 0
𝑑 (4𝑥 3 + 2𝑥 2 − 16𝑥 + 10) = 𝑑𝑥 𝑑 3 𝑑 𝑑 𝑑 =4 𝑥 + 2 𝑥 2 − 16 𝑥 + 10 = 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 = 12𝑥 2 + 4𝑥 − 16
Demonstrations in Chaptrer-3
Exponential and logarithmic differentiation.
𝑑
𝑑
𝑑 𝑑𝑥
Sum Rule: 𝑑𝑥 [(𝑓(𝑥) + 𝑔(𝑥)] = 𝑑𝑥 𝑓(𝑥) + Constant Multiple Rule: Constant Rule:
𝑑 𝑑𝑥
Exponent:
𝑑 𝑑𝑥
𝑐𝑥 𝑛 = 𝑐
𝒅 𝒙 𝒆 𝒅𝒙
Natural Logarithm:
𝑑 𝑑𝑥
𝑥𝑛
𝑔(𝑥)
𝑑 4 (𝑥 − 3𝑒 𝑥 ) = 4𝑥 3 + 3𝑒 𝑥 𝑑𝑥 𝑑 4 4 (𝑥 − 4 ln 𝑥 ) = 4𝑥 3 + 𝑑𝑥 𝑥
= 𝒆𝒙 𝑑
𝑑𝑥
1
ln 𝑥 = 𝑥
2.3. Product and Quotient Rules Several rules have been developed for finding the derivatives without having to use the definition directly. These rules simplify the process of differentiation. The Product Rule is defined as the product of the first function and the derivative of the second function plus the product of the derivative of the first function and the second function: 𝑑 𝑑 𝑑 [𝑓(𝑥)𝑔(𝑥)] = [ 𝑓(𝑥)] [ 𝑔(𝑥)] + [𝑓(𝑥)] [ 𝑔(𝑥)] 𝑑𝑥 𝑑𝑥 𝑑𝑥 One of the simplest demonstrations is based on the application of the limit derivative definition: Add and subtract
[𝑓(𝑥 + ℎ)][ 𝑔(𝑥 + ℎ)] − [𝑓(𝑥)][𝑔(𝑥)] ℎ→0 ℎ lim
𝑓(𝑥)𝑔(𝑥 + ℎ)
[𝑓(𝑥 + ℎ)][ 𝑔(𝑥 + ℎ)] + [𝑓(𝑥)𝑔(𝑥 + ℎ)] − [𝑓(𝑥)][ 𝑔(𝑥 + ℎ)] − [𝑓(𝑥)][𝑔(𝑥)] ℎ→0 ℎ lim
Group factoring, apply limit when h towards 0 and get: the result:
[𝑔(𝑥 + ℎ)][ 𝑓(𝑥 + ℎ) − 𝑓(𝑥)] + 𝑓(𝑥)[𝑔(𝑥 + ℎ) − 𝑔(𝑥)] = ℎ→0 ℎ lim
𝑓(𝑥 + ℎ) − 𝑓(𝑥) 𝑔(𝑥 + ℎ) − 𝑔(𝑥) + [𝑓(𝑥)] lim = ℎ→0 ℎ→0 ℎ ℎ = [ 𝑔(𝑥)][𝑓 ′(𝑥) ] + [𝑓(𝑥)][𝑔′(𝑥) ]
[𝑔(𝑥)] lim
Similar to product rule, the Quotient Rule is a way of differentiating division case. It is defined as the quantity of the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator all over the denominator squared. Suppose that:
𝑤(𝑥) =
𝑓(𝑥) 𝑔(𝑥)
So 𝑓 = 𝑤𝑔
𝑓(𝑥) = 𝑤(𝑥)𝑔(𝑥)
Apply product rule
𝑓 ′ (𝑥) = 𝑤(𝑥)𝑔′ (𝑥) + 𝑤 ′(𝑥) 𝑔(𝑥) 53
Business Calculus Notebook
𝑓(𝑥) ′ 𝑓(𝑥) ′ 𝑓 ′ (𝑥) − 𝑔 (𝑥) 𝑓 ′ (𝑥) − 𝑔 (𝑥) ′ (𝑥) ′ (𝑥) 𝑓 − 𝑤(𝑥)𝑔 𝑔(𝑥) 𝑔(𝑥) 𝑤 ′ (𝑥) = = = 𝑔(𝑥) 𝑔(𝑥) 𝑔(𝑥)
Solve for 𝑤 ′(𝑥) And simplify
′(
𝑤 𝑥) =
𝑔(𝑥)𝑓′ (𝑥)−𝑓(𝑥) 𝑔′ (𝑥) 𝑔2 (𝑥)
or
𝑑
[
𝑓(𝑥)
𝑑𝑥 𝑔(𝑥)
]=
[
𝑑 𝑑 𝑓(𝑥)][ 𝑔(𝑥)]−[𝑓(𝑥)][ 𝑔(𝑥)] 𝑑𝑥 𝑑𝑥 𝑔(𝑥)2
Example (product rule) 1 𝑑 𝑑 𝑑 [(𝑥 2 + 3)(𝑥 − 5)] = [ (𝑥 2 + 3)[ (𝑥 − 5)] + [(𝑥 2 + 3)] [ (𝑥 − 5)] = 3𝑥 2 − 10𝑥 + 10 𝑑𝑥 𝑑𝑥 𝑑𝑥
Example (quotient rule) 2: 𝑑 𝑑 (𝑥 − 5) [ (𝑥 2 + 3)] − [ (𝑥 − 5)] (𝑥 2 + 3) 𝑥 2 − 10𝑥 − 3 𝑑 (𝑥 2 + 3) 𝑑𝑥 𝑑𝑥 = ( )= (𝑥 − 5)2 (𝑥 − 5)2 𝑑𝑥 (𝑥 − 5)
Homework Topic 2.2 and 2.3. In exercises 63 through 70 differentiate using derivative properties 63)
y = −5 0
64)
𝑓(𝑥) = 𝑥 2 − 5𝑥 + 10 2𝑥−5
65)
𝑓(𝑥) = 3𝑥 + 4 3
66)
𝑦 = 6 + (𝑥 + 1)2 2𝑥+2
67)
𝑓(𝑥) =
8 √𝑥
−
4 3
𝑥2 68)
𝑓(𝑥) =
3𝑥 2 − 𝑥 + 2𝑥 6 𝑥 12 𝑥 5 + 3
69)
70)
𝑓(𝑥) =
1 2 3 + 2+ 3 𝑥 𝑥 𝑥
5⁄ 6
𝑓(𝑥) = (3𝑥)
−
1 4 9 − 3− 4 2 𝑥 𝑥 𝑥 5 6
2 √3 𝑥 54
Juan J. Prieto-Valdés
In exercises, 71 – 76 find equation of the tang. line to the graph of the given function at the indicated point. 71)
y = −x 3 + x 2 𝑎𝑡 𝑥 = −1, 0, +1
𝑑𝑦 𝑑𝑥
72)
= 2 𝑥 − 3 𝑥2
𝑦 = (𝑥 + 2)2 + 3; 𝑎𝑡 𝑥 = −2, 1, 𝑎𝑛𝑑 2
𝑑 [(𝑥 𝑑𝑥
73)
𝐹𝑜𝑟 𝑥 = −1, 𝑦 = −5𝑥 − 3 𝐹𝑜𝑟 𝑥 = 0, 𝑦 = 0. etc.
+ 2)2 + 3] = 2 𝑥 + 4
Next; write the equation of the line
1 3 𝑦 = ( 𝑥) + 2 ; 𝑎𝑡 𝑥 = 1 3
3 𝑑 1 𝑥) [( 𝑑𝑥 3
𝑥2
+ 2]= 9 Next; write the equation of the line 55
Business Calculus Notebook
74)
𝑦 = 3 − √𝑥; 𝑎𝑡 𝑥 = 6
𝑑 𝑑𝑥[3−√𝑥]
75)
= −2
1 √𝑥
Next; write the equation of the line
𝑦 = −2𝑥 2 − 4𝑥 + 2 at 𝑥 = 1, 𝑥 = 0, 𝑥 = 1
𝑑 (−2𝑥 2 𝑑𝑥
76)
− 4𝑥 + 2)=−4 𝑥 − 4
If the volume of production (𝑦) in a specific manufacturing factory is represented by 𝑦 = 0.5𝑥 3 − 24𝑥 + 250; Where 𝑥 = ℎ𝑜𝑢𝑟𝑠 𝑎𝑓𝑡𝑒𝑟 8 𝐴𝑀 (use 𝑥 ≥ 0) 𝑑𝑦
a) Calculate the volume of production and the rate of change 𝑑𝑥 at 8 AM (x=0) and at 4 PM (x=8). b) Calculate when (at what time?) the production is minimal.
0.5𝑥 3 − 24𝑥 + 250 3 𝑥2 − 24 = 0 2 𝑥 = −4 𝑜𝑟 𝑥 = 4
56
Juan J. Prieto-Valdés
In exercises 77 - 82 differentiate given expression using product rule 77) (𝑥 + 2)(𝑥 − 5)
2𝑥−3 78) 2(𝑥 2 − 5𝑥 + 10)( 2𝑥 + 6) 12 𝑥 2 − 16 𝑥 − 20 79)
1 (3𝑥 + 4)( + 2𝑥) 𝑥 12 𝑥 + 8 −
4 𝑥2
80) (3 − 𝑥)2 (𝑥 3 ) 5 𝑥 4 − 24 𝑥 3 + 27 𝑥 2 81)
8 (𝑥 √𝑥
1
+ 𝑥) 4 √𝑥
82)
(2𝑥)6 (
−
12 5
𝑥2
3𝑥 2 − 𝑥 ) 𝑥
1344 𝑥 6 − 384 𝑥 5 In exercises 83 -90 through 81 differentiate given expression using quotient rule. 83) −𝑥 3 + 𝑥 2 (𝑥 + 2)
4 𝑥 − 5 𝑥2 − 2 𝑥3 𝑥2 + 4 𝑥 + 4 84)
(𝑥 + 2)2 (
5 ) 𝑥−1
5 𝑥 2 − 10 𝑥 − 40 𝑥2 − 2 𝑥 + 1 57
Business Calculus Notebook
85)
1 3 (3 𝑥) + 2 𝑥2 1 4 − 3 27 𝑥
86)
√𝑥 (𝑥 + 2)
3−
− 87)
1 2 √𝑥 (𝑥 + 2)
+
√𝑥 (𝑥 + 2)2
−2𝑥 2 − 4𝑥 + 2 𝑥 2 − 25
𝑦=
𝑑𝑦 4 (25 + 24 𝑥 + 𝑥 2 ) = (−25 + 𝑥 2 )2 𝑑𝑥 4 𝑥 2 + 96 𝑥 + 100 𝑥 4 − 50 𝑥 2 + 625 88) If the volume of production (𝑦) in a specific manufacturing factory is represented by
𝑦=
12𝑥 3 −24𝑥+240
𝑥+1
. Where 𝑥 = ℎ𝑜𝑢𝑟𝑠 𝑎𝑓𝑡𝑒𝑟 8 𝐴𝑀 (use 𝑥 ≥ 0)
Calculate the volume of production and the rate of change of the production at 8 AM (𝑥 = 0) and them, at 4 PM (𝑥 = 8). Calculate when (at what time?) the production is minimal (give the expression, no numeric value is required).
12𝑥 3 − 24𝑥 + 240 𝑥+1 3 𝑑𝑦 12 (2 𝑥 + 3 𝑥 2 − 22) = (𝑥 + 1)2 𝑑𝑥 𝑦=
58
Juan J. Prieto-Valdés
89) Find all points on the graph of 𝑦 = 𝑥 2 +7𝑥 −11 where the tangent line is a horizontal line: 𝑥 2 −7𝑥+11
𝑑𝑦(𝑥) 𝑑𝑥
When
𝑑𝑦 𝑑𝑥
= 0;
2 (22−7 𝑥)𝑥 (𝑥 2 −7 𝑥+11)2
2 (22−7 𝑥)𝑥
= (𝑥 2 −7 𝑥+11)2
= 0 so 𝑥 = 0 𝑜𝑟 𝑥 =
22 7
22
Point ONE (0, 𝑦1 ) , and SECOND ( 7 , 𝑦2 ); where 𝑦1 and 𝑦2 can be found from the original equation: For example, for the first point: 𝑦=
02 +7(0)−11 02 −7(0)+11
= −1; Horizontal line 𝑦 = 1
90) It is estimated that the population of certain type of bird 𝑥 years from now will be: 24
𝑦 = −𝑥 3 + 200𝑥 + 500 + 𝑥+1. Determine if the population will be maximal in any moment. Use Wolfram Alpha or Microsoft Mathematics to visualize your answer. Consider logic real domain of the function (only positive time and population values). Check your answer graphically.
.
−𝑥 3 + 200𝑥 + 500 + −3 𝑥 2 + 200 − 𝑀𝑎𝑥: 𝑦 = 1591.28,
24 𝑥+1
24 =0 (𝑥 + 1)2 𝑥 = 8.1591 59
Business Calculus Notebook
2.4. The Chain Rule Definition of the Chain Rule or differentiation of a composite function: If 𝒚 = 𝒚(𝒖) is a function of 𝒖 and 𝒖 = 𝒖(𝒙) is a function of 𝒙, then the derivative of 𝒚 with respect to 𝒙 equals the derivative of 𝒚 with respect to 𝒖 times the derivative of 𝒖 with respect to 𝒙: With other words: the chain rule is used to differentiate a function of a function. 𝑑 𝑑 𝑑 𝑑 𝑑𝑓(𝑢) 𝑑𝑢 𝑓(𝑢) = [𝑓(𝑔(𝑥))] = ( 𝑓(𝑔(𝑥))) ( 𝑔(𝑥)) 𝑜𝑟 𝑐ℎ𝑎𝑛𝑔𝑖𝑛𝑔 𝑡ℎ𝑒 𝑣𝑎𝑡𝑖𝑎𝑏𝑙𝑒: 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑢 𝑑𝑥
Example 1: Given: 𝒉(𝒙) =(𝒙𝟐 − 𝟓𝒙 + 𝟓)𝟑 , find 𝒉′ (𝒙). 𝑔(𝑥) = 𝑥 2 − 5𝑥 + 5, while 𝑓(𝑥) = 𝑥 3 ; 𝑑𝑓 𝑑𝑥
= 3(𝑥 2 − 5𝑥 + 5)2
and
𝑑𝑔 = 2𝑥 𝑑𝑥 2 (2𝑥
ℎ′ (𝑥) = 3(𝑥 2 − 5𝑥 + 5)
𝑢 = (𝑥 2 − 5𝑥 + 5), 𝑑𝑢/𝑑𝑥 = 𝑢′ = 2𝑥 + 5 −5
𝑑 𝑓(𝑢) 𝑑𝑥
=
𝑑𝑢3 𝑑𝑢 𝑑𝑢 𝑑𝑥
= 3𝑢2 𝑢′ = 3(𝑥 2 − 5𝑥 + 5)2 (2𝑥 + 5)
− 5)
Example 2: 𝒚 = √𝒙 + 𝒆𝒙 𝑢 = 𝑥 + 𝑒𝑥 𝑑𝑢 = 1 + 𝑒𝑥 𝑑𝑥 𝑑𝑦 𝑑(√𝑢) 𝑑𝑢 = 𝑑𝑥 𝑑𝑥 𝑑𝑥
y x ex (x ex ) 2 1 1 y ' ( x e x ) 2 1 (1 e x ) 2 1
=
1 + 𝑒𝑥 2√𝑥 + 𝑒 𝑥
Example 3: 𝒚 = (𝟏 − 𝟑𝒙)𝟑 (𝟐𝒙𝟐 + 𝟑)𝟓 𝑑𝑦 = (1 − 3𝑥)3 5(2𝑥 2 + 3)4 (4𝑥) 𝑑𝑥 + (2𝑥 2 + 3)5 3(1 − 3𝑥)2 (−3)
1.
Apply the product rule
2.
Apply the General Power Rule for derivatives and use Chain Rule with one of the terms, and.
= −(1 − 3 𝑥)2 (2 𝑥 2 + 3)4 (78 𝑥 2 − 20 𝑥 + 27)
3.
Simplify as possible.
1.
Apply the product rule
2.
Apply the General Power Rule for derivatives and use Chain Rule with one of the terms, and.
3.
Simplify as possible.
Example 4: 𝒚 = 𝒙𝟐 (𝐥𝐧(𝒙𝟐 )) 𝑑𝑦/𝑑𝑥 = (2𝑥(2ln(𝑥) + 2 𝑥 2 (1/𝑥) = 2𝑥(2 ln(𝑥) + 1)
60
Juan J. Prieto-Valdés
61
Business Calculus Notebook
2.5. Implicit Differentiation. An implicit function is a function that is defined implicitly (inside the equation) by an implicit equation; means: the variables are not isolated or represented independently to directly associate one of the variables (the value) to the others (the arguments). Implicit differentiation is the procedure of differentiating an implicit equation with respect to the desired variable x while treating the other variables as unspecified functions of 𝑥. Let’s see the following examples:
Example 1:
Differentiating implicitly instead gives
𝒙𝒚 = 𝟏
𝑑 (𝑥𝑦) 𝑑𝑥
can be solved for 𝑦:
Applying product rule, we get:
1
𝑦 = 𝑥 = 𝑥 −1 , 𝑥
and differentiated directly to get: 𝑑𝑦 1 =− 2 𝑑𝑥 𝑥 In this case direct differentiation is notably easer. However, this example is a good one to illustrate the implicit differentiation procedure (on the right)
Example 2: 𝒙𝟐 − 𝟒𝒙𝒚 + 𝟔𝒚𝟐 = 𝟏𝟖 ,
𝑑𝑥
+𝑦
Solving for 𝑑𝑦 𝑑𝑥
=−
(𝑑𝑥) 𝑑𝑥
𝑑𝑦 𝑑𝑥 𝑦 𝑥
or
=0
𝑥
(𝑑𝑦) 𝑑𝑥
+𝑦 = 0
and expressing only in 𝑥 values or
𝑑𝑦 𝑑𝑥
=−
1 𝑥
𝑥
=−
𝑑𝑦 𝑑𝑦 + 12𝑦 =0 𝑑𝑥 𝑑𝑥
𝑑𝑦 (4𝑥 + 12𝑦) = 0 2𝑥 − 4𝑦 − 𝑑𝑥
1. Take the derivative of each element. 2. Note that in the second term 4𝑥𝑦 you have to apply the product rule as you have two variables 𝑥 𝑎𝑛𝑑 𝑦. 𝑑𝑦 , 𝑑𝑥 𝑑𝑦 for 𝑑𝑥
3. Factor to isolate the furthermore, solve
𝑑𝑦 2𝑥 − 4𝑦 = 𝑑𝑥 4𝑥 + 12𝑦
Example 3: 𝒆𝒙𝒚 = 𝒆𝟑𝒙 − 𝒆𝟒𝒚 , Find 𝒚′ 𝑒
𝑥𝑦 (𝑥𝑦 ′
+ (1)𝑦) = 𝑒
3𝑥 (3)
1. Use exponential differentiation, 4𝑦
− 4𝑒 𝑦′
𝑒 𝑥𝑦 (𝑥𝑦 ′ ) + 𝑒 𝑥𝑦 (𝑦) = 𝑒 3𝑥 (3) − 4𝑒 4𝑦 𝑦 ′ 𝑒 𝑥𝑦 (𝑥𝑦 ′ ) + 4𝑒 4𝑦 𝑦 ′ = −𝑒 𝑥𝑦 (𝑦) + 𝑒 3𝑥 (3) 𝑦 ′ (𝑥𝑒 𝑥𝑦 + 4𝑒 4𝑦 ) = 3𝑒 3𝑥 − 𝑦𝑒 𝑥𝑦 𝑦′ =
62
3𝑒 3𝑥 − 𝑦𝑒 𝑥𝑦 𝑥𝑒 𝑥𝑦 + 4𝑒 4𝑦
1 𝑥2
𝒅𝒚
𝑑𝑥 𝑑𝑥 𝑑𝑦 𝑑𝑦 − 4 [𝑦 + 𝑥 ] + 12𝑦 =0 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 2𝑥 − 4𝑦 − 4𝑥
(𝑑𝑦)
Find 𝒅𝒙:
𝑑(𝑥 2 ) 𝑑(4𝑥𝑦) (6𝑦 2 ) 𝑑(18) − + = 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 2𝑥
𝑑
= 𝑑𝑥 (1) = 0
2. chain rule and 3. product rule. 4. Them collect 𝑦 ′ and 5. Solve for 𝑦 ′
Juan J. Prieto-Valdés
63
Business Calculus Notebook
2.6. High Order Derivatives (second, third, …): The second derivative, or second order derivative, is the derivative of the derivative of a function. The derivative of the function may be denoted by 𝑓’’(𝑥) 𝑜𝑟
𝑑 2 𝑓(𝑥) . 𝑑𝑥 2
Because the derivative of function is defined as a function
representing the slope of function, the double derivative is the function representing the slope of the first derivative function. Also, it can be explained from the physical point of view using the rate of change. In mechanics for example: the rate of change of the distance during the time is de velocity, while the rate of change of the velocity in time is the acceleration. Furthermore, the third derivative is the derivative of the derivative of the derivative of a function and can be represented as: 𝑓 ′′′(𝑥) 𝑜𝑟
𝑑 3 𝑓(𝑥) . 𝑑𝑥 3
This is read as “𝑓 𝑡𝑟𝑖𝑝𝑙𝑒 𝑝𝑟𝑖𝑚𝑒 𝑜𝑓 𝑥", or "The third derivative of ". This can
continue as long as the resulting derivative is itself differentiable, with the fourth derivative, the fifth derivative, and so on. Any derivative beyond the first derivative can be referred to as a higher order derivative.
Example 1: 𝒇(𝒙) = 𝟑𝒙𝟒 − 𝟓𝒙𝟑 + 𝟒𝒙 − 𝟐 𝑓 ′ (𝑥) = 12𝑥 3 − 15𝑥 2 + 4
Example 2: 𝒚 = −𝒆−𝟐𝒙 + 𝐥 𝐧 𝟒𝒙
𝑓 ′′′ (𝑥) = 72𝑥 − 30
1 𝑥 ′′ −2𝑥 𝑦 = −4𝑒 − 𝑥 −2
𝑓 𝐼𝑉 (𝑥) = 72
𝑦 ′′′ = +8𝑒 −2𝑥 + 2𝑥 −3
𝑓 ′′ (𝑥) = 36𝑥 2 − 30𝑥
𝑦 ′ = +2𝑒 −2𝑥 +
𝑓 𝑉 (𝑥) = 0
Homework Topics 2.4, 2.5, and 2.6. In exercises 91-99 differentiate using the chain rule 91)
y =(
𝑥+2 2 ) 4
𝑥 1 + 8 4
92) 𝑓(𝑥) =(𝑥 2 − 5𝑥 + 10)3
(6 𝑥 − 15) (𝑥 2 − 5 𝑥 + 10)2 93) 𝑓(𝑥) = (2𝑥 + 3)5
10 (2 𝑥 + 3)4 94) 𝑦 = 6 + (𝑥 + 1)3 3 (𝑥 + 1)2
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95) 𝑦 = 6𝑥 + (𝑥 + 1)3 3 (𝑥 + 1)2 + 6 96)
𝑥 3 𝑓(𝑥) = ( ) 𝑥−3 9 𝑥2 (𝑥 − 3)4
97)
3𝑥 2 − 𝑥 ( ) 𝑥
1⁄ 2
+
3 4 3 2 √3 𝑥 − 1
98)
2𝑥 2 + 5𝑥 √ 𝑥−1
𝑑𝑦(𝑥) 5 + 4 𝑥 − 𝑦2 = − 𝑑𝑥 2𝑦−2𝑥𝑦
99) Find the equation of the line tangent to the circle at 𝑥 = −3 𝑦 2 + (x − 2)2 = 25; , at 𝑥 = 5
Check your answer in class In exercises 100 – 106 differentiate implicitly (use Wolfram Alpha if needed) 100) 1 3 𝑥 + 2𝑦 2 + 𝑥𝑦 + 8 = 0 3
𝑑𝑦(𝑥) 𝑥2 + 𝑦 = − 𝑑𝑥 𝑥+4𝑦
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101) (𝑥 + 2)2 + (𝑦 − 3)2 = 16
𝑑𝑦(𝑥) 2 + 𝑥 = 𝑑𝑥 3−𝑦 102)
𝑦 + 1 = √𝑥 (𝑦 2 )
𝑑𝑦 = 𝑑𝑥
𝑦2 1
2𝑥 2 + 4𝑥𝑦 or
𝑑𝑦 𝑦4 = 𝑑𝑥 2 + 2 𝑦 − 4 𝑥 𝑦 3 103) −2𝑥 2 − 4𝑥 + 4𝑦 3 + 2𝑦 = 7
𝑑𝑦(𝑥) 2 (1 + 𝑥) = 𝑑𝑥 1 + 6 𝑦2 104) Project question: (3𝑥𝑦 2 + 2)4 = 3𝑥 − 4𝑦
𝑑𝑦(𝑥) 3 (−1 + 32 𝑦 2 + 144 𝑥 𝑦 4 + 216 𝑥 2 𝑦 6 + 108 𝑥 3 𝑦 8 ) = − 𝑑𝑥 4 (1 + 48 𝑥 𝑦 + 216 𝑥 2 𝑦 3 + 324 𝑥 3 𝑦 5 + 162 𝑥 4 𝑦 7 )
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105) Find the equation of the tangent line to the function 𝑥𝑦 2 = 16 at 𝑥 = 4
At 𝑥 = 4, 𝑦 = 2; on the 1st quadrant. so, we are looking for the derivative at (4,2) point
𝑚=
𝑑𝑦(𝑥) 𝑑𝑥
𝑦
1
= − 2 𝑥 at (4.2) 𝑚 = − 4 𝑦 = 𝑚𝑥 + 𝑏 1
Substituting: 2 = (− 4) (4) + 𝑏, so 𝑏 = 3 Finally, we can write the equation of the line: 1 𝑦 =− 𝑥+3 4 106) Find the equation of the tangent line to the function 𝑥 3 𝑦 + 2𝑥𝑦 3 = 12 at the point (2,1):
𝑑𝑦/𝑑𝑥 = −(3 𝑥 2 𝑦 + 2 𝑦 3 )/(𝑥 3 + 6 𝑥 𝑦 2 ) At 𝑥 = 2 𝑎𝑛𝑑 𝑦 = 1 𝑑𝑦 = 𝑚 = −0.7 𝑑𝑥 𝑦 = 𝑚𝑥 + 𝑏 1 = −0.7(2) + 𝑏 2.4 = 𝑏 𝑦 = −0.7𝑥 + 2.4 67
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In exercises 107 - 109 find the indicated high order derivative 107)
1
𝑦 = 5𝑥 4 + 3𝑥 2 + 𝑥 2 ; find
𝑑𝑦 3 𝑑𝑥 3
𝑦 ′ = 20 𝑥 3 + 6 𝑥 + 𝑦 ′′ = 60 𝑥 2 + 6 − 𝑦 ′′′ = 120 𝑥 +
1 2 √𝑥 1 3
4 𝑥2 3 5
8 𝑥2 108) 𝑓(𝑥) = 3𝑥 7 − 2𝑥 2 + 3 ; find 𝑑𝑦2 𝑑𝑥 2
𝑓 ′ (𝑥) = 21 𝑥 6 − 4 𝑥 𝑓 ′′ (𝑥) = 126 𝑥 5 − 4
109) 𝑦 = 𝑥 (𝑥 2 + 2); find 𝑑𝑦3 √ 𝑑𝑥 3
3
5 𝑥2 1 𝑦′ = + 2 √𝑥 15 √𝑥 1 − 3 4 2 𝑥2 15 3 = + 8 √𝑥 4 𝑥 52
𝑦 ′′ = 𝑦 ′′′
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PRACTICING QUIZ-2. Derivatives 1. (15 points) Find the derivative of the 𝑓(𝑥) = 5𝑥 2 + 2𝑥 − 5 using the limit definition of the derivative (no other method will be accepted).
2.
(20 points) The curve 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 passes through the point (2, 12) and it is tangent to the line 𝑦 = 4𝑥 at the origin. Find a, b, and c. Show the graph and full work. Tip: you have two points (2, 12) and (0, 0)
3. (15 points) Find the derivative using the product rule: 𝑦 = 2(𝑥 2 − 5𝑥 + 10)( 2𝑥 + 6)
4. (15 points) Find the derivative using quotient rule: −2𝑥 2 − 4𝑥 + 2 𝑦= 𝑥 2 − 25
5. (15 points) Find the derivative using chain rule 𝑦 = 2𝑥(𝑥 2 + 3𝑥)5
6. (20 points) Find the equation of the line tangent to the circle 𝑦 2 + (𝑥 − 2)2 = 25 at (5, y), using 𝑥 = 5 and only positive ordinate 𝑦 > 0 . Show work and graph. You can use graphic utility to show your graph. Not for calculation. Tip: use implicit differentiation.
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2.7. Real Case Examples (Marginal Analysis and Approximations) The marginal analysis is the examination of the situation when we add one unit as an independent variable to the system. This technique is well observed using derivatives, it allows to study the effect on a cost, profit, and/or income ... by increasing the production, sales, and/or specific service in one unit, lets describe the concept using the cost of production 𝑥 units (similarly can be calculated for other functions: Profit, Revenue, etc.). 𝐶(𝑥 + 1) − 𝐶(𝑥) ℎ→0 ℎ
𝑀𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝐶𝑜𝑠𝑡 ≈ 𝐶 ′ (𝑥) = lim
𝑀𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝐶𝑜𝑠𝑡 =
𝐶(𝑥+1)−𝐶(𝑥) ℎ
at
ℎ = 1, 𝑀𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝐶𝑜𝑠𝑡 ≈ 𝐶(𝑥 + 1) − 𝐶(𝑥)
Example (Marginal Profit): Let’s suppose that the profit of producing some specific item is given by the function: 𝑃(𝑥) = 80𝑥 + 0.1𝑥 2 + 5000, also supouse that we want to know the profit of producing the 301th unit. Therefore, 𝑃(𝑥)′ = 80 + 0.2𝑥 The marginal profit at 𝑥 = 300 is 𝑃′ (300) = 80 + 0.2(300) = 140 Follow the ideas/concepts from above, $140.00 represents an estimate of the change in profit from 𝑥 = 300 to to 𝑥 = 301. Means, that the production of 301st item will increase the profit by $140.00. It is important to compare this approximate value to the exact change in profit as we move to the 301st item. Remember that the derivative gives us a point on the tangent line, which is not exactly the exact difference of the profit function from 𝑥 = 300 to 𝑥 = 301 as follow: 𝑃(301) − 𝑃(300) = [80(301) + 0.1(301)2 + 5000] − [80(300) + 0.1(300)2 + 5000] = 140.1 We can see that the estimate from the derivative is very close to the exact value. Example (Business situation) The daily cost to produce 𝑥 number of some specific product is given by 𝐶(𝑥) = 800 + 10𝑥 − 0.03𝑥 2 + 0.000005𝑥 3 , 𝑓𝑜𝑟 1 ≤ 𝑥 ≤ 10000 While the demand function for this specific product is given by, 𝑝(𝑥) = 20 − 0.005𝑥, 𝑓𝑜𝑟 1 ≤ 𝑥 ≤ 10000 Determine the marginal profit when the 251th item is sold. Considering the profit equal to the Revenue [𝑅(𝑥) = 𝑥 ∙ 𝑝(𝑥)] minus the cost [𝐶(𝑥)], we can write the Profit as 𝑃(𝑥) = 𝑥 𝑝(𝑥) − 𝐶(𝑥) = 𝑥(20 − 0.005𝑥) − (800 + 10𝑥 − 0.03𝑥 2 + 0.000005𝑥 3 ) = −5.0 × 10−6 𝑥 3 + 0.03 𝑥 2 + (20 − 0.005 𝑥)𝑥 − 10 𝑥 − 800 = −5.0 × 10−6 𝑥 3 + 0.025 𝑥 2 + 10 𝑥 − 800. The Marginal Profit will be given by: 𝑑𝑃(𝑥) 𝑑𝑥
At 𝑥 = 250,
70
𝑑𝑃(𝑥) 𝑑𝑥
= = −0.000015 𝑥 2 + 0.05 𝑥 + 10.
= 21.5625, which is the profit of producing the 251th item.
Juan J. Prieto-Valdés
To succeed in these topics, you should understand Graphically and numerically, the difference between instantaneous (virtual) rate of change and the rate of change when the coordinate increases in one unit (marginal analysis).
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Homework Topic 2.7. In exercises, 110 - 116 solve word problems. 110)
An automobile salesperson is paid $100 a week, plus $200 for every sale he makes. Let 𝑥 be the number of autos he sold in a week, and 𝑦 his weekly income in dollars. Model this situation algebraically
𝑦 = 200𝑥 + 100 111)
Suppose $10,000 is deposited into an account that earns 5% interest compounded annually and no further deposits or withdrawals are made. Let 𝐴(𝑡) denote the amount in the account 𝑡 years after the initial deposit (actual money). Find a formula for 𝐴(𝑡) and graph it. Evaluate A(t) after 12 years.
𝑟 𝑛𝑡 0.05 (1)(12) 𝐴(𝑡) = 𝑃𝑜 (1 + ) = 10000 (1 + ) = 17958.56 𝑛 1 112)
The concentration of oil diminishes 1% for every 0.5 miles from specific point in the Caribbean Sea where the concentration reached 20 % due to a platform leak. a) Find a model for the concentration distribution after any number of miles. b) How much of oil can be found 5 miles away from the leak? c) How far we can still find 0.5% of oil in the sea water?
𝐶 = 20 −2x Where 𝑥- represent the miles away from the leak 113)
The price (in cents) for a pound of sugar at the international market can be modeled by the (fictitious) function: 𝑦 = 0.0052𝑥 2 – 0.12𝑥 + 6.25 cents, where 𝑥 are years after 1960. a) Find the average rate of change of the price of sugar from 1980 to 1990. b) Estimate the rate of increase the sugar price in 2000 and in 2020. [0.0052(30)2 – 0.12(30)−0.0052(20)2 + 0.12(20)] 10
= 0.14
In 40 yr. a new function will be required/modeled. As a source of practice, we will use the same function 𝑦 ′ = 0.0104 𝑥 + 0.12, (𝑎𝑡 𝑥 = 40) 𝑦 ′ = 0.536 (𝑎𝑡 𝑥 = 60) 𝑦 ′ ≅ 0.744 Interesting fact (intermediate value theorem) 𝑦 ′ = 0.0104 𝑥 + 0.12, (𝑎𝑡 𝑥 = 25, 𝑚𝑒𝑑𝑖𝑎 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 20 𝑎𝑛𝑑 30), 𝑦 ′ = 0.14 72
Juan J. Prieto-Valdés
114)
A student spends: 𝑬 (𝒙) = 𝒙 + 0.25 (𝑥)𝟐 kilo dollars for x terms for the related school fees. At the present time, it has been taken 5 semesters. Find the cost of the fees, if the student takes an extra term (6th term). Use E(6) - E(5), and Marginal Analysis. Compare your results. If the difference is more than expected, explain the result.
𝑬 (𝒙) = 6 + 0.25 (6)𝟐 − 5 + 0.25 (5)𝟐 = 3.75 𝑦 ′ = 0.5𝑥 + 1 { 𝑎𝑡 𝑥 = 5} ≅ 3.5 115)
If The total annual cost (in dollars) incurred by the publishing company in producing 𝑥 pocketbooks is 𝐶(𝑥) = 2100 + 2.1 𝑥 – 0.00035 𝑥 2 , find the marginal cost of producing the 1001 book Use marginal analysis to ESTIMATE the cost incurred in producing the 1001-st book!
𝑦′ = − 116)
7𝑥 + 2.1 {𝑎𝑡 𝑥 = 1001] ≅ $1.40 10000
The weekly cost to produce x specific product is given by 𝐶(𝑥) = 8,000 + 100𝑥 − 0.03𝑥 2 + 0.000005𝑥 3 , 𝑓𝑜𝑟 1 ≤ 𝑥 ≤ 10000 While the demand function for this specific product is given by, 𝑝(𝑥) = 200 − 0.005𝑥, 𝑓𝑜𝑟 1 ≤ 𝑥 ≤ 10000 Determine the marginal cost, marginal revenue and marginal profit when 2500th item is sold. Assume that the company sells exactly what they produce.
Solution: Let’s get all functions we need (Revenue and Profit) 𝑅(𝑥) = 𝑥(200 − 0.005𝑥) = 200𝑥 − 0.005𝑥 2 𝑃(𝑥) = (200𝑥 − 0.005𝑥 2 ) − (8,000 + 100𝑥 − 0.03𝑥 2 + 0.000005𝑥 3 )= = −0.000005 𝑥 3 + 0.025 𝑥 2 + 100 𝑥 − 8000 Now, all the marginal functions are, 𝐶 ′ (𝑥) = 100 − 0.06𝑥 + 0.000015𝑥 2 {𝑎𝑡 𝑥 = 2500} = 43.75 𝑅 ′ (𝑥) = 200 − 0.01𝑥 {𝑎𝑡 𝑥 = 2500} = $175.00 𝑃′ (𝑥) = −
3 𝑥2 𝑥 525 + + 100 {𝑎𝑡 𝑥 = 2500} = 200000 20 4 73
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2.8.
Increasing and Decreasing Functions (critical points) A function is increasing on an interval if for any 𝑥1 < 𝑥2
implies 𝑓(𝑥1 ) < 𝑓(𝑥2 )
A function is decreasing on an interval if for any 𝑥1 < 𝑥2
implies 𝑓(𝑥1 ) > 𝑓(𝑥2 )
Theorem: Let f be a differentiable function on the interval (𝑎, 𝑏) then If 𝑓′(𝑥) > 0 for 𝑥 in (𝑎, 𝑏), then f is increasing there. If 𝑓′(𝑥) < 0 for 𝑥 in (𝑎, 𝑏), then 𝑓 is decreasing there. If 𝑓′(𝑥) = 0 for 𝑥 in (𝑎, 𝑏), then f is constant.
Definition of Maximum and Minimum 𝑓(𝑥) has a maximum at 𝑥 = 𝑐 if 𝑓(𝑥) < 𝑓(𝑐) for every 𝑥 in the domain we are working on [𝑥1 , 𝑥2 ]. 𝑓(𝑥) has a minimum at 𝑥 = 𝑐 if 𝑓(𝑥) > 𝑓(𝑐) for every 𝑥 in the domain we are working on[𝑥1 , 𝑥2 ].
Fermat’s Theorem If 𝑓(𝑥) has a relative extreme at 𝑥 = 𝑐 and 𝑓 ′ (𝑐) exists, then 𝑥 = 𝑐 is a critical point (Maximum or Minimum) if 𝑓 ′ (𝑐) = 0.
Relative vs Absolute Relative and absolute Maximum or Minimum are slight different as can be observed on the graph:
Quick Example: Given the function 𝑦 = 2𝑥 3 − 9𝑥 2 − 60𝑥 − 24, locate the x position of the extrema (max. and min.) if exist. Differentiate the function and make it equal to zero, them solve the equation for 𝑥: 𝑦 ′ = 6 𝑥 2 − 18 𝑥 − 60 = 6(𝑥 2 − 3𝑥 − 10) = (𝑥 − 5)(𝑥 + 2) = 0. Resulting that 𝑥 = −2 𝑜𝑟 𝑥 = 5 are the solutions. These x values are the x values for the Max and Min (graph on the right). At the extrema the tangent line to the graph has the slope equal to zero; horizontal line). The function is increasing from −∞ to 𝑥 = −2 (in this interval the derivative of the function always takes positive value. From 𝑥 = −2 𝑡𝑜 𝑥 = 5, decreasing function (the derivative of the function takes negative value. The function increases after 𝑥 = 5, having positive derivative.
To succeed in these topics, you should understand How to construct the graph of the function and derivative of the function. The concept of “extrema” and how to locate Maximum, and Minimum, what are absolute and relative Maximum & Minimum. 74
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2.9. Rolle's Theorem and The Mean Value Theorem 2.9.1.
Rolle’ s Theorem: Let 𝐟 be a continuous and differentiable function defined on a closed interval [𝐚, 𝐛]. If 𝐟(𝐚) = 𝐟(𝐛), then there is at least one number 𝐱 = 𝐜 in the open interval (𝐚, 𝐛) for which 𝐟(𝐜) = 𝟎.
2.9.2.
The Mean Value Theorem: If 𝐟 is continuouson the closed interval [𝐚, 𝐛]and differentiable on the open interval (𝐚, 𝐛), then there exist a number 𝒄 in (𝒂, 𝒃) such that: 𝑓 ′ (𝑐) =
𝑓(𝑏) − 𝑓(𝑎) 𝑏−𝑎
Proof: Let’s write the equation of the line passing thought the points 𝑃1 𝑎𝑛𝑑 𝑃2 : 𝑓(𝑏) − 𝑓(𝑎) 𝑦=[ ] (𝑥 − 𝑎) − 𝑓(𝑎) 𝑏−𝑎 Now. the difference between 𝑓(𝑥) 𝑎𝑛𝑑 𝑦 can be written as 𝑔(𝑥): 𝑓(𝑏) − 𝑓(𝑎) 𝑔(𝑥) = 𝑓(𝑥) − 𝑦 = 𝑓(𝑥) − [ ] (𝑥 − 𝑎) − 𝑓(𝑎) 𝑏−𝑎 Further by evaluating 𝑔(𝑥)𝑖𝑛 𝑎 𝑎𝑛𝑑 𝑎𝑙𝑠𝑜 𝑖𝑛 𝑏, we get the 𝑔(𝑎) = 𝑔(𝑏) = 0: So we can consider that there is a point between 𝑎 𝑎𝑛𝑑 𝑏 (constant level), were 𝑔′ (𝑐) = 0. In such case; 𝑔′ (𝑐) = 𝑓 ′ (𝑐) − 𝑦 = 𝑓 ′ (𝑐) − Resulting that 𝑓 ′ (𝑐) =
𝑓(𝑏) − 𝑓(𝑎) 𝑏−𝑎
Example: 1 Find the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 of 𝑓(𝑥) = 𝑥 2 − 5𝑥 + 6 , and show that there is a point belonging to the interval formed by the two intercepts where 𝑓 ′ (𝑐) = 0 𝑓(𝑥) = 𝑥 2 − 5𝑥 + 6 = (𝑥 − 2)(𝑥 − 3) = 0 Our interval will be [2, 3] 𝑓 ′ (𝑥) = 2𝑥 − 5 = 0,
𝑠𝑜 𝑥 =
5 2
Example 2. Verify that the function 𝑓(𝑥) = 𝑥 3 − 3𝑥 + 5, 𝑎𝑡 − 1 ≤ 𝑥 ≤ 1 satisfy the conditions of the Mean Value Theorem: 𝑓(−1) = 7 𝑎𝑛𝑑 𝑓(1) = 3 𝑓 ′ (𝑥) = 3𝑥 2 − 3 𝑎𝑛𝑑 𝑓 ′(𝑐) =
𝑓(1) − 𝑓(−1) = −2 = 1 − (−1)
𝑓 ′ (𝑐) = 3𝑐 2 − 3 = −2 1 √3 𝑐=√ =± ≈ ± 0.577 3 3
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𝑓(𝑏)−𝑓(𝑎) 𝑏−𝑎
= 0.
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Homework Topic 2.8 and 2.9. In exercises 117-125 determine the values of x where the function is increasing or decreasing (write the intervals), locate the maximum and minimum if exist, verify your answer using Wolfram Alpha: 117) 𝑦 = 2𝑥 3 + 3𝑥 2 − 12𝑥 − 18 𝑥
2 (2𝑥
+ 3) − 6(2𝑥 + 3) = 0 2
(2𝑥 + 3)(𝑥 − 6) = 0
6(𝑥 + 2)(𝑥 − 1) = 0 𝑥 = −2 (𝑀𝑎𝑥), 𝑥 = 1 (𝑀𝑖𝑛) 𝑓(−4) < 𝑓(−2), 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
(2𝑥 + 3)(𝑥 + √6)(𝑥 − √6) = 0
𝑓(−2) > 𝑓(1), 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
3 𝑥=− , −√6, 𝑎𝑛𝑑 + √6 2 First derivative Test:
𝑓(1) < 𝑓(4), 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
𝑦 ′ = 6 (𝑥 2 + 𝑥 − 2 ) = 0
Later we will study more advanced methods to verify if the critical point is a Max or a Min.
118) 𝑦 = 𝑥 2 − 5𝑥 + 10
119) 𝑦 = 3𝑥 + 4
120) 𝑦 = 6 + (𝑥 + 1)3
𝑑𝑦 = 3 (𝑥 + 1)2 𝑑𝑥 78
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121) 𝑦 = 6𝑥 + (𝑥 + 1)3
𝑑𝑦 = 3 (𝑥 + 1)2 + 6 𝑑𝑥 122)
𝑦=
𝑥 𝑥−3
𝑑𝑦 3 =− 2 𝑑𝑥 𝑥 −6𝑥+9 123)
𝑦=
3𝑥 3 − 𝑥 3 + 𝑥 4
𝑑𝑦 =6𝑥 𝑑𝑥 124)
𝑦=
3𝑥 4 − 2𝑥 3 3 + 𝑥 4
𝑑𝑦 = 9 𝑥 2 − 4𝑥 − 𝑑𝑥
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125) 𝑦 = −2𝑥 2 − 4𝑥 + 2
𝑑𝑦 = −4 𝑥 − 4 𝑑𝑥
2.10. Concavity and Point of Inflexion A function 𝑓 ( 𝑥 ) is called convex in an interval ( 𝑎, 𝑏), if the function 𝑓 ( 𝑥 ) graph is placed in this interval lower than a tangent line, going through any point ( 𝑥0 , 𝑓 ( 𝑥0 )), where 𝑥0 ∈ (𝑎, 𝑏). A function 𝑓 ( 𝑥 ) is called concave in an interval (𝑎, 𝑏), if the function 𝑓 ( 𝑥 ) graph is placed in this interval higher than a tangent line, going through any point ( 𝑥0 , 𝑓 ( 𝑥0 )), where 𝑥0 ∈ (𝑎, 𝑏).
Sufficient condition of concavity (convexity) of a function: Let a function 𝑓 ( 𝑥 ) be twice differentiable in an interval ( 𝑎, 𝑏 ), then: if 𝑓 ′′ ( 𝑥 ) > 0 for any 𝑥 ∈ ( 𝑎, 𝑏 ), then the function 𝑓 ( 𝑥 ) is concave in the interval ); if 𝑓 ′′ ( 𝑥 ) < 0 for any 𝑥 ∈ ( 𝑎, 𝑏 ), then the function 𝑓 ( 𝑥 ) is convex (concave down) in the interval ( 𝑎, 𝑏 ). If a function changes a convexity to a concavity or vice versa at passage through specific point, then this point is called an inflexion point. That if the second derivative 𝑓 ′′ (𝑥) exists in an inflexion point𝑥0 , then 𝑓 ′′ (𝑥0 ) = 0. Quick Example: 3
Given the function 𝑦 = 2𝑥 − 9𝑥 2 − 60𝑥 − 24, to locate the x position of the extrema (max. and min.) we have to differentiate the function and make the obtained derivative function equal to zero, them solve the equation for 𝑥: as follow: 𝑦 ′ = 6 𝑥 2 − 18 𝑥 − 60 = 0. Resulting that 𝑥 = −2 𝑜𝑟 𝑥 = 5 are the solutions. These x values are the max. and min which can be identified using the second derivative test: 𝑦 ′′ = 12 𝑥 − 18 At 𝑥 = −2, 𝑦’’ < 0, so this in a maximum (function is concave down). At 𝑥 = 5, 𝑦” > 0, so in this point is located the minimum (function is concave up). The second derivative will be equal to zero at: 12 𝑥 − 18 = 0 3 2
𝑥 = , which is the x position of the inflexion point (black point)
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Juan J. Prieto-Valdés
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Homework Topic 2.10. In exercises 126-135 find the exact position of maximum, minimum and point of inflexion if exist. Note that only partial solutions are included. For full solution use Wolfram Alfa or Microsoft Mathematics 126)
𝑦 = 2𝑥 3 + 3𝑥 2 − 12𝑥 − 18 𝑥
2 (2𝑥
+ 3) − 6(2𝑥 + 3) = 0
(2𝑥 + 3)(𝑥 2 − 6) = 0
(𝑥 + 2)(𝑥 − 1) = 0 𝑥 = −2 (𝑀𝑎𝑥), 𝑥 = 1 (𝑀𝑖𝑛) Because: Second Derivative Test: 𝑦 ′′ = 2𝑥 + 1
(2𝑥 + 3)(𝑥 + √6))(𝑥 − √6) = 0 3
𝑥 = − 2 , −√6, 𝑎𝑛𝑑 + √6 First derivative Test: 𝑦 ′ = 6 (𝑥 2 + 𝑥 − 2 ) = 0
′′ ′′ 𝑦𝑎𝑡 𝑥=−2 < 0, 𝑎𝑛𝑑 𝑦𝑎𝑡 𝑥=−1 > 0
𝑦 ′′ = 2𝑥 + 1 = 0, 1 𝑥 = − (𝑖𝑛𝑓𝑙𝑒𝑥𝑖𝑜𝑛 𝑝𝑜𝑖𝑛𝑡) 2
Follow exercise 108
Repeat the procedures (make your own calculation)
127) y = 16x 4 − 40x 2 + 25
𝑦 ′ = 64 𝑥 3 − 80 𝑥 = 0 𝑥 = 0 𝑜𝑟 𝑥 = 𝑦 ′′ = −80 + 192 𝑥 2 = 0, so: 𝑥 =
82
√5 √5 𝑜𝑟 𝑥 = − 2 2 √15 6
𝑜𝑟 𝑥 = −
√15 6
Juan J. Prieto-Valdés
128) 𝑦 = 25𝑥 3 + 15𝑥 2 − 10𝑥 − 6
Solve: 25𝑥 3 + 15𝑥 2 − 10𝑥 − 6 = 0; 𝑥=−
3 √10 √10 𝑜𝑟 𝑥 = 𝑜𝑟 𝑥 = − 5 5 5 First derivative test
𝑦 ′ = 5 (−2 + 6 𝑥 + 15 𝑥 2 ) = 0 √(13) 1 3 𝑥 = ± 5 5 Second Derivative test: 𝑦 ′′ = 30 (1 + 5 𝑥) = 0 1 𝑥=− 5 129) Graph: 𝑦 = −6𝑥 3 + 15𝑥 2 + 12𝑥 − 30
Solve: −6𝑥 3 + 15𝑥 2 + 12𝑥 − 30 = 0 5 𝑜𝑟 𝑥 = √2 𝑜𝑟 𝑥 = −√2 2 Apply First and Second Derivative Tests. 𝑥=
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130) 𝑦 = 16𝑥 4 + 40𝑥 2 + 25
Solve: 16𝑥 4 + 40𝑥 2 + 25 = 0 . 𝑥=
√5𝑖 √5𝑖 𝑜𝑟 𝑥 = − 𝑜𝑟 2 2 𝑥=
√5𝑖 2
𝑜𝑟 𝑥 = −
𝑀𝑖𝑛 𝑎𝑡 𝑥 = 0, 𝑦 ′ = 16 𝑥 (5 + 4 𝑥
√5𝑖 2 2)
Apply Second Derivative Tests to verify that this is a Minimum. 131) 𝑦 = 16𝑥 4 − 40𝑥 2 + 25
Solve: 16x 4 − 40x 2 + 25 = 0 𝑥=
84
√5 √5 √5 √5 𝑜𝑟 𝑥 = − 𝑜𝑟 𝑥 = 𝑜𝑟 𝑥 = − 2 2 2 2 Apply First and Second Derivative Tests.
Juan J. Prieto-Valdés
132) 𝑦 = 6𝑥 4 − 140𝑥 2 + 125
Solve: 6𝑥 4 − 140𝑥 2 + 125 = 0 5 √166 35 5 √166 35 + 𝑜𝑟 𝑥 = + 6 3 6 3 Apply First and Second Derivative Tests.
𝑥=−
133) y = −25𝑥 3 + 15𝑥 2 + 10𝑥 − 6
1st derivative test:
𝑑𝑦 (−25𝑥 3 𝑑𝑥
+ 15𝑥 2 + 10𝑥 − 6) = −75 𝑥 2 + 30 𝑥 + 10
Solve for 𝑥: −75 𝑥 2 + 30 𝑥 + 10 = 0 Critical points (Max & Min) are located at: 𝑥=− 2nd derivative test:
𝑑𝑦 (−75 𝑥 2 𝑑𝑥
1 √39 + 15 5
𝑜𝑟 𝑥 =
1 √39 + 15 5
+ 30 𝑥 + 10) = 30 − 150 𝑥
Solve for 𝑥: 30 − 150 𝑥 = 0 1
The inflexion point is located at: 𝑥 = 5 Finally, to complete the graph, find the intercepts: 𝑿𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕𝒊𝒐𝒏 ; Solve for 𝑥 the original equation (𝑓(𝑥) = 0); −25𝑥 3 + 15𝑥 2 + 10𝑥 − 6 = 0 3
𝑥 = 5 𝑜𝑟 𝑥 =
√10 5
𝑜𝑟 𝑥 = −
√10 5
and 𝒀𝒊𝒏𝒕.{𝒂𝒕 𝒙=𝟎)=−𝟔
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134) y = −3𝑥 4 + 24𝑥 3 + 12𝑥 2
Solutions: 𝑥 = 0, 𝑥 = 4 ± 2√5 First derivative test: 𝑦 ′ = −12 𝑥 (−2 − 6 𝑥 + 𝑥 2 ) = 0 𝑥 = 0, 𝑥 = 3 ± √11 Second derivative test 𝑦 = 24 + 144 𝑥 − 36 𝑥 2 = 0 14 𝑥 =2±√ 3
135) 𝑦 = −𝑥 4 + 2𝑥 3 + 12𝑥 2
Solve: −𝑥 4 + 2𝑥 3 + 12𝑥 2 = 0 𝑥 = 0 𝑜𝑟 𝑥 = 0 𝑜𝑟 𝑥 = 1 − √13 𝑜𝑟 𝑥 = √13 + 1
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Juan J. Prieto-Valdés
PRACTICING QUIZ-3. Application of Derivatives 1.
The weekly cost to produce x specific product is given by 𝐶(𝑥) = 7,000 + 100𝑥 − 0.02𝑥 2 + 0.000005𝑥 3 , 𝑓𝑜𝑟 1 ≤ 𝑥 ≤ 10000 While the demand function for this specific product is given by, 𝑝(𝑥) = 200 − 0.005𝑥, 𝑓𝑜𝑟 1 ≤ 𝑥 ≤ 10000 Determine the marginal cost, marginal revenue and marginal profit when 2500th item is sold. Assume that the company sells exactly what they produce.
2.
The concentration of oil diminishes 1% for every 0.25 miles from specific point in the Caribbean Sea where the concentration reached 25 % due to a platform leak. a) Find a model for the concentration distribution after any number of miles. b) How much of oil can be found 5 miles away from the leak? c) How far we can still find 0.20% of oil in the sea water?
3.
For 𝑦 = 2𝑥3 + 3𝑥2 − 12𝑥 − 18 , determine the values of x where the function is increasing or decreasing (write the intervals), locate the maximum and minimum if exist.
4.
Verify that the function (𝑥) = 𝑥3 − 3𝑥 + 5, 𝑎𝑡 − 1 ≤ 𝑥 ≤ 1 satisfy the conditions of the Mean Value Theorem. Use the graph on the right to interpreter your work.
5.
For the function 𝑦 = 4𝑥3 + 6𝑥2 − 24𝑥 − 36, find the exact position of maximum, minimum and point of inflexion if exist.
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CHAPTER-3. EXPONENTIAL AND LOGARITHMIC FUNCTIONS. OPTIMIZATIONS. 3.1. Remembering Exponential and Logarithmic Functions (an Algebra Topic): 3.1.1.
Inverse Functions.
In Algebra, every operation is paired with its "opposite" (inverse): Addition with Subtraction, Multiplication with Division, Powers with Roots, and Reciprocals with Reciprocals. In functional terms, inverse function is the function obtained by expressing the dependent variable of one function as the independent variable of another; in other words; we interchange x with y to find the inverse function If f expresses a functional relationship between x and y, so that y f ( x) , and if f is invertible, then f
relationship, x f
1
1
We have Inverted our functions
expresses the reverse
( y) .
Examples: Quadratic Function 𝑦 = 𝑥 2 − 2
Linear Function y 2 x 3 Inverse substitution: x 2 y 3 Solving for y : x 2 y 3
y 12 x 3 The graphs of a function and its inverse function are symmetric with respect to the line y = x Inverse substitution: 𝑥 = 𝑦 2 − 2 Solving for y : 𝑦 = √𝑥 + 2 , valid only for 𝑥 ≥ 0
3.1.2.
Exponential Functions.
An exponential function is a mathematical expression in which the variable is located on the exponent. For example: 𝑦 = 2𝑥 This simplest equation, it is read as “y equals 2 to the x power”, the graph is displayed on the left. If 𝑦 = 2−𝑥 , we can plot the graph point by point, providing 𝑥 −values and calculating its corresponding 𝑦; or, we can apply transformation, by reflecting the graph towards axis y (on the right) 88
Juan J. Prieto-Valdés
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3.1.3. Logarithmic Functions. Giving exponential function 𝑦 = 𝑎 𝑥 ; its inverse will be: 𝑥 = 𝑎 𝑦 To solve for y, we can use graphic method (on the right). Using traditional procedures, we cannot solve the inverse exponential function. However, a new method was adopted by representing the solution as follow: 𝑦 = log 𝑎 𝑥 If x and a are real positive numbers such that 𝑎 ≠ 1, then 𝑦 = log 𝑎 𝑥 is called the logarithmic function with base a of x. 𝑦 = 𝑙𝑜𝑔𝑎 𝑥
and 𝑥 = 𝑎 𝑦 are equivalent.
Examples: y = log 2 𝑥
→
𝑥 = 2𝑦
𝑦 = log10 10 = 1
→
10 = 101
y = log 4 𝑥
→
𝑥 = 4𝑦
𝑦 = log10 100 = 2
→
100 = 102
y − 1 = log 3 𝑥
→
𝑥 = 3(𝑦−1)
𝑦 = log 5 25 = 2
→
25 = 52
→
2 = √8
log 8 2 =
1 3
log 25 5 =
3
Further representation: log10 (… ) = log(… ) and
3.1.4.
1 2
→
1
5 = √25 = (25)2
log e (… ) = ln(… ), named natural logarithm (e=2.71828…)
Properties of Logarithms:
1
𝑙𝑜𝑔𝑎 𝑎 = 1 because 𝑎 = 𝑎1
5
𝑙𝑜𝑔𝑏 (𝑥𝑦) = 𝑙𝑜𝑔𝑏 𝑥 + 𝑙𝑜𝑔𝑏 𝑦.
2
𝑙𝑜𝑔𝑎 1 = 0 because 1 = 𝑎0
6
𝑥 𝑙𝑜𝑔𝑏 ( ) = 𝑙𝑜𝑔𝑏 𝑥 − 𝑙𝑜𝑔𝑏 𝑦. 𝑦
3
𝑙𝑜𝑔𝑎 𝑎 𝑥 = 𝑥 because 𝑎 𝑥 = 𝑎 𝑥
7
𝑙𝑜𝑔𝑏 (𝑥 𝑛 ) = 𝑛 𝑙𝑜𝑔𝑏 𝑥.
4
𝑎𝑙𝑜𝑔𝑎 𝑥 = 𝑥 because 𝑙𝑜𝑔𝑎 𝑥 = 𝑙𝑜𝑔𝑎 𝑥
8
Example (Condense):
1 log 2(𝑥)4 2
𝑙𝑜𝑔𝑏 𝑥 = 𝑙𝑜𝑔𝑎 𝑥
+ 4 log 2 (√𝑥 − 2) − log 2 (𝑥 − 2) =
= log 2 (√(𝑥)4 ) + log 2 (√(𝑥 − 2)4 ) − log 2 (𝑥 − 2) =
𝑙𝑜𝑔𝑎 𝑏
Apply property # 3, then properties 5 and 6 and simplify.
(𝑥 − 2)2 log 2 [(𝑥) ] = log 2 [𝑥 2 (𝑥 − 2)] (𝑥 − 2) 2
Example (Expand): log
𝑥 3 (𝑦−2) √(𝑥+2)5
=
Apply property # 6, 5
= log[x3 (𝑦 − 2) − log √(𝑥 + 2)5 = log(𝑥3 ) + log(𝑦 − 2) − 2 log(𝑥 + 2))= 5 = 3log(𝑥) + log(𝑦 − 2) − log(𝑥 + 2) 2 90
then properties 5 and 7
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3.1.5.
Exponential and Logarithmic Equations Introductory Exponential Equation Examples:
Example 1. Same Base 𝑥
5 =5
Example 3. Creating same Base −3𝑥+4
2𝑥−3
1 4𝑥 2 = ( ) 16
𝑥 = 2𝑥 − 3 3=𝑥
4
Example 2. Creating same Base
𝑥2
3𝑥 = 8 In this example (3𝑊ℎ𝑎𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 = 8)
1 −3𝑥+4 = ( 2) 4
So, we apply logarithms: log 3𝑥 = log 8
4𝑥 2 = (4−2 )−3𝑥+4 𝑥 = 6𝑥 − 8
𝑥 log 3 = log 8 log 8 0.903089987 … 𝑥 = = log 3 0.477121255 …
𝑥 2 − 6𝑥 + 8 = 0
𝑥 = 1.892789261 …
(𝑥 − 4)(𝑥 − 2) = 0
In this example we can use log or ln.
𝑥 = 4, 𝑥 = 2
No whole exponent was found:
4
32𝑥−3 = 27
𝑥2
=
(4)6𝑥−8
2
32𝑥−3 = 33 , So: 2𝑥 − 3 = 3, 2𝑥 = 6, 𝑥=3
Example 4. Using log when there is not whole exponent using same base:
31.892789261 = 8 Introductory Logarithmic Equation Examples:
Example 1. Equal arguments log a (𝑥 − 2) = log a (3𝑥 − 8)
Example 2. Applying properties of logarithm:
Example 3. Applying properties of logarithm:
𝑥 − 2 = 3𝑥 − 8
log 2 (𝑥 − 2) + log 2 (𝑥) = 3
log 3 (𝑥 + 5) − log 3 (𝑥 −3) = 2
Apply product rule
Apply quotient rule
log 2 [𝑥(𝑥 − 2)] = 3
𝑥+5 ]=2 𝑥−3 Apply definition property:
6 = 2𝑥, so, 𝑥 = 3 Previous example using quotient rule and logarithmic definition: log a (𝑥 − 2) − log a (3𝑥 − 8) = 0 𝑥−2 log 𝑎 =0 3𝑥 − 8 𝑥−2 = 𝑎0 = 1 3𝑥 − 8 𝑥 − 2 = 3𝑥 − 8. . . 𝑠𝑜 𝑥 = 3
3.1.6.
Apply definition property: [𝑥(𝑥 − 2)] = 23 = 8
log 3 [
𝑥+5 𝑥−3
𝑥 2 − 2𝑥 − 8 = 0
= 32 =9
(𝑥 − 4)(𝑥 + 2) = 0
(𝑥 − 5) = 9(𝑥 − 3)
𝑥 = 4 𝑎𝑛𝑑 𝑥 = −2
𝑥 + 5 = 9𝑥 − 27
Reject 𝑥 = −2
32 = 8𝑥, 𝑠𝑜 𝑥 = 4
Compound Interest. Periodic and Continuous Compounding Interest:
The Actual Amount 𝐴(𝑡) for compound interest is a power law function in terms of time 𝑡 as follow: 𝑟 𝑛𝑡
𝐴(𝑡) = 𝑃0 (1 + 𝑛)
.
Where:
𝑡 = Total time in years, 𝑛 = Number of compounding periods per year (note that the total number of compounding periods is 𝑛 × 𝑡), and 𝑟 = Nominal annual interest rate expressed as a decimal. e.g.: 6% = 0.06 As 𝑛, the number of compounding periods per year, increases without limit, we have the case known as continuous compounding, in which case the effective annual rate approaches an upper limit (follow lecture demonstration). The amount after 𝑡 periods of continuous compounding can be expressed in terms of the initial amount 𝑃0 as: 𝐴(𝑡) = 𝑃0 𝑒 𝑟𝑡 92
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Homework Topic 3.1 136) Given 1 𝑓(𝑥) = 4𝑥 − 12, Write the inverse function: 2 𝑥 = 4𝑦 − 12,
9
𝑓 −1 (𝑥) =
. 137) Given 𝑦 = 𝑥 2 − 2, write and graph the inverse function (use x>0):
𝑥+12 4
𝑦 = 𝑥2 − 2
𝑦 = √𝑥 + 2
138) Given 𝑦 = 𝑥 2 + 2𝑥 − 15, graph the inverse function (use 𝑥 > 0).
139)
5
Given 𝑦 = 𝑥+2 , write the inverse function: y=
140)
1
Write the inverse function of 𝑓(𝑥) = 𝑥+1 + 4
𝑓 1 (𝑥) = 94
5 −2 x
1 −1 𝑥−4
Juan J. Prieto-Valdés
Exponential Functions. 141)
Graph: 𝑦 = 2𝑥−3
First, use table of value, secondly, transformations: 𝑥
2𝑥−3
𝑦 (𝑥, 𝑦)
-3 -2 -1 0 1 2 3 142)
Graph: 𝑦 = 3𝑥
𝑥
3𝑥
𝑦 (𝑥, 𝑦)
-3 -2 -1 0 1 2 3 95
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143)
1 𝑥
Graph: 𝑦 = (2)
𝑥
1 𝑥 ( ) 2
𝑦 (𝑥, 𝑦)
-2 -1 0 1 2 144)
145)
1 𝑥 3
Graph: 𝑦 = ( )
1 𝑥 𝑥
Evaluate: (1 + ) , For: 𝑥 = 1, 10, … , 1000000000. After calculating write the 𝑒 constant definition.
1 1
𝑥 = 1, (1 + 1) = 1 10
𝑥 = 10, (1 + 10) = 100
1
𝑥 = 100, (1 + 100) 1
=
1000
𝑥 = 1000, (1 + 1000)
1
= 1000000
𝑥 = 10000000, (1 + 10000000)
96
=
Juan J. Prieto-Valdés
Logarithmic Functions and Properties of Logarithms. 146) Write in exponential form: log 2 16 = 4 24 = 16 147) Explain the representation log 𝑥 and ln 𝑥. Give 2 examples:
148) log10 1000 = 3, further log 1000 = 3 103 = 1000 149) log 5 1 = 0 50 = 1 150)
log 5
1 = −2 25
5−2 =
1 1 = 2 5 25
151) Write in logarithmic form: 53 = 125 log 5 125 = 3 152) 82 = 64 log 8 64 = 2 153) √9 = 3 log 9 3 =
1 2
154) 34 = 81 log 3 81 = 4 155) Write in exponential form and solve: log ( 1 ) = 𝑦 √5 25
𝑦
𝑦
1
√5 = (25) = 5−2, so 52 = 5−2 , 𝑦 2
= −2, 𝑎𝑛𝑑 𝑦 = −4
156) Graph: 𝑦 = log 2 𝑥
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157) Graph: 𝑦 = log 2 (𝑥 − 1)
In the following examples, expand the logarithmic expression: 158)
𝑥3 log 2 ( 2 ) 𝑦𝑧 3log 2 𝑥 − log 2 𝑦 − 2 log 2 𝑧
159)
3 (𝑥 + 9)9 log 5 ( √ ) 5
3log 5 (𝑥 + 9) − 160)
log (
1 3
√𝑥 3 ) 2𝑥 + 4 3 log 𝑥 − log(2𝑥 + 4) 2
161) ln[(𝑒 2𝑥+3 )(𝑥 + 9)]
(2𝑥 + 3) + ln(𝑥 + 9) Write the logarithmic expression as a single logarithm (condense): 162) log 5 70 − log 5 25
log 5
70 14 = log 5 25 5
163) 4 log 5 (27) + log 5(12) use formula change-of base. log(12) 4 log(27) log( 12 × 274 ) + = log(5) log(5) log 5 98
Juan J. Prieto-Valdés
164) 1 log(𝑥 + 3) + 4 log(√𝑥 + 3) − log(16) 2 5
(𝑥 + 3)2 log 16 165) log(𝑥 − 5)3 + log∛(𝑥 + 2) − log(5 − 𝑥)
1
log (−(𝑥 − 5)2 (𝑥 + 2)3 ) 166) log(𝑥 − 5)3 − 2 log(𝑥 + 2) − log 3√𝑥
log ( 167)
3log(𝑥) − 2 log(𝑥 + 2) − log
(𝑥 − 5)3 (𝑥 + 2)2 3√𝑥
)
(𝑥 + 2) 2
log
2𝑥 3 (𝑥 + 2)3
Exponential and Logarithmic Equations. Solve the following Exponential Equations 168)
1 𝑥 643−𝑥 = ( ) 8 𝑥≈6
169) 5(𝑥+2) = 125
𝑥=1 170) 33𝑥−2 = 81
𝑥=2 171) 5𝑥 = 56
𝑥 = log 5 56
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Business Calculus Notebook
172)
3𝑥 =
1 81 𝑥 = −4
173) 7𝑥 = 25 𝑥 = 2 log 7 5 174) 2(𝑥−3) = 3(𝑥+1)
𝑥 ≈ −7.8380451654058 175) 𝑒 2𝑥 + 4𝑒 𝑥 − 12 = 0
𝑥 ≈ 0.6931471805599
Solve the following Logarithmic Equations: 176) log 2 (3𝑥 + 2) = log 2(𝑥 − 1) 𝑥 = −
3 2
(assuming a complex-valued logarithm) 177) log 3 (𝑥 − 1) = 2
x = 10 178) log(√𝑥 + 1) = 2
𝑥 = 9999 179)
1 ln(6) − 2 = ln ( ) 𝑥 𝑥 =
𝑒2 6
180) log 5 (10 − x) + log 5 𝑥 = 2
𝑥 = 5 100
Juan J. Prieto-Valdés
Compound Interest. 181) Suppose you want to invest $10,000 in a governmental bond plan. The government bank pays 4.0 % and the earnings are reinvested every 6 months. Determine the amount you will have in 10 years.
𝐴 = 10000 (1 +
0.04 2×10 ) = 14,859.5 2
182) Suppose you want to open an account with $5,000 in a bank, which pays 3.5% yearly. Determine how long it will take to double your money.
10000 = 5000 (1 +
0.035 𝑡 )= 1
20.1488 years
183) Find the principal needed now to get $1000.00 in 2 years at 5% compounded monthly.
0.05 12×2 1000 = 𝑃0 (1 + ) 12 𝑃0 = 905.025
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184) Susan wants to invest $30,000.00 in saving account for her retirement; she can invest at 5,2% simple interest (compounded yearly), or 4.5% compounded continuously for 20 years. Which option results more convenient for her?
𝐴5.2
0.052 1 ×20 = 30000 (1 + ) = 82686.8 1 𝐴4.5 = 30000(𝑒)0.045 ×20 = 73788.1
185) If $5,000.00 is deposited in an account paying 3.5% interest compound annually, and at the same time 4,000.00 is deposited in an account paying 5% compounded also annually, When the two accounts will have the same balance.
𝐴 = 5000(1 + 0.035)𝑡 𝐴 = 4000(1 + 0.05)𝑡 {𝐴 = 8524.48,
𝑎𝑡
𝑡 = 15.5082}
Additional Homework Topic 3.1.5-plus: The following three (176-178) exercises are included to remain you an important application of exponential functions studied before in College Algebra: Exponential Growth and Decay: The exponential growth at time can be written as: 𝑁 = 𝑁0 𝑒 𝑟𝑡 Where: 𝑁0 - is the starting population, 𝑁 - is the population after a certain time, 𝑡 , has elapsed, and e - is the base of natural logarithms (Nipper constant = 2.71828...). 𝒓 is the rate of increase (rate of growth), always positive. On the right, plotted dots are values known from census and historical estimates of world population, while the continuous line is the graphic solution of the previous equation. The exponential Decay or simple Decay (graphically represented also on the right) is the same exponential process but the constant < 0 , so the amount decreases in time.
102
Juan J. Prieto-Valdés
Exponential Growth and Decay exercises: 186)
The population of Canada in 2010 was approximately 34 million with annual growth rate of 0.804%. Determine the approximately population in Canada in 2014, and in 2020. Verify for 2014 by checking the current population in Canada using available info on internet.
𝑁2014 = 34𝑒 0.00804×4 = 35.1112 𝑁 = 34 𝑒 0.00804×10 = 𝑁 = 36.8465 187)
The half-life of Plutonium-239 is 24,000 years. If 100 grams are present now, how long will it take until only 1 gram of the original sample remains?
1 = 𝑒 24000𝑘 2 ln(2)
𝑘 = − 24000=-0.00002888... 1 = 100𝑒 −0.00002888𝑡 𝑡= 188)
25000000 ln(10) 361
≅ 𝑡 = 159459.
Suppose that at the start of an experiment there are 10,000 bacteria. A growth inhibitor and a lethal pathogen are introduced into the colony. After two hours 2,000 bacteria are dead. If the death rates are exponential, (a) how long will it take for the population to drop below 1,000? (b) How long will it take for two-thirds of the bacteria to die?
8000 = 10000𝑒 2𝑘 4 ln ( ) 5 ≅ −0.11157 … 𝑘= 2 1000 = 10000𝑒 𝑡(−0.11157)
𝑡=
100000 ln(10) ≅ 20.6380.. 11157
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Business Calculus Notebook
3.2. Differentiation of Exponential Functions To differentiate an exponential function like this: 𝑓(𝑥) = 𝑒 𝑥 , we go back to the derivative definition: 𝑒 𝑥+ℎ −𝑒 𝑥
𝑓(𝑥+ℎ)−𝑓(𝑥) 𝑓 = lim = 𝑙𝑖𝑚 ℎ = ℎ ℎ→0 ℎ→0 𝑥 ℎ ℎ 𝑥 ℎ 𝑥 𝑒 (𝑒 −1) (𝑒 −1) 𝑒 𝑒 −𝑎 𝑙𝑖𝑚 ℎ = 𝑙𝑖𝑚 = 𝑒 𝑥 𝑙𝑖𝑚 ℎ = ℎ ℎ→0 ℎ→0 ℎ→0 𝑥 (1) ′ (𝑥)
It is easy to demonstrate that 𝑙𝑖𝑚
ℎ→0
(𝑒 ℎ −1) ℎ
representation of: 𝑦 =
= 1 using graphic
(𝑒 𝑥 −1) 𝑥
𝑒
Because 𝑙𝑖𝑚
ℎ→0
(𝑒 ℎ −1) ℎ
= 1,
𝑤𝑒 𝑐𝑎𝑛 𝑤𝑟𝑖𝑡𝑒 𝑑 𝑥 (𝑒 ) = 𝑒 𝑥 𝑑𝑥
In general form, when 𝑓(𝑥) = 𝑎 𝑥 , we can write: 𝑎 𝑥+ℎ − 𝑎 𝑥 𝑎 𝑥 𝑎ℎ − 𝑎 𝑥 𝑎 𝑥 (𝑎ℎ − 1) = 𝑙𝑖𝑚 = 𝑙𝑖𝑚 = 𝑎 𝑥 × 𝑐𝑜𝑛𝑠𝑡 ℎ→0 ℎ→0 ℎ→0 ℎ ℎ ℎ (𝑎 ℎ −1) 𝑑 𝑙𝑖𝑚 ℎ = 𝑐𝑜𝑛𝑠𝑡. So: 𝑑𝑥 (𝑎 𝑥 ) = 𝑎 𝑥 × 𝑐𝑜𝑛𝑠𝑡.
𝑓 ′ (𝑥) = 𝑙𝑖𝑚
ℎ→0
On the right, quick numeric examples to demonstrate that the 𝑐𝑜𝑛𝑠𝑡 = ln 𝑎 𝑑 (𝑎 𝑥 ) = 𝑎 𝑥 ln 𝑎 = ln 𝑎 (𝑎 𝑥 ) So:
𝑎
𝑎𝑥 − 1 = 𝑥→0 𝑥
2
= 0.693147 = ln 2
3
= 1.098612 = ln 3
4
= 1.386294 = ln 4
lim
𝑑𝑥
More evidences using Wolfram Alpha: 𝑎=2
𝑎=3
𝑎=4
𝑎 = 10
Examples: EXAMPLE 1
EXAMPLE 2
Finding The relative rate of change of an investment: If the value of 𝑃0 -dollars investment doubles every year, its value after 𝑡 years is given by 𝑣(𝑡) = 𝑃0 (2𝑡 ) Find the instantaneous percentage rate of change of the increment of 𝑃0 and its relatively rate of change: 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒: 𝑣
′ (𝑡)
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒𝑙𝑦 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒:
𝑡
= 𝑃𝑜 (2 )(ln 2)
𝑣 ′ (𝑡) ln 2) = (𝑃𝑜 (2𝑡 )( ) 𝑣(𝑡) 𝑃𝑜 (2𝑡 )
𝑣 ′ (𝑡) = ln 2 ≈ 0.693 = 69.3% 𝑣(𝑡)
104
𝑓(𝑥) =
𝑥2 𝑒𝑥 + 1
Use the quotient rule
𝑓 ′ (𝑥) =
(𝑒 𝑥 + 1)
=
𝑑 2 𝑑 𝑥 (𝑥 ) − 𝑥 2 (𝑒 + 1) 𝑑𝑥 𝑑𝑥 = 𝑥 2 (𝑒 + 1)
(𝑒 𝑥 + 1)2𝑥 − 𝑥 2 (𝑒 𝑥 ) = (𝑒 𝑥 + 1)2 =
2𝑥𝑒 𝑥 + 2𝑥 − 𝑥 2 𝑒 𝑥 (𝑒 𝑥 + 1)2
Juan J. Prieto-Valdés EXAMPLE 3
EXAMPLE 4
𝑓(𝑥) = 𝑥 3 𝑒 𝑥
𝑓(𝑥) = ( ) or 𝑓(𝑥) = 2−𝑥
𝑓(𝑥) = 35𝑥−3
Use the product rule 𝑑 𝑥 𝑑 3 𝑥 ′ (𝑥) (𝑒 ) + (𝑥 )𝑒 𝑓 = 𝑥3 𝑑𝑥 𝑑𝑥
𝑑 −𝑥 𝑑 (2 ) = (2−𝑥 )(𝑙𝑛 2) ( (−𝑥)) 𝑑𝑥 𝑑𝑥
Use chain rule 𝑑(5𝑥 − 3) 𝑓 ′ (𝑥) = [(𝑙𝑛3)35𝑥−3 ] = 𝑑𝑥
3 𝑥
2 𝑥
𝑥 2 (𝑥
𝑥 𝑒 + 3𝑥 𝑒 = 𝑒 𝑥
1 𝑥 2
+ 3)
1 𝑥 1 = (2−𝑥 )(−𝑙𝑛 2) = ( ) (𝑙𝑛 ( )) 2 2
EXAMPLE 5
(5𝑙𝑛3)(35𝑥−3 )
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Business Calculus Notebook
3.3. Differentiation of Logarithmic Functions To differentiate a logarithmic function like this 𝑦 = ln(𝑥), we will consider the equivalent function in exponential format: 𝑒 𝑦 = 𝑥. Now, applying the chain rule and the property of exponents, we take the derivative of both sides: 𝑑 𝑦 𝑒 𝑑𝑥
𝑑
=>
= 𝑑𝑥 𝑥
𝑑𝑦
𝑒 𝑦 𝑑𝑥 = 1
𝑑𝑦 𝑑𝑥
𝑠𝑜:
1
= 𝑒𝑦 𝑑𝑦 𝑑𝑥
Substituting back our original equation of 𝑥 = 𝑒 𝑦 , we find that 𝑑 1 ln(𝑥) = 𝑑𝑥 𝑥
1
= 𝑥 , we get:
ln(𝑥)
For the general case: log 𝑏 (𝑥) we apply the formula base change: log 𝑏 (𝑥) = In this case: Examples: EXAMPLE 1
ln(𝑏)
𝑑 𝑑 ln(𝑥) 1 𝑑 1 log 𝑏 (𝑥) = ( ln(𝑥) = )= 𝑑𝑥 𝑑𝑥 ln(𝑏) ln(𝑏) 𝑑𝑥 𝑥 ln(𝑏)
EXAMPLE 2
EXAMPLE 3
𝑓(𝑥) = 𝑥𝑙𝑛𝑥
𝑓(𝑥) = ln(𝑥 5 + 1)
Use product rule
Use chain rule
𝑙𝑛𝑥 3 𝑓(𝑥) = 2 𝑥 use quotient rule, first apply properties:
2
1 𝑓 ′ (𝑥) = 𝑥 ( ) + 𝑙𝑛𝑥 𝑥 = 1 + ln 𝑥
𝑓 ′ (𝑥) =
(𝑓(𝑥) =
1 𝑑 5 (𝑥 + 1) 5 𝑥 + 1 𝑑𝑥
𝑓 ′ (𝑥) =
𝑓 ′ (𝑥) =
5𝑥 4 𝑥5 + 1
=
2 𝑙𝑛𝑥 3 𝑥2
)
2 𝑑 𝑑 (𝑙𝑛𝑥) − 𝑙𝑛𝑥 𝑥 2 ] [𝑥 2 4 3𝑥 𝑑𝑥 𝑑𝑥
2 1 (𝑥 2 ( ) − 3𝑥 4 𝑥
2𝑥𝑙𝑛𝑥) =
2 (1 − 3𝑥 3
2𝑙𝑛𝑥)
EXAMPLE 4. Finding maximum concentration of chemical: The concentration of some soluble chemical in water from the leak (maritime accident) is 10
𝑦 = (𝑒 −2𝑥+5)+1, where 𝑥 𝑖𝑠 𝑡ℎ𝑒 𝑟𝑎𝑑𝑖𝑢𝑠 from the place of the
given by accident.
10 (𝑒 −2𝑥+5 )
Show that the concentration never will exceed 10 %.
𝐶’(𝑥) = −10 [(𝑒−2𝑥+5 ) + 1]−2 (
𝑑 𝑑𝑥
𝑒−2𝑥+5 + 1) = −10[(𝑒−2𝑥+5 ) + 1]−2 (𝑒−2𝑥+5 )(−2) =
𝐶 ′ (𝑥) = 20 Small changes after 𝑥 = 5: 𝐶 ′ (5) = 1.006,
(𝑒−2𝑥+5 ) (𝑒−2𝑥+5 + 1)2
,
𝐶(10) = 1,0000, and
EXAMPLE 5: Logarithmic Differentiation. 5
√𝑥−2
10
lim
𝑥→∞ (𝑒 −2𝑥+5 )+1
= 10
Using the chain rule in both sides 𝑓 ′ (𝑥) 1 2 = − 𝑓(𝑥) 5(𝑥 − 2) 𝑥 − 3
5
√𝑥−2
𝑓(𝑥) = (𝑥−3)2 ln 𝑓(𝑥) = ln (𝑥−3)2
Resulting:
𝑓 ′ (𝑥) = 𝑓(𝑥) (
5
ln 𝑓(𝑥) = ln √𝑥 − 2 − ln(𝑥 − 3)2 = 1 = ln(𝑥 − 2) − 2 ln(𝑥 − 3) = 5
5
𝑓 106
𝑦=
′ (𝑥)
=
1 2 − ) 5(𝑥 − 2) 𝑥 − 3
1 2 √𝑥 − 2 ( − ) (𝑥 − 3)2 5(𝑥 − 2) 𝑥 − 3
+1
Juan J. Prieto-Valdés
Homework Topic 3.2 and 3.3 Differentiate given expression 189)
𝑓(𝑥) = 𝑒 4𝑥 4 𝑒4 𝑥
190)
𝑓(𝑥) = 2𝑒 5𝑥+2 10 𝑒 5 𝑥+2
191)
𝑓(𝑥) = 3𝑒 5𝑥
2 +2𝑥+3
3(10 𝑥 + 2) 𝑒 5 𝑥 192)
𝑓(𝑥) = 𝑒 √2𝑥+3
2 +2 𝑥+3
𝑒 √2 𝑥+3 √2 𝑥 + 3
193)
2
2
𝑓(𝑥) = 𝑒 𝑥
194)
2 𝑒𝑥 − 2 𝑥
𝑓(𝑥) = 5𝑥 ln(5) · 5𝑥
195)
𝑓(𝑥) = 3𝑥+2 ln(3) · 3𝑥+2
196)
𝑓(𝑥) = 3𝑥
2 +2
2 ln(3) 𝑥 · 3𝑥 197)
𝑓(𝑥) = 𝑒 𝑥
2 +2
2 +2
(ln(𝑥)) 2
2 𝑥 ln(𝑥) 𝑒
𝑥 2 +2
𝑒 𝑥 +2 + 𝑥
In exercises from 198 to 201 find an equation for the tangent line to 𝒚 = 𝒇(𝒙)at the specified point: 198) 𝑦 = 𝑥𝑒 𝑥 at 𝑥 = 0
Use 𝑒 𝑥 𝑥 format 𝑦′ = 𝑥 𝑒 𝑥 + 𝑒 𝑥 107
Business Calculus Notebook
199) 𝑦 = 𝑥𝑒 −𝑥 at 𝑥 = 0
𝑦′ =
1−𝑥 𝑒𝑥
200) 𝑦 = 𝑒 −𝑥 (𝑥 + 2) at 𝑥 = 1
𝑦′ =
−(𝑥 + 2) + 1 𝑒𝑥
201) 𝑦 = (𝑥 + 5)4𝑥 at 𝑥 = 0.5
2 ln(2) 𝑥 · 4𝑥 + 10 ln(2) · 4𝑥 + 4𝑥 In exercises 202 – 209 differentiate the given function 202)
𝑓(𝑥) = ln 𝑥 3 3 𝑥
203)
𝑓(𝑥) = ln 𝑥 2 2 𝑥
204)
𝑓(𝑥) = ln 𝑥 −2 −
108
2 𝑥
Juan J. Prieto-Valdés
205)
𝑓(𝑥) = ln(𝑥 2 + 1) 2𝑥 +1
𝑥2 206) 𝑓(𝑥) = ln √(𝑥 2 + 1)3
3𝑥 +1
𝑥2 207)
𝑓(𝑥) = log 2 𝑥 2
2 ln(2) 𝑥 208) y=
2 log( 𝑥) + log(𝑥 2 + 1) log 4
1 + 2 𝑥2 (𝑥 + 𝑥 3 ) ln(2) 209)
𝑦 = √(𝑥 2 + 1) log 2 𝑥
1 + 𝑥 2 + 𝑥 2 ln[𝑥] 𝑥 √[1 + 𝑥 2 ]ln[2]
In exercises from 210 to 217, find an equation for the tangent line to 𝒚 = 𝒇(𝒙)at the specified point: 210) 𝑓(𝑥) = ln 𝑥 at 𝑥 = 3
𝑦′ = 𝑥
1 𝑥
The equation of the line will be 𝑦 = −1 + 3 + 𝐿𝑜𝑔[3]
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Business Calculus Notebook
211) 𝑦 = ln 𝑥 2 at 𝑥 = 1 𝑥
− ln(𝑥 2 ) + 2 𝑥2 The equation of the line: 𝑦 = −2 + 2 𝑥 𝑦′ =
212) 𝑦 = ln 𝑥 2 + 𝑥 3 at x=0 and at x=1. Explain your results 2
1 𝑥 The equation of the line: 𝑦 = −3 + 4 𝑥 𝑦′ = 3 𝑥2 +
In exercises from 213 to 216, use logarithmic differentiation to find the derivative: 213)
1
𝑦 = (𝑥 + 3)2 (𝑥 − 5)2 𝑦 ′ = √𝑥 − 5 (2 𝑥 + 6) +
(𝑥 + 3)2 2 √𝑥 − 5
Or 𝑦 ′ = 214)
2 √−5 + 𝑥
1
𝑦 = 𝑒 2𝑥 (𝑥 − 1)2 𝑦′ =
215)
(3 + 𝑥)(−17 + 5 𝑥)
𝑒 2 𝑥 (−3 + 4 𝑥) 2 √−1 + 𝑥
1
(𝑥 − 1)3 𝑒 5𝑥 𝑦′ =
𝑒5 𝑥 3 (𝑥
3
2 − 1)3
+ 5 √𝑥 − 1 𝑒 5 𝑥
or 𝑦 ′ =
𝑒 5 𝑥 (15 𝑥−14) 2
3 (𝑥−1)3
216) 𝑦 = log 2 (√𝑥 + 2) 1 2 ln(2) (𝑥 + 2)
110
Juan J. Prieto-Valdés
3.4. L’ Hospital Rule / Revisiting Limits In its simplest form, this rule states that for functions f and g which are differentiable on I interval: If:
𝑓′ (𝑥) 𝑔 𝑥→𝑐 ′ (𝑥)
lim 𝑓(𝑥) = lim 𝑔(𝑥) = 0 𝑜𝑟 ± ∞ and lim
𝑥→𝑐
𝑥→𝑐
exist, and 𝑔′ (𝑥) ≠ 0
for all 𝑥 in 𝐼 with 𝑥 ≠ 𝑐, where 𝑥 ∈ 𝐼, then 𝑓 (𝑥) 𝑓 ′(𝑥) lim = lim ′ 𝑥→𝑐 𝑔 (𝑥) 𝑥→𝑐 𝑔 (𝑥) The differentiation of the numerator and denominator often simplifies the quotient and/or converts it to a determinate form, allowing the limit to be evaluated more easily. To demonstrate this rule, we will use an extension (Cauchy) of the Mean Value Theorem: Let 𝑓 and 𝑔 be differentiable on [𝑎, 𝑏], Suppose that 𝑔’(𝑥) ≠ 0 in [𝑎, 𝑏]. Ten there is at least one point 𝑐 in [𝑎, 𝑏] such that: 𝑓 ′ (𝑥) 𝑓(𝑎) − 𝑓(𝑏) = 𝑔′ (𝑥) 𝑔(𝑏) − 𝑔(𝑎) 0 0
To proof L’Hospitale Rule for the
as 𝑥 → 𝑐 +, where 𝑐 is finite. (the case 𝑐 − can be proven in a similar way).
These two cases are used to prove the Rule. 𝐼𝑓 𝑓(𝑥) → 0 𝑎𝑛𝑑 𝑔(𝑥) → 0
𝑎𝑠 𝑥 → 𝑐 + 𝑎𝑛𝑑
𝑓′ (𝑥) 𝑔′ (𝑥)
→ 𝐿 (𝑓𝑖𝑛𝑖𝑡𝑒 𝐿𝑖𝑚𝑖𝑡)
𝑤ℎ𝑒𝑟𝑒 𝑔′ (𝑥) ≠ 0 𝑎𝑛𝑑 𝑓 ′ 𝑎𝑛𝑑 𝑔′ 𝑒𝑥𝑖𝑠𝑡 𝑎𝑛𝑑 𝑎𝑟𝑒 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑜𝑛 (𝑐, 𝑐 + ℎ] 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑒, 𝑤ℎ𝑒𝑟𝑒 𝑎𝑙𝑠𝑜 𝑓(𝑐) = 0 𝑎𝑛𝑑 𝑔(𝑐) = 0
Indeterminate Forms and an Example Function
Follow previous condition, we can write: 𝑓 ′ (𝑐) 𝑓(𝑐 + ℎ) − 𝑓(𝑐) 𝑓(𝑐 + ℎ) = = 𝑔′ (𝑐) 𝑔(𝑐 + ℎ) − 𝑔(𝑐) 𝑔(𝑐 + ℎ) Observe that: lim+
𝑥→0
𝑓 ′ (𝑐) 𝑓 ′ (𝑥) = lim 𝑥→𝑐 + 𝑔′ (𝑥) 𝑔′ (𝑐) 𝑥→𝑐
While lim+
ℎ→0
0 0
3𝑥 2 − 3 𝑥→−1 𝑥 − 1
3𝑥 2 − 1 𝑥→∞ 𝑥 2 − 1
0
𝑓(𝑐 + ℎ) − 𝑓(𝑐) 𝑓(𝑥) = lim+ 𝑔(𝑐 + ℎ) − 𝑔(𝑐) 𝑥→𝑐 𝑔(𝑥)
1
Which represents what we wanted to proof: 𝑓 (𝑥) 𝑓 ′(𝑥) = lim ′ 𝑥→𝑐 𝑔 (𝑥) 𝑥→𝑐 𝑔 (𝑥) lim
00
How L’Hôpital’s Rule works?
lim
lim
lim 𝑒 −𝑥 √𝑥
𝑥→∞
1 𝑥 lim (1 + ) 𝑥→∞ 𝑥 lim (sin 𝑥) 𝑥
𝑥→0+
1 1 lim+ ( + ) 𝑥→1 ln 𝑥 𝑥−1
If the indeterminate form is either 0/0 or ∞/∞ you can immediately apply the rule by taking the derivative of both the numerator and denominator 1 0 lim ( 𝑥)𝑥 separately and reevaluate the limit of the quotient. If one of the two 𝑥→∞ indeterminate forms appear again, apply the rule again, else the limit can be found. The Table son the right shows the most common indeterminate forms that you can solve using L’Hôpital’s Rule. Also, there are several determinate forms that do not require (and cannot use) L’Hôpital’s rule. They are:
.
0 0
0 111
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112
Juan J. Prieto-Valdés
Homework Topic 3.4. In exercises from 217 to 220, apply L’ Hospital’s Rule. Analyze the answer when provided. 217)
5𝑥 4 − 4𝑥 2 − 1 20𝑥 3 − 8𝑥 12 = lim = 𝑥→1 𝑥→1 𝑥−1 1 1
218)
𝑒𝑥 𝑒𝑥 𝑒𝑥 = lim = lim =∞ 𝑥→∞ 𝑥 𝑥→∞ 2𝑥 𝑥→∞ 2
219)
1 ln 𝑥 lim 𝑥 ln 𝑥 = lim+ = lim+ 𝑥 = lim+ ln|−𝑥| = 0 1 1 𝑥→0+ 𝑥→0 𝑥→0 𝑥→0 − 2 𝑥 𝑥
220)
lim
lim
1
1
lim 𝑥 𝑥 = (𝑢𝑠𝑒 𝑙𝑜𝑔𝑎𝑟𝑖𝑡𝑚𝑖𝑐 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 ln 𝑦 = ln 𝑥 𝑥 =
𝑥→∞
1 ln 𝑥 lim ln 𝑦 = lim = lim 𝑥 = 0, 𝑠𝑜 lim 𝑦 = 1, 𝑥→∞ 𝑥→∞ 𝑥 𝑥→∞ 1 𝑥→∞ lim ln(𝑦)
𝑎𝑛𝑑 lim 𝑦 = 𝑒 ln(𝑦) = 𝑒 𝑥→∞ 𝑥→∞
221)
1 ln 𝑥) 𝑥
𝑎𝑠 𝑦 = 𝑒 ln(𝑦) ,
= 𝑒0 = 1
𝑥2 − 9 𝑥→3 3 − 𝑥 lim
−6 222)
ln( 𝑥) 𝑥→1 𝑥 − 1 lim
1 113
Business Calculus Notebook
3.5. Optimization and Business Applications Examples. 3.5.1. Application to Volume optimization An open rectangular box with square base is to be made using 98 sqf of material. What dimensions will result in a box with the largest possible volume? Solution: Let variable 𝑥 be the length of one edge of the square base and variable 𝑦 the height of the box. The total surface area of the box is given to be 𝑉𝑜𝑙𝑢𝑚𝑒 = (𝑎𝑟𝑒𝑎 𝑜𝑓 𝑏𝑎𝑠𝑒) + 4 (𝑎𝑟𝑒𝑎 𝑜𝑓 𝑜𝑛𝑒 𝑠𝑖𝑑𝑒)
4𝑥𝑦 = 96 − 𝑥2 𝑓𝑟𝑜𝑚 𝑤ℎ𝑒𝑟𝑒 𝑦 =
96 = 𝑥 2 + 4(𝑥𝑦), 𝑠𝑜
96 − 𝑥2 24 𝑥 = − 4𝑥 𝑥 4
We wish to MAXIMIZE the total VOLUME of the box 𝑉 = (𝑙𝑒𝑛𝑔𝑡ℎ) (𝑤𝑖𝑑𝑡ℎ) (ℎ𝑒𝑖𝑔ℎ𝑡) = (𝑥) (𝑥) (𝑦) = 𝑥 2 𝑦. Before we differentiate the right-hand side, we will write it as a function of x only. Substitute for y getting 24 𝑥 𝑥3 − ) = 24𝑥 − 𝑥 4 4 Now differentiate this equation and making it equal to zero: 1 𝑉 ′ = 24 − ( ) 3𝑥 2 = 0, 𝑟𝑒𝑠𝑢𝑙𝑡𝑖𝑛𝑔 𝑥 = √32 = 5.6568 ft 4 (Reject negative solution) 𝑉 = 𝑥2 𝑦 = 𝑥2 (
𝑦=
96 − 𝑥 2 24 √32 = − = 2√2 = 2.8294 4𝑥 4 √32 24
Finally, we can verify the total area: (
−
√32
√32 ) √32 4
+ √32√32 = 96
3.5.2. Application to Analytic Geometry: Find the point (x, y) on the graph of 𝒚 = √𝒙 nearest the point (4, 0). Let (x, y) represent a randomly point on the graph. We wish to MINIMIZE the Distance between points (x, y) and (4, 0), 𝐷 = √(𝑥 − 4)2 + 𝑦 2 Before we differentiate the right-hand side, we will write it as a function of x only. Substitute for y getting: 2
𝐷 = √(𝑥 − 4)2 + (√𝑥)
Now differentiate this equation using the chain rule, and make the derivative equal to zero: 2𝑥 − 7 𝐷′ = =0 2√(𝑥 − 4)2 + 𝑦 2 2𝑥 − 7 = 0
or 𝑥 =
7 2
7 2
and 𝑦 = √ ≅ 1.87, so 𝐷 = √
15 2
≅ 1.94
Which is the shortest possible distance from (4, 0) to the graph of 𝑦 = √𝑥 . 114
Juan J. Prieto-Valdés
115
Business Calculus Notebook
3.5.3. Application to evacuation situation. Find the faster path to evacuate workers from the electronuclear plant to the safe refuge in case of emergency. This is a classical exercise, you can find many different variants in all calculus books, but we will use a Microsoft Word solution. Consider that maximum driving speed through the land-rout is 65 miles per hours. On the bridge cannot be more than 18 miles per hour, and the ferry can average a speed of 45 miles per hour This is a special fast ferry designed for evacuation purposes. Under these moving conditions, find the correct position to dock the ferry to avoid passing over the bridge. Consider that Consider only one bus and one ferry are required to evacuate all personal, and that the people changing-time from the bus to the ferry is despicable.
C
B A
To analyze the problem, we can use the diagram on the right where are represented the distances using the rectangle of 50 x 18 miles. The dock point will be located in B-point. Therefore, the distances will be: (𝐴 → 𝐵) = 50 − 𝑥 and (𝐵 → 𝐶) = √𝑥 2 + 18. The velocity of the buss (ferry) can be expressed as the ̅̅̅̅)/𝑡1 and 𝑣𝑓𝑒𝑟𝑟𝑦 = (𝐵𝐶 ̅̅̅̅ )/𝑡2 . In such case, the total time (T) will be: distance divided by the time. 𝑣𝑏𝑢𝑠 = (𝐴𝐵 𝑇 = 𝑡1 + 𝑡2 =
50 − 𝑥 √𝑥 2 + 18 50 − 𝑥 √𝑥 2 + 18 + = + 𝑣𝑏𝑢𝑠 𝑣𝑓𝑒𝑟𝑟𝑦 65 45
We are looking for minimal time, that is to say:
𝑑𝑇 𝑑𝑥
= 0.
Solve the problem using Microsoft Mathematics in your Word document and CLICK at Mathematics Tab
Firs use command “Compute”, then “derivative”, and 50 − 𝑥 √𝑥 2 + 18 + 65 45 make it equal to zero, to evaluate 𝑥 at the optimal (minimum) position, then use “solve” command and the answer: 𝑇=
116
Juan J. Prieto-Valdés
𝑑 √(𝑥 2 + 18) 50 − 𝑥 𝑥 1 27 ( + )= − = 0 𝑓𝑟𝑜𝑚 𝑤ℎ𝑒𝑟 𝑥 = = 4.07 2 𝑑𝑥 45 65 2 √11 45 √(𝑥 + 18) 65 Now use “Graph” command to visualize your function and the approximate position of the minimum.
This is prettiest and smart solution using technology. The dock point must be located a 50 − 𝑥 = 46 𝑚𝑖𝑙𝑒𝑠 from the electro nuclear plant to guarantee faster evacuation.
3.5.4. Application in Electrical Engineering: Ohm’s Law states the inverse proportionality of the Current Intensity to Voltage 𝑰 =
𝑽 𝑹
Find the rate of change of 𝑰 with respect to 𝑹 at 𝑽 = 𝟏𝟐 𝒗𝒐𝒍𝒕𝒔: Solution: 𝑑𝐼 12 =− 2 𝑑𝑅 𝑅 The minus indicates that the current decreases as the resistance 𝑅 in the wire increases.
117
Business Calculus Notebook
3.5.5. More about Related Rate of Change in time. Related rates of change problems will always give you the rate of one quantity that’s changing and ask you to find the rate of something else that’s changing as a result. The rate of change is usually with respect to time but can be associated to any other variable/parameter. In algebra, we find the rate of change of the function between two points (𝑓(𝑎), 𝑎) 𝑎𝑛𝑑 (𝑓(𝑏), 𝑏) by calculating the slope 𝑚 of the secant line passing through these two points: 𝑓(𝑏) − 𝑓(𝑎) . 𝑏−𝑎 In differential calculus we evaluate the instantaneous rate of change at one point, which is the slope of the tangent line to the curve at a certain point (𝑥0 ), or instant if considering the time on the axis 𝑥. Means that the instantaneous rate of change is the derivative of the curve/function at the given point, in this case at 𝑥0 = 𝑎 . 𝑚=
The Differentiation with respect to time or one of the other variables requires application of the chain rule, since most problems involve several variables. 𝑑 𝑑 𝑑 𝑔(𝑥)) [𝑓(𝑔(𝑥))] = ( [𝑓(𝑔(𝑥)]) ( 𝑑𝑥 𝑑𝑔 𝑑𝑥 Example: Water is poured into a conical container at the rate of 𝟏𝟎 𝒄𝒎^𝟑/𝒔𝒆𝒄. The cone points directly down, and it has a height of 30 cm and a base radius of 10 cm. How fast is the water level rising when the water is 4 cm deep? The water forms a conical shape into the cone; its height, base radius, and volume increase as water is poured into the container. This means that we actually have three things varying with time: the water level h, the radius r of the circular top surface of water, and the volume of water V. 1
1
The volume of a cone is given by 𝑉 = 3 (𝐵𝑎𝑠𝑒 𝐴𝑟𝑒𝑎)(ℎ𝑖𝑔ℎ) = 3 𝜋𝑟 2 ℎ. 𝑑𝑉
We know 𝑑𝑡 = 10
𝑐𝑚3 𝑠𝑒𝑐
𝑑ℎ
, and we want to know 𝑑𝑡 . Note that we have a 3𝑟𝑑 variable 𝑟.
We will take the derivative of the Volume function with respect to the time. To perform this operation, we have to express the volume as a function of one variable, we select h because we are looking for the rate of change of h ,which is dh/dt. Because the dimensions of the cone of water must have the same proportion as those of the container, we can use this proportion to express r as a function of h . Follow the similar 𝑟 ℎ
triangles relationship we can write: =
10 , 30
𝑠𝑜 𝑟 =
1 3
ℎ 1
Follow this relationship we will write the Volume as a function of 𝑟 , by replacing 𝑟 = 3 ℎ 1
1
2
1
𝑉 = 3 𝜋 (3 ℎ) ℎ = 27 𝜋ℎ3 . Now we can get the derivative of the Volume with respect to the time using the chain rule:
𝑑𝑉 𝑑𝑡
Solving for dh/dt:
𝑑
1
𝑑ℎ
= 𝑑ℎ (27 𝜋ℎ3 )( 𝑑𝑡 ) = 𝑑ℎ 𝑑𝑡
=
𝑑𝑉 𝑑𝑡
27
(3𝜋ℎ2 ) =
3𝜋 27
𝑑𝑉 𝑑𝑡
𝑑ℎ
ℎ2 ( 𝑑𝑡 ) 9
(𝜋ℎ2 ). Plug in ℎ = 4 𝑎𝑛𝑑
𝑑𝑉 𝑑𝑡
= 10
𝑑ℎ 9 90 = 10 ( 2 ) = = 1.79 𝑐𝑚/𝑠𝑒𝑐 𝑑𝑡 𝜋4 16𝜋 118
𝑐𝑚3 , 𝑠𝑒𝑐
we get:
Juan J. Prieto-Valdés
119
Business Calculus Notebook
Homework Topic 3.5. Try to answer before reading provided step-by-step solution (when included). Bring your questions to the class. All this exercise will be discussed in class. 223) Find the number of units that must be produced and sold in order to yield the maximum profit, were the Revenue and cost functions are: 𝑅(𝑥) = 70𝑥 − 0.5𝑥 2 𝑎𝑛𝑑 𝐶(𝑥) = 4𝑥 + 9 𝑃(𝑥) = 𝑅(𝑥) − 𝐶(𝑥) 𝑊ℎ𝑒𝑟𝑒: 𝑃(𝑥) = 𝑃𝑟𝑜𝑓𝑖𝑡 Function, 𝐶(𝑥) = 𝐶𝑜𝑠𝑡 𝐹𝑢𝑛𝑐𝑡𝑖𝑜𝑛, and 𝑅(𝑥) = 𝑥 𝑝(𝑥) → 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 = (𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑖𝑡𝑒𝑚 𝑠𝑜𝑙𝑑)(𝑝𝑟𝑖𝑐𝑒), the price is given by the demand function Solution: 𝑃(𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = 70𝑥 − 0.5𝑥 2 − 4𝑥 − 9 𝑃′ (𝑥) = 66 − 𝑥 = 0, 𝑟𝑒𝑠𝑢𝑙𝑡𝑖𝑛𝑔 𝑥 = 66 𝑢𝑛𝑖𝑡𝑠 224) Find the maximum profit given the following revenue and cost functions: 𝑅(𝑥) = 100𝑥 − 𝑥 2 1 𝐶(𝑥) = 𝑥 3 − 3𝑥 2 + 104𝑥 − 12 3 Where x is in thousands of units and R(x) and C(x) are in thousands of dollars.
𝑥 = 2 𝑡ℎ𝑜𝑢𝑠𝑎𝑛𝑑, 𝑃(𝑥) = $9,333.33 225) Find the number of units that must be produced and sold in order to yield the maximum profit, given the following equations for revenue and cost: 𝑅(𝑥) = 25𝑥 − 0.5𝑥 2 𝐶(𝑥) = 8𝑥 + 8.
𝑥 = 17 120
Juan J. Prieto-Valdés
226)
Find the maximum profit given the following revenue and cost functions: 𝑅(𝑥) = 50𝑥 – 𝑥 2 𝐶(𝑥) = 0.3333 𝑥 3 − 6𝑥 2 − 25𝑥 − 36 where x is in thousands of units and R(x) and C(x) are in thousands of dollars
𝑥=15, 𝑃(𝑥)=$1,161 in thousand units 227) An appliance company determines that in order to sell x special microwave oven, the price per oven must be 𝑝 = 600 − 0.4𝑥 It also determines that the total cost of producing x dishwashers is given by 𝐶(𝑥) = 2500 + 0.2𝑥 2 . What is the maximum profit?
𝑃(𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = = 𝑥(600 − 0.4𝑥) − (2500 + 0.2𝑥 2 ) 𝑃′ (𝑥) = 600 − 1.2𝑥 = 0 resulting 𝑥 = 500 and 𝑃𝑚𝑎𝑥 (𝑥) = (500)[600 − 0.4(500)] − [2500 + 0.2(500)2 ] = 146.500.00 𝑑𝑜𝑙𝑙𝑎𝑟𝑠 228) An appliance company determines that in order to sell x dishwashers, the price per dishwasher must be 𝑝 = 660 − 0.5𝑥 It also determines that the total cost of producing x dishwashers is given by 𝐶(𝑥) = 5000 + 0.5𝑥2. What is the maximum profit?
𝑃𝑚(𝑥)=$103,900.00 121
Business Calculus Notebook
229) A hotel has 250 units. When the hotel charges $100 per day for a room, all rooms are occupied. For every increase of x dollars in the daily room rate, there are x rooms vacant. Each occupied room costs $30 per day to service and maintain. What should the hotel charge per day in order to maximize daily profit?
𝑅(𝑥) = 𝑥 𝑝(𝑥) = = (250 − 𝑥)(100 + 𝑥), 𝐶(𝑥) = (250 − 𝑥)30 𝑃(𝑥) = 25000 − 100𝑥 + 250𝑥 − 𝑥 2 − 7500 + 30𝑥 𝑃′ (𝑥) = −2𝑥 + 180 = 0, 𝑟𝑒𝑠𝑢𝑙𝑡𝑖𝑛𝑔 𝑥 = 90
𝑎𝑛𝑑
𝑡ℎ𝑒 𝑝𝑟𝑖𝑐𝑒 (100 + 90) = 190.00 𝑑𝑜𝑙𝑙𝑎𝑟𝑠 230) A hotel has 250 units. All rooms are occupied when the hotel charges $80 per day for a room. For every increase of 5x dollars in the daily room rate, there are 3x rooms vacant. Each occupied room costs $15 per day to service and maintain. What should the hotel charge per day in order to maximize daily profit?
𝑥=$35.10,𝑎𝑛𝑑 𝑡ℎ𝑒 𝑃𝑟𝑖𝑐𝑒=$115.10 231) If the price charged for a pencil is p cents, then x thousand pens 𝑥
will be sold in a certain retail store, where 𝑝 = 60 − 200 How many pens must be sold to maximize revenue? 𝑥2 , 200 𝑥 𝑅 ′ (𝑥) = 60 − , 100 𝑎𝑡 𝑅 ′ (𝑥) = 0 𝑟𝑒𝑠𝑢𝑙𝑡𝑠
𝑅(𝑥) = 𝑥𝑝(𝑥) = 60𝑥 −
𝑥 = 6 𝑡ℎ𝑜𝑢𝑠𝑎𝑛𝑑 𝑝𝑒𝑛𝑠 232) If the price charged for a chocolate tablet is p dollar, then 𝑥 hundred tablets will be sold in a certain cafeteria, where 𝑝 = 90 –
𝑥 120′
How many tablets must be sold to maximize revenue?
𝑥 = 5,400 𝑡𝑎𝑏𝑙𝑒𝑠
122
Juan J. Prieto-Valdés
233) The management of a stadium is trying to determine what price to charge for tickets during the national baseball competition. At a price of $10 per ticket, it averages 50,000 people per game. For every increase of $1, it loses 5,000 people. Every person at the game spends an average of $5 on concessions. What price per ticket should be charged in order to maximize revenue?
S = sales,
p = price/ticket,
R = revenue Revenue will be $p per ticket + $5 for concessions per ticket, so P+5 total per ticket. 𝑅 = 𝑆(𝑝 + 5) 𝑆 = 50,000 − 5,000(𝑝 − 10) notice that, in the above equation, a price of 10 yields sales of 40,000 and each $1 increase drops sales by 5,000, so our minimum price is 10. this equation can be simplified to: 𝑆 = 50,000 − 5,000𝑝 + 50,000 𝑆 = 100,000 − 5,000𝑝 Now, simply combine the two equations to find revenue in terms of price: 𝑅 = (100,000 − 5,000𝑝)(𝑝 + 5) using the foil method on the above equation yields 𝑅 = 100,000𝑝 + 500,000 − 5,000𝑝2 − 25,000𝑝 So 𝑅 = −5,000𝑝2 + 75,000𝑝 + 500,000 Now all we have to do is set the derivative equal to zero to find the price that will yield maximum revenue: ′ 𝑅 = −5000(−15 + 2𝑝) = 0 𝑝 = 7.5 Final answer: the best price is $7.50
234) A Theater company is trying to determine what price to charge for tickets. At a price of $25 per ticket, it averages 800 people per presentation. For every increase of $5, it loses 10 people. Every person at the theater spend an average of $12 on concessions. What price per ticket should be charged to maximize revenue?
𝑝 = $36.50
123
Business Calculus Notebook
235) A bookstore has an annual demand for 48,000 copies of a winner in a competition book. It costs $0.50 to store one copy for one year, and it costs $50 to place an order. Find the optimum number of copies per order. This is an inventory-cost situation. Let x stand for the Full Lot Size. 𝑥
The carrying costs is equal to 0.50 (2) 48000 ) 𝑥
while the reorder cost is 50 (
That makes this the cost function: 2,400,000 . 𝑥 2400,000 𝑎𝑛𝑑 𝐶 ′ (𝑥) = 0.25 − . 𝑥2 Equal to zero when 2,400,000 𝑥2 = , 𝑥 = 3,098 0.25 The store should order in lots of 3,098 books per order. 𝐶(𝑥) = 0.25𝑥 +
236) A national auto sales company has an annual demand for 15,500 autos. It costs $400 per year to have parked one auto and $2,500 to place an order. Find the optimum number of autos per order that the dealer must request.
𝑥 = 440 𝑎𝑢𝑡𝑜𝑠 𝑝𝑒𝑟 𝑜𝑟𝑑𝑒𝑟 124
Juan J. Prieto-Valdés
237) A trough of water is 10 meters in length, its ends are in the shape of isosceles triangles whose width is 5 meters, and height is 2 meters. If water is being pumped in at a constant rate of 6 𝑚3 /𝑠𝑒𝑐. At what rate is the height of the water changing when the water has a height of 120 cm? At what rate is the width of the water changing when the water has a height of 120 cm?
Check your answer in class. 238) A company has determined the demand curve for their product is 𝑞 = √8000 − 𝑝2 where p is the price in dollars, and q is the quantity in millions. If transport conditions are driving the price up 2
$ , 𝑤𝑒𝑒𝑘
, find the rate at which demand changes when the
price is $40.
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239) The manager of a garden store wants to build a 900 square foot rectangular enclosure on the store’s parking lot in order to display some equipment. Three sides of the enclosure will be built of redwood fencing, at a cost of $6 per running foot. The fourth side will be built of bricks, at a cost of $35 per running foot. Find the dimensions of the least costly such enclosure.
Check your answer in class. 240) The price p (in dollars) and demand x for a product are related by: 𝑥 2 + 2𝑥𝑝 + 25𝑝2 = 75,000. If the price is increasing at a rate of $2.50 per month when the price is $40, find the rate of change of demand.
Check your answer in class. 241) Repeat example 3.5.3 (Application to evacuation situation)
considering same conditions except the maximum speed of the buss. Use 𝒗𝒃𝒖𝒔𝒔 = 𝟓𝟎 𝑴𝒊𝒍𝒆𝒔/𝒉𝒐𝒖𝒓
Check your answer in class.
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PRACTICING QUIZ-4. Optimization, Exponential and Logarithmic Functions. 1. (20 Points) Two post, one 15 feet high and the other 30. Stand 40 feet apart. They are to be stayed by two wires attached to a single stake, running from ground level to the top of each post. Where the stake should be placed to use the least amount of wire.
2. (20 points) Find the height and the radius of the least expensive closed cylinder which has a volume of 1000 cubic inches. Assume that the material cost is free, but the welding cost $0.80 per inch, to weld the top and bottom onto the cylinder and to join the side as per the figure on the right.
3. (10 Points) Find the horizontal asymptote: 𝑓(𝑥) = 3𝑥 − √9𝑥 2 − 2𝑥 + 2
4. (50 points) Evaluate the limit:
lim (𝑒 𝑥 √𝑥)
𝑥→∞
𝑏 𝑥 lim (1 + ) 𝑥→∞ 𝑥
lim+ (
𝑥→1
1 1 − ) ln(𝑥) 𝑥 − 1
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PROJECT-3. Quick Example of a Business Profit Optimization Optimization of a Refresh Vending Business (VB) using automated vending machines.
I.
General Information:
The ads are alluring: “Make $500 an Hour in the Vending Business!” “Earn Money While You Sleep in a Vending Machine Business!” But they sound way too good to be true. Can vending machines really be a viable way to earn cash? Retrieved from: http://quickbooks.intuit.com/r/money/can-you-really-make-money-with-vending-machines/ The National Automatic Merchandising Association reports that 18 percent of vending-machine operators make between $1 million and $5 million a year. But proceed with caution: The Better Business Bureau warns of scams, and a search of “vending machines” on the Federal Trade Commission’s website unearths dozens of fines and lawsuits. In the present project will be used the type of machines offered by REFRESHMENTSERVICES.COM for 12 OZ. CAN and 16.9 OZ. PLASTIC BOTTLE. The VB plans to work with medium capacity machine of 500 cans with 7 different flavors, all of them at the same price.
II.
Specific Information/Data:
The main objective of the owner of the VB is to maximize the profit. As per his request, during the first month of operation, was investigated the business situation, collecting the following data when selling the refreshes at different prices:
Item price (𝒑 − $)
Number of Items sold (𝒙 − 𝒏𝒖𝒎𝒃𝒆𝒓)
0.90
2165
0.70
3271
0.50
4987
1. The number of sold items decreases significantly with price 0.40 6187 increasing. The Table above displays the monthly average data (the item is referred to a soda can): 1.1. The student must learn how to collect statistical data, in this case to investigate how the price influences into selling activity. In economics, the demand curve is the graph depicting the relationship between the price of a certain commodity and the amount of it that consumers are willing and able to purchase at any given price. It is a graphic representation of a demand schedule. 1.2. Using the collected data (Table above), the student must create a function from the scratch (point by point), such activity is named regression analysis. Regression analysis is a statistical process to estimate the relationships among variables. It includes many techniques for modeling and analyzing several variables, when the focus is on the relationship between a dependent variable and one or more independent variables (function). 1.3. Apply technology to create the demand function 𝑝(𝑥). Using the collection of points (data), linear, polynomial, or exponential regressions (functions) can be created using Microsoft Excel, Graphic Calculators (Ti83/84 or similar) Microsoft Mathematics, Wolfram Alpha, or special data processing regression software. Any of these methods are very well described in the HELP command of each software or device. Also, the regression can be created manually by graphic-differentiation method, differentiating up to linear regression to evaluate the constant term, then integrate as required. We will use Wolfram Alpha Go to https://www.wolframalpha.com/examples/RegressionAnalysis.html and check all possible operations to create the function that best fits our data. Let’s test different options to select the most convenient regression to simulate the following points selected from our previous Table: (𝑥, 𝑝) ≡ (2165, 0.9); (3271, 0.7); (4987, 0.5); (6187, 0.4)
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Juan J. Prieto-Valdés Screenshot shows the regression function which best fits previously selected points.
For verification purposes we can graph the discovered function using Microsoft Mathematics as it is displayed on the right side 𝑦 = −2.83132 × 10−12 𝑥 3 + 5.22896 × 10−8 𝑥 2 − 0.000401463 𝑥 + 1.55281
This regression (function) can be confirmed for approximately 100 < 𝑥 < 7000 as the Demand Function 𝑝(𝑥).
2. The VB pays $0.17 per item (soda can) wholesales price. The acquisition cost of sodas can be expressed as: 𝐶𝑖 (𝑥) = 0.17 𝑥
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3. The VB pays $75.00 lease per month per machine. Also, pays a technical service insurance of $5.00 per 720
machine per month and a $720.00 annual obligation ( 12 = 60 𝑚𝑜𝑛𝑡ℎ𝑙𝑦) to the institution where the vending machines are installed (it includes: license, energy consumption, commissions, etc.) Operating cost and other required fees can be expressed as 𝐶𝑜𝑝 (𝑛) = 80 𝑛 + 60 Each machine (𝑛) has a capacity of 500 items, and as per signed contract with the Refresher Producer Company, Vending Machines will be loaded as needed one time per week. Lot size 𝑥 of each reorder can vary depending on the number of sodas sold per week. However, for the calculation we will consider that machines are loaded in full four time per month, later we can reject this assumption. Under previous condition, the number of working machines can be written as follow: 𝑥 𝑛= 500(4) So, the operating cost as a function of 𝑖𝑡𝑒𝑚𝑠 (𝑥) can by written as: 80 𝐶𝑜𝑝 (𝑥) = ( ) 𝑥 + 60 = 0.04𝑥 + 60 500(4) 4. For keeping the machines in normal working conditions, the owner of the VB pays to his employee (only one) a salary of $100.00 base salary plus 10 % of the total sales: 𝐶𝑆 (𝑥) = 0.1 ∙ 𝑅(𝑥) + 100, were 𝑅(𝑥) is the revenue (total sales). 𝐶𝑠 (𝑥) = 0.10 𝑥(−2.83132 × 10−12 𝑥 3 + 5.22896 × 10−8 𝑥 2 − 0.000401463 𝑥 + 1.55281) + 100 III.
Analysis
For the primary analysis we will represent the profit function as the Revenue minus the Total Cost: 𝑃(𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = 𝑥 ∙ 𝑝(𝑥) − [𝐶𝑖 (𝑥) + 𝐶𝑜𝑝 (𝑥) + 𝐶𝑠 (𝑥)] 𝑃(𝑥) = 𝑥(−2.83132 × 10−12 𝑥 3 + 5.22896 × 10−8 𝑥 2 − 0.000401463 𝑥 + 1.55281) − 0.17 𝑥 − 0.04𝑥 − 60 − 0.10𝑥(−2.83132 × 10−12 𝑥 3 + 5.22896 × 10−8 𝑥 2 − 0.000401463 𝑥 + 1.55281) − 100 Use Wolfram Alpha to consolidate (simplify) previous equation, differentiate, and solving at
𝑑𝑦 𝑔𝑥
= 0, resulting:
𝑃(𝑥) = 0.9 (−2.83132 × 10−12 𝑥 3 + 5.22896 × 10−8 𝑥 2 − 0.000401463 𝑥 + 1.55281)𝑥 − 0.21 𝑥 − 160 𝜕𝑃(𝑥) 1187529 3613167 𝑥 882387 𝑥 2 657 𝑥 3 = − + − 𝜕𝑥 1000000 5000000000 6250000000000 64457567495020 max{𝑃(𝑥) = 0.9 (−2.83132 × 10−12 𝑥 3 + 5.22896 × 10−8 𝑥 2 − 0.000401463 𝑥 + 1.55281)𝑥 − 0.21 𝑥 − 160}
max: 𝑷(𝒙) ≈ 𝟏𝟐𝟏𝟓. 𝟕𝟒 𝒂𝒕 𝒙 ≈ 𝟑𝟏𝟎𝟑. 𝟑𝟒 To graphically visualize this result, we also can use Microsoft Mathematics add-in (see Graph 𝑷(𝒙) & 𝒙, 𝑵𝒐. 𝟏): The graph provides the same solution with the selected precision in the graphic software setting.
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Graph 𝑷(𝒙) & 𝒙, 𝑵𝒐. 𝟏
Graph 𝑷(𝒙) & 𝒙, 𝑵𝒐. 𝟐
As we can see from the graphic, the most convenient number of items to be sold is approximately 3100. The owner of the VB will have a profit of $1,230 per month. Such conditions will be suitable when selling the items at: 𝑝(𝑥) = −2.83132 × 10−12 𝑥 3 + 5.22896 × 10−8 𝑥 2 − 0.000401463 𝑥 + 1.55281 𝐴𝑡 𝑥 = 3100, 𝑝(𝑥) = $0.73 Such conditions can be reached using 𝑥 3100 𝑛= = = 1.5 ≈ 2 𝑣𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑎𝑐ℎ𝑖𝑛𝑒𝑠 500(4) 2000 Because we have a 0.5 extra vending machine, to stimulate the business growth, the owner can sell the items at lower price scarifying some profit, for example at $ 0.60 per item, to sell approximately 4000 items per month to get a profit of $1,168 monthly: 𝑝(𝑥) = −0.000000000000698194𝑥 3 + 0.0000000247729𝑥 2 − 0.000293727𝑥 + 1.42551 𝐴𝑡 𝑥 = 4000, 𝑝(𝑥) = $0.60 The “Profit versus Item” graph can be used to display such situation (see Graph 𝑷(𝒙) & 𝒙, 𝑵𝒐. 𝟐): In optimal conditions the employee will receive 𝐶𝑠 (𝑥) = 0.10 𝑥(−2.83132 × 10−12 𝑥 3 + 5.22896 × 10−8 𝑥 2 − 0.000401463 𝑥 + 1.55281) 𝐴𝑡 𝑥 = 3100,
𝐶𝑠 (𝑥) = $325.00
Which is a good salary to attend a couple of vending machines, means, for working a couple of hours one day per week. This is a $40.00/hour salary. However, the owner can implement a bonus stimulus to guarantee the quality of work. If decreasing the price to $0.60 per item, under hypothetical assumption that 4000 items will be sold, the employee will receive a monthly salary of $341.00 + Bonus, under previously established condition of 10% of sales: 𝐴𝑡 𝑥 = 4000,
IV.
𝐶𝑠 (𝑥) = $341.00
Conclusion
Similar to this project, the student has to present a real case business example and discus it with Lecturer Professor. The student also can present a modification of the present project, including vending machines with different products having completely different prices and considering different inventory situation. It is strongly recommended to create your own example based on your work experience or needs.
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CHAPTER-4. INTEGRATION 4.1. Indefinite Integration and Simple Differential Equation Integration is the algebraic method of finding the integral for a function at any point on the graph. The integral is usually called the anti-derivative, because integrating is the reverse process of differentiating. However, the integration method was developed not only to find the inverse process of taking the derivative but trying to solve the area and distance problem among other applications. At first glance; as the process of differentiation is used to find the slope at any point on the graph, the process of integration can be used to find the area under the curve. The General Power Rule for Integration (indefinite integral) and other rules for common functions: The general power rule is the exact opposite of the power rule for differentiation. If 𝑓(𝑥) = 𝑐𝑥 𝑛 , the integral of that function will be: ∫ 𝑓(𝑥)𝑑𝑥 =
𝑐𝑥 𝑛+1 +𝑐 𝑛+1
Examples: Additional examples at the lecture ∫ 𝑥 2 𝑑𝑥 =
+ 𝑐 and ∫ 𝑥𝑑𝑥 =
𝑥2 2
+𝑐
3 ∫ 𝑑𝑥 = 3 ln 𝑥 + 𝑐 𝑥
1
The Logarithmic Rule: ∫ 𝑥 𝑑𝑥 = 𝑙𝑛|𝑥| + 𝑐 𝑎𝑡 𝑥 ≠ 0
1 ∫ 𝑒 3𝑥 𝑑𝑥 = 𝑒 3𝑥 + 𝑐 3
1
The Exponential Rule: ∫ 𝑒 𝑘𝑥 𝑑𝑥 = 𝑘 𝑒 𝑘𝑥 + 𝑐 The Constant Rule: ∫ 𝑘𝑓(𝑥)𝑑𝑥 = 𝑘 ∫ 𝑓(𝑥)𝑑𝑥
∫(4𝑥 2 + 2𝑥 − 3)𝑑𝑥 =
The Sum Rule: ∫[𝑓(𝑥) + 𝑔(𝑥)]𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 + 𝑔(𝑥)𝑑𝑥
= 4 ∫ 𝑥 2 𝑑𝑥 + 2 ∫ 𝑥𝑑𝑥 − 3 ∫ 𝑑𝑥 =
The Difference Rule: ∫[𝑓(𝑥) − 𝑔(𝑥)]𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 − 𝑔(𝑥)𝑑𝑥
4.2.
𝑥3 3
4 = 𝑥 3 + 𝑥 2 − 3𝑥 + 𝑐 3
Integration by substitution
The substitution method technic uses the Chain Rule “in reverse”. This method works when the integrand is of the form 𝑓(𝑢(𝑥))𝑢′ (𝑥). If 𝐹(𝑢) is an antiderivative of 𝑓(𝑢), ,them by the Chain Rule: ∫ 𝑓(𝑢(𝑥))𝑢′ (𝑥)𝑑𝑥 = 𝐹(𝑢(𝑥)) + 𝑐 Before proceeding to examples we have to discuss the procedure and notation: A differential is a symbol such as 𝑑𝑢
𝑑𝑥 𝑜𝑟 𝑑𝑢 that occurs when using the notation 𝑑𝑥 and ∫ 𝑓(𝑥)𝑑𝑥. We treat differentials as a symbol that do not represent actual quantities, but we operate them algebraically as if they were elements of an equation where we may 𝑑𝑢
cancel 𝑑𝑥. For example: 𝑑𝑢 = 𝑑𝑥 𝑑𝑥, equivalently 𝑑𝑢 = 𝑢′ (𝑥)𝑑𝑥 Example Evaluate ∫ 2𝑥(𝑥 2 + 9)5 𝑑𝑥 𝑢 = 𝑥 2 + 9, and the differential 𝑑𝑢 = 2𝑥𝑑𝑥 1 ∫(𝑥 2 + 9)5 2𝑥𝑑𝑥 = ∫ 𝑢5 𝑑𝑢 = 𝑢6 + 𝑐 6 1 6 1 2 𝑢 + 𝐶 = (𝑥 + 9)6 + 𝑐 6 6
Step 1: Chose the function 𝑢 and compute 𝑑𝑢 Step 2: Rewrite the integral in terms of 𝑢 and evaluate Step 3: Express the final result in terms of 𝑥
Additional examples will be reviewed during the lecture
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Homework Topic 4.1 and 4.2. Find the indicated integral. Check your answer by differentiation 242)
∫ 2𝑑𝑥 2𝑥 + 𝐶
243)
∫(2𝑥 + 𝑝)𝑑𝑥 𝑥2 + 𝑝 𝑥 + c
244)
∫(2𝑥 3 + 3𝑥 − 5)𝑑𝑥 𝑥4 3 𝑥2 + −5𝑥+c 2 2
245)
2
∫(𝑥 3 + 4𝑥)𝑑𝑥 5
3 𝑥3 2 𝑥2 + +c 5 246)
∫(2𝑒 𝑥 + 𝑥 2 )𝑑𝑥 2 𝑒𝑥 +
247)
∫(
𝑥 2 + 4𝑥 − 12 √𝑥
𝑥3 +𝑐 3
) 𝑑𝑥 5
3
2 𝑥2 8 𝑥2 + − 24 √𝑥 + 𝑐 5 3 248)
∫ (2 𝑥 2 +
1 1 + 5 ) 𝑑𝑥 2 𝑥 𝑥 2 𝑥3 1 1 − − +𝑐 3 𝑥 4 𝑥4
249)
(∫ 2 𝑒 −𝑥 +
1 1 + ) 𝑑𝑥 3 𝑥 2 √𝑥 5 −
250)
∫ (2 ln(2−𝑥 ) +
2 3 1 − − +𝑐 2 𝑥 𝑒 𝑥 2 𝑥3
1 ) 𝑑𝑥 𝑥3 1 1 −2 (− 𝑥 2 ln(2) − 𝑥 ln2−𝑥 − )+𝑐 2 2 𝑥2
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Juan J. Prieto-Valdés
251)
∫ 𝑥 −2 (2𝑥 +
1 ) 𝑑𝑥 𝑥3
2 ln(𝑥) −
1 +𝑐 4 𝑥4
252) Find the function whose tangent has slope 2𝑥 2 + 1 for any value of 𝑥 and whose graph passes though the point (3, 25) 2 𝑥3 𝑦= +𝑥+4 3
253) Find the function whose tangent has slope 4𝑥 + 2 for any value of 𝑥 and whose graph passes though the point (1, 3)
𝑦 = 2 𝑥2 + 2 𝑥 − 1
254) An object is moving straight way at 50 miles per hour (22.352 meter/sec) If the driver is forcing to apply 𝑚 the brakes and the car stops reducing its speed with an acceleration of 12 2 how far does the car can 𝑠𝑒𝑐 travel before applying he brakes. 𝑣(0) = −12𝑡 + 22.35 𝑣(𝑡) − 12𝑡 + 22.35 = 0 149 𝑡= 80 𝑠 = −12 𝑡 2 + 22,35𝑡 + 𝑐 𝑆(0) = 0 = 𝑐 149 2 149 𝑠 = −6 ( ) + (22.35) 80 80 𝑠 = 20.84 𝑚𝑡𝑠 255) Marginal cost of producing certain cosmetic product is 2𝑞 2 − 40𝑞 + 266.66 dollar per 100 units. The total cost of producing the firs 200 units is 600. What is the total cost of production 100000 unit? 2𝑞 2 − 40𝑞 + 266.66 2 𝑞3 − 20 𝑞 2 + 266.66𝑞 + 𝑐 3 2×8 𝑐 = 600 − + 20 × 4 − 267 × 2 3 𝐶𝑜𝑠𝑡 =
𝑐 = 4223 2 ×8 𝐶𝑜𝑠𝑡 = − 20 × 4 + 267 × 2 + 4223 3 = 140473 135
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256) Solve a differential equation
𝑑𝑦 𝑑𝑥
4𝑥
= 𝑦3
𝑦4 − 2 𝑥2 + 𝑐 = 0 4 4 𝑦 = √8 𝑥 2 − 𝑐 257)
Solve a differential equation
𝑑𝑦 𝑑𝑥
=
9𝑥 2 2
4𝑦 3 5
12 𝑦 3 − 3 𝑥3 + 𝑐 = 0 5 3
5 𝑥3 5 𝑐 5 𝑦=( − ) 4 12 258) A biomass is growing at the rate of 𝑀’ = 1.5𝑒 0.55𝑡 grams/hour. How much the biomass change will be at the second hour if at the beginning of our calculation the Mass was 200 Kilograms?
𝑀=
1.5𝑒 0.55𝑡 + 197 0.55
1.5 𝑒 0.55𝑡 +197×0.55
𝑀= 0.55 𝑀 = 30 𝑒1.111 + 197
Find the indicated integral. Apply substitution method. 259)
∫(𝑥 2 + 1)𝑥𝑑𝑥 𝑥4 𝑥2 + +𝑐 4 2
260)
2 ∫ ( 𝑥 3 + 5𝑥) (2𝑥 2 + 5)𝑑𝑥 3
2
2 𝑥3 ( + 5 𝑥) 3 +𝑐 2 261)
1 ∫ (𝑥 4 + 3𝑥 2 )(2𝑥 3 + 3𝑥 − 5)𝑑𝑥 2
𝑥8 3 𝑥6 𝑥5 9 𝑥4 5 𝑥3 + − + − +𝑐 8 4 2 8 2 136
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262)
∫(
ln(𝑥) + 2) 𝑑𝑥 𝑥
2𝑥 + 263) ∫(2 𝑒 𝑥 +
log 2 (𝑥) + 𝑐 2
𝑥3 + 4)(2𝑒 𝑥 + 𝑥 2 )𝑑𝑥 3 2 𝑥3 𝑒 𝑥 𝑥6 4 𝑥3 + 2 𝑒2 𝑥 + 8 𝑒 𝑥 + + +𝑐 3 18 3
264) ∫(
𝑥 2 + 4𝑥 − 12 √𝑥
) 𝑑𝑥
5
3
2 𝑥2 8 𝑥2 + − 24 √𝑥 + 𝑐 5 3 265) ∫ ((
2 𝑥3 1 1 − + 2) (2 𝑥 2 + 2 )) 𝑑𝑥 3 𝑥 𝑥
2 𝑥6 4 𝑥3 2 𝑥2 2 1 + − − + +𝑐 9 3 3 𝑥 2 𝑥2 266) (∫ 2 𝑒 −𝑥 +
1 1 + 3 ) 𝑑𝑥 2 𝑥 √𝑥 5
− 267) ∫
2 3 1 − − +𝑐 2 𝑒𝑥 𝑥 2 𝑥3
1 1 (2 ln(𝑥 𝑥 ) + 3 ) 𝑑𝑥 2 𝑥 𝑥
For x>0, = ln2 𝑥 −
1 4𝑥 4
+𝐶
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4.3. Riemann Integration / Riemann Sum. Before integration was developed, we could calculate approximately the area under the curve towards axis x, by dividing the space into rectangles and adding the areas. The approximation could be improved by increasing the number of rectangles under the curve. The sum of the areas of all rectangles will be a good approximation, better when they are very think. The area of these rectangles can be calculated by multiplying length times width The following expression is the Mathematical representation of Riemann Sum to calculate the area under the graph. After discovering calculus, the partition of the rectangles was considered infinitely large, so it was required to consider the basis of the rectangles Δ𝑥 = Δ𝑥𝑖 ≅ 0. Further we will solve the Riemann Sum follow a limit exercise: 𝑏
∑𝑛𝑖=1 𝑓(𝑥𝑖 )Δ𝑥
∫𝑎 𝑓(𝑥)𝑑𝑥 = lim (∑𝑛𝑖=1 𝑓(𝑥𝑖 )Δ𝑥) in a closed interval [𝑎, 𝑏] 𝑛→∞
Example 1: Calculate the area under the curve 𝑦 = 𝑥 2 on the interval [1, 3]. Consider only 6 subintervals with end points on the right. So, each subinterval will be Δ𝑥
=
3−1 6
1
= ; 3
And the total area (Summa) will be: 𝑛
1 4 5 6 7 8 9 𝐴𝑟𝑒𝑎 = ∑ 𝑓(𝑥)Δ𝑥 = [𝑓 ( ) + 𝑓 ( ) + 𝑓 ( ) + 𝑓 ( ) + 𝑓 ( ) + 𝑓 ( )] 3 3 3 3 3 3 3 𝑖=1
1 16 25 36 49 64 81 271 = ( + + + + + )= 3 9 9 9 9 9 9 27 Example 2: Estimate the area between 𝑓(𝑥) = 𝑥 3 − 5𝑥 2 + 6𝑥 + 5 and the axis 𝑥 on [0, 4] using 𝑛 = 5 subintervals.
Δ𝑥 =
4−1 5
= 0.8. So, the end points of the subintervals are 0, 0.8, 1.6, 2.4, 3.2, and 4’
A. Let us first look at using the right endpoints for the function height. .
B. Now let us look at left endpoints for the function height.
𝑛
𝐴𝑟𝑒𝑎 (𝐶𝑎𝑠𝑒 𝐴) = ∑ 𝑓(𝑥)Δ𝑥 = 0.8[𝑓(0.8) + 𝑓(1.6) + 𝑓(2.4) + 𝑓(3.2) + 𝑓(4) = 28.96 𝑖=1 𝑛
𝐴𝑟𝑒𝑎 (𝐶𝑎𝑠𝑒 𝐵) = ∑ 𝑓(𝑥)Δ𝑥 = 0.8[𝑓(0) + 𝑓(0.8) + 𝑓(1.6) + 𝑓(2.4) + 𝑓(3.2) = 25.56 𝑖=1
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Juan J. Prieto-Valdés
C. Finally, let us look at the midpoints for the heights of each rectangle.
In this case, the mid points are located at 0.4, 1.2, 2, 2.8, and 3.6. 𝑛
𝐴𝑟𝑒𝑎 (𝐶𝑎𝑠𝑒 𝐶) = ∑ 𝑓(𝑥)Δ𝑥 = 0.8[𝑓(0.4) + 𝑓(1.2) + 𝑓(2) + 𝑓(2.8) + 𝑓(3.6) = 25.12 𝑖=1
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Homework Topic 4.3 Find the indicated Riemann Sum 268) Find the area under the graph of 𝑓(𝑥) = 2𝑥 + 2 on the intevale 0 < 𝑥 < 2. Use the method of your choice.
269) Find the area under the graph of 𝑓(𝑥) = 2𝑥 2 + 2 on the interval [1, 4]. Use 10 subintervals with endpoint on the right.
4.4.
Fundamental Theorem of Calculus and Definite Integral
The notation on the left side denotes the definite integral of 𝑓(𝑥) from 𝑎 to 𝑏. When we calculate the integral from an interval [a,b], we plug 𝑎 in the integral function and subtract it from 𝑏 in the integral function: F denotes the integrated function. This accurately calculates the area under any continuous function. 140
Juan J. Prieto-Valdés 𝑏
∫𝑎 𝑓(𝑥)𝑑𝑥 = lim (∑𝑛𝑖=1[𝑓(𝑥𝑖 )](Δ𝑥) 𝑛→∞
Example:
3 ∫1 (2𝑥 2
𝑏
Equivalently to: ∫𝑎 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑏) − 𝐹(𝑎)
2 3
− 8𝑥 + 10)𝑑𝑥 = ( 𝑥 3 − 4𝑥 2 + 10𝑥) │ 31 = 12 −
20 3
=
16 3
Properties of Definite Integral PROPERTY 1: The integral of a constant times a function is the constant times the integral of the function
∫ 𝑘 𝑓(𝑥)𝑑𝑥 = 𝑘 ∫ 𝑓(𝑥)𝑑𝑥
PROPERTY 2: The integral of a sum is the sum of the integrals.
∫ 𝑓(𝑥)𝑑𝑥 + 𝑔(𝑥)𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 + ∫ 𝑔(𝑥)𝑑𝑥
PROPERTY 3: For any number 𝑐 between 𝑎 and 𝑏, the integral from a to b is the integral from a to c plus the integral from c to b. (𝑎 < 𝑐 < 𝑏)
∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 + ∫ 𝑓(𝑥)𝑑𝑥
𝑏
𝑏
𝑎
𝑎
𝑏
𝑏
𝑎
𝑏
𝑎 𝑏 𝑎
𝑎
𝑐 𝑎
𝑏 𝑐
Example 1: 2 1 2 1 1 14 ∫ (𝑥 2 + 1)𝑑𝑥 = ( 𝑥 3 + 𝑥) │ = (23 ) + 2 − ( (03 ) + (0)) = 3 0 3 3 3 0
Example 2. 3 2 3 2 2 16 ∫ (2𝑥 2 − 8𝑥 + 10)𝑑𝑥 = ( 𝑥 3 − 4𝑥 2 + 10𝑥) │ = (33 ) − 4(32 ) + 10(3) − + 4 − 10 = 3 1 3 3 3 1
Example 3: 4 1 4 1 1 88 ∫ (𝑥 2 + 2)𝑑𝑥 = ( 𝑥 3 + 2𝑥) │ = (43 ) + 2(4) − ( (03 ) + 2(0)) = 3 0 3 3 3 0
Example 4: 0 1 0 1 1 88 ∫ (𝑥 2 + 2)𝑑𝑥 = ( 𝑥 3 + 2𝑥) │ = (03 ) + 2(0) − ( (−43 ) + 2(−4)) = 3 −4 3 3 3 −4
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Example 5: 4 1 4 1 1 176 ∫ (𝑥 2 + 2)𝑑𝑥 = ( 𝑥 3 + 2𝑥) │ = (43 ) + 2(4) − ( (−43 ) + 2(−4)) = 3 −4 3 3 3 −4
Example 6: −1
∫
(
−10
3 𝑒 −𝑥
−
1 1 −1 1 1 ) 𝑑𝑥 = (3𝑒 𝑥 − ln|𝑥|) │ = 3𝑒 −1 − ln|−1| − (3𝑒 −20 − ln|−20|) = 0.105 2𝑥 2 −10 3 3
4.5. Application of Integrals (Average Function Value). The average value of a function 𝑓(𝑥) over the interval [a, b] is given by: If 𝑓(𝑥) is a continuous function on [a,b] then there is a number c in [a,b] such that This theorem is telling us that there is a number a < c < b such that If this function is continuous, then somewhere in [a,b] the function will take on its average value
1
𝑏
𝑓𝑎𝑣𝑒 (𝑥) = 𝑏−𝑎 ∫𝑎 𝑓(𝑥)𝑑𝑥 ; 𝑏
∫𝑎 𝑓(𝑥)𝑑𝑥 = 𝑓(𝑐)(𝑏 − 𝑎). 𝑓𝑎𝑣𝑒 (𝑥) = 𝑓(𝑐), 𝑏 1 ∫ 𝑓(𝑥)𝑑𝑥 = 𝑓(𝑐) 𝑏−𝑎 𝑎
Example: Determine the number c that satisfies the Mean Value Theorem for Integrals for the function 𝑓(𝑥) = 𝑥 2 + 3𝑥 + 1 on the interval [1,4]. The function is a polynomial, it is continuous on the given interval, means that we can use the Mean Value Theorem. 4
∫ (𝑥 2 + 3𝑥 + 1)𝑑𝑥 = (4 − 1)(𝑐 2 + 3𝑐 + 1) 1
1 3 4 [ 𝑥 3 + 𝑥 2 + 𝑥] = 3(𝑐 2 + 3𝑐 + 1) 3 2 1 1 3 3 2 1 3 1 [ 4 + 4 + 4] − [ + + 1] = 46 3 2 3 2 2 93 = 3 𝑐2 + 9 𝑐 + 3 2 3
𝑐 = −2 ±
√67 2
= 2.5926 … Reject negative value as it is out of the interval
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Juan J. Prieto-Valdés
Homework Topic 4.4 and 4.5. . Solve the following definite integrals 270)
3
∫ 2𝑑𝑥 1
4 271)
4
∫(2𝑥 + 5)𝑑𝑥 0
36 272)
10
∫ (2𝑥 3 + 3𝑥 − 5)𝑑𝑥 3
5061 273)
5
2
∫ (𝑥 3 + 4𝑥) 𝑑𝑥
274)
𝑒
12
∫ 3
276)
121 18 + ln ( ) 4 8 (ln(𝑥))2
𝑥
2
𝑑𝑥 (ln(8))3 (ln(2))3 − = 3ln2 − 3ln2 = 0 3 3
24.5
∫
10𝑥(5𝑥 2 − 7) 𝑑𝑥
12
278)
𝑒6 𝑒3 2 − + 2 𝑒𝑒 − 2 𝑒𝑒 3 3
2𝑥 𝑑𝑥 𝑥−1
∫
277)
5
𝑒2
∫ (2𝑒 𝑥 + 𝑥 2 )𝑑𝑥
275)
5
3 · 5 3 3 · 33 − + 32 ≅ 37.028 5 5
3
0
∫ 𝑥 2 √𝑥 + 2 𝑑𝑥 −2
135314625 32
3 2 (2 + 𝑥)2 (32 − 24 𝑥 + 15 𝑥 2 )│ 0 = 1.72399 −2 105
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4.6. Application of Integrals (Area Under and Between Curves). The definite integral can be used to find the area between a graph curve and the ‘𝑥’ axis, between two given 𝑥 values. This area is called the ‘area under the curve’ regardless of whether it is above or below the ‘𝑥’ axis. 𝑏
𝐴𝑟𝑒𝑎 = ∫ 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑏) − 𝐹(𝑎) 𝑎
When the graph line is below the ‘𝑥’ axis, the definite integral is negative. 𝑏
𝐴𝑟𝑒𝑎 = − ∫ 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑎) − 𝐹(𝑏) 𝑎
Sometimes part of the graph is above the ‘x’ axis and part is below, then it is necessary to calculate several integrals. When the area of each part is found, the total area can be found by adding the parts: 𝑐
𝑏
𝐴𝑟𝑒𝑎 = ∫ 𝑓(𝑥)𝑑𝑥 − ∫ 𝑓(𝑥)𝑑𝑥 𝑎
𝑐
Property 3 in Topic 2.5 has particular application when the function is defined in different ways over subintervals 𝑏
𝑐
𝑏
𝑏
∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 + ∫ 𝑓(𝑥)𝑑𝑥 𝑎
𝑎
𝑐
∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 − ∫ 𝑓(𝑥)𝑑𝑥
𝑐
𝑎
𝑎
Another particular application when the area of a region is bounded by two graphs Let 𝑓(𝑥)and 𝑔(𝑥) be continuous functions and suppose that over the interval [𝑎, 𝑏]. Then the area of the region between the two curves, will be: 𝑏
𝑏
𝑏
∫ 𝑓(𝑥)𝑑𝑥 − 𝑔(𝑥)𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 − ∫ 𝑔(𝑥)𝑑𝑥 𝑎
𝑎
𝑎
Example: 1: Determine the area of the region bounded by 𝑦 = 𝑥 2 + 2 and 𝑦 = 2𝑥 + 5 First solve the system: (−1,3) 𝑎𝑛𝑑 (3,11); the integral limits are -1 and 3. Them, solve both integrals subtracting 𝑎 ∫𝑏 (𝑢𝑝𝑒𝑟
𝑎
𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛)𝑑𝑥 − ∫𝑏 (𝑙𝑜𝑤𝑒𝑟 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛)𝑑𝑥: 3
3
∫ (2𝑥 + 5) 𝑑𝑥 − ∫ (𝑥 2 + 2)𝑑𝑥 = 𝐴= 144
−1 3 ∫−1(−𝑥 2 +
2𝑥 + 3)𝑑𝑥 =
−1 32 3
𝑏
= 10.667 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠
𝑐
Juan J. Prieto-Valdés
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Example 2: Determine the area of the region bounded by 𝑓(𝑥) = 2𝑥 2 + 8, 𝑥 = −3,
𝑔(𝑥) = 4𝑥 + 14, and
𝑥=6
Recommendations: This example has to be solved manually; however, you can use Wolfram Alpha or Microsoft Word Mathematics (graph on the right), to plot the functions 𝑓(𝑥)𝑎𝑛𝑑 𝑔(𝑥), as well as the bounding lines located at 𝑥 = −3 𝑎𝑛𝑑 𝑥 = 6. In addition, you can use the software to solve the system and find the intercept points that separate the enclosed areas between the two functions. After you create the three integrals, you also can verify your numeric solution using any one of the recommended software. Proceed as follow: 1. Solve the system: 2𝑥 2 + 8 = 4𝑥 + 14 and determine intercepts: They are at (−1, 10) 𝑎𝑛𝑑 (3, 26) 2. Select the integration intervals: They are (−3 → −1), (−1 → 3), 𝑎𝑛𝑑 (3 → 6) 3. Write the integrals: −1
3
6
∫ [𝑓(𝑥) − 𝑔(𝑥)] 𝑑𝑥 + ∫ [𝑔(𝑥) − 𝑓(𝑥)]𝑑𝑥 + ∫ [𝑓(𝑥) − 𝑔(𝑥)] 𝑑𝑥 = −3
−1
−1
3
3
6
∫ [(2𝑥 2 + 8) − (4𝑥 + 14)] 𝑑𝑥 + ∫ [(4𝑥 + 14) − (2𝑥 2 + 8)]𝑑𝑥 + ∫ [(2𝑥 2 + 8) − (4𝑥 + 14)] 𝑑𝑥 = −3
−1
3
−1
3
6
∫ [2𝑥 2 − 4𝑥 − 6] 𝑑𝑥 − ∫ [2𝑥 2 − 4𝑥 − 6]𝑑𝑥 + ∫ [2𝑥 2 − 4𝑥 − 6] 𝑑𝑥 = −3
−1 3
3
2(
3
𝑥 𝑥 𝑥3 −1 3 6 − 𝑥 2 − 3 𝑥) │ − 2 ( − 𝑥 2 − 3 𝑥) │ + 2 ( − 𝑥 2 − 3 𝑥) │ = −3 −1 3 3 3 3 2(
(−1)3 (−3)3 − (−12 ) − 3 (−1)) − 2 ( − (−3)2 − 3 (−3)) 3 3
(3)3 (−1)3 −( − (3)2 − 3(3)) + 2 ( − (−1)2 − 3 (−1)) + 3 3 (6)3 (3)3 290 2( − (6)2 − 3 (6)) − 2 ( − (3)2 − 3(3)) = 3 3 3
Even Functions 𝒂
Another important property of the definite integrals is given by the Even and Odd characteristics of the functions. On the right can be observed this property.
146
𝒂
∫ 𝒇(𝒙)𝒅𝒙 = 𝟐 ∫ 𝒇(𝒙)𝒅𝒙 −𝒂
𝟎
Odd Functions 𝒂
∫ 𝒇(𝒙)𝒅𝒙 = 0 −𝒂
Juan J. Prieto-Valdés
To succeed in this topic, you should understand The definitions of Riemann Sum and the process of solving it when the limit of the length of the intervals towards zero (∆𝑥 → 0). You should understand how one shape (area) is subtracted from another and why the sign of the numeric value of the area can result negative while the reality is only positive. Also, pay attention to symmetry properties of integrals when calculating symmetric intervals of even and odd functions. Follow lecture explanation.
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Homework Topic 4.6 Find the area enclosed between the following two functions over the provided interval Step-by-step solutions at Prof. Prieto-Valdes Web-page 279)
𝑦1 = 𝑥 2 + 2𝑥 + 15 𝑎𝑛𝑑 𝑦2 = 3𝑥 + 5,
𝑎𝑡 [−2, 2]
2
∫−2(𝑥 2 + 2 𝑥 + 15 − 3 𝑥 − 5) 𝑑𝑥 ≈ 45.333
280)
𝑦1 = 𝑥 3 − 2𝑥 − 𝑥 + 2 𝑎𝑛𝑑 𝑌2 = 2𝑥,
𝑎𝑡 [1, 3]
2
∫ (2 𝑥 − 𝑥 3 + 2 𝑥 + 𝑥 − 2)𝑑𝑥 1 3
+ ∫ (𝑥 3 − 2 𝑥 − 𝑥 + 2 − 2 𝑥)) 𝑑𝑥 = 2
148
15 2
Juan J. Prieto-Valdés
281)
𝑦1 = |𝑥| 𝑎𝑛𝑑 𝑦2 = 𝑥 2 − 4, 𝑎𝑡 [−2, 2]
2
2 ∫ (4 + 𝑥 − 𝑥^2) 𝑑𝑥 = 44/3 ≈ 14.667 0
282) 𝑦1 =
𝑥2
2𝑥 − 4𝑥 − 12
𝑎𝑛𝑑 𝑦2 = 𝑥 2 − 2𝑥 𝑎𝑡[−1, 1] You can use Wolfram Alpha and study step-by-step solution.
283)
𝑦1 = 𝑥 4 − 2𝑥 2 𝑎𝑛𝑑 𝑦2 = 𝑥 3 𝑎𝑡 [−1, 2] You can use Wolfram Alpha and study the solution.
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PRACTICING QUIZ-5. Integrals and its applications 1. Calculate the area under the curve 𝑦 = 𝑥 2 on the interval [1, 3]. Consider only six subintervals with ends on the right. Show graph (Fig. 1). 2. Integrate: ∫(4𝑥 2 + 2𝑥 − 3)𝑑𝑥 2
3. Evaluate: ∫0 (𝑥 2 + 1)𝑑𝑥 −1
3
1
4. Evaluate: ∫−10 (𝑒 −𝑥 − 2𝑥) 𝑑𝑥 5. A car is moving straight way at 50 miles per hour (22.352 meter/sec). If the driver is force to apply the brakes 𝑚 and the car stops reducing its speed with an acceleration of 12 𝑠𝑒𝑐 2 how far does the car can travel before applying he brakes.
6. Find the function whose tangent has slope 2𝑥 2 + 1 for any value of 𝒙 and whose graph passes though the point (3, 25).
7. Find the area between 𝑥 3 − 3𝑥 + 2 and 𝑦 = 𝑥 + 4 (Fig.2)
8. Solve the Integral
∫
∫
1 1 (2 ln(𝑥 𝑥 ) + 3 ) 𝑑𝑥 𝑥2 𝑥 𝑥𝑑𝑥 √4𝑥 2 + 9
8 (ln(𝑥))2
∫
2
𝑥
𝑑𝑥 =
∫(𝑒 3𝑥 + 2𝑥 2 )𝑑𝑥
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Juan J. Prieto-Valdés
PROJECT – 4. Optimization of the Material Cost. Complete this Project using technology (Microsoft Mathematics, Wolfram Alpha or Wolfram APP) When completing this project, pay attention to the following: (1)- The optimization operation as a process of locating the Maximum or Minimum of the function (depending of maximization or minimization objectives) (2). How to differentiate real case functions and how to evaluate the optimum conditions numerically and graphically (in both cases: manually and using digital technology). (3). How to evaluate areas using analytic geometry and calculus and how to solve real case integrals.
Your answer has to satisfy the following criteria: Repeat all operations showing step by step hand written solution for each operation. Also, you can propose your own project (preferably for extra credit points)
The Problem: For storing 150,000 liters of toxic chemical substance is recommended a type of tank exhibited in Figure-1. Environmental regulations require that such type of tanks remain suspended not less than 2 meters above the ground. Our project is dedicated to the optimization of the material cost to build such type of tank. We need to build a tank with a minimum of material consumption to store a maximum volume, in this case 150000 liters.
Figure-1. View of the Tank The engineer constructor, based on his available technology, have decided to build the tank as per the scheme exhibited in Figure-2. Using two semi-spheres of diameter 𝑑 connected by a tube of length 𝑙 The cost of the material to conform the spherical part is approximately $250 per square meter (including the forming process), while the cost of the material for the tube is $150 per square meter (also including forming process). The cost of the legs’ material is 200 per square meter 0.25 m wide. This is a special isolating and antivibration composite material that can be molded before draying. Due to thermal and possible seismic conditions, the tank should rest on the legs through special reinforced pad which cost will not be calculated as it is provided (The pad is illustrated in Figure-3).
Figure-2. Simplified Construction Schema
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Analysis Based on previous information, the total cost of the material to construct the tank can be expressed as the sum of the following parts: 1. Cost of the material for the spherical part of the tank 𝐶𝑠𝑓 𝐶𝑠𝑓 = (𝑇𝑜𝑡𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 2 𝑠𝑒𝑚𝑖𝑠𝑝ℎ𝑒𝑟𝑒𝑠) × (𝑢𝑛𝑖𝑡 𝑐𝑜𝑠𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙) (Two semi-spheres are equal to one sphere) 𝐶𝑠𝑓 = (𝜋𝑑2 ) × (250)
(1)
2. Cost of the material for the cylindrical part of the tank 𝐶𝑐𝑦 𝐶𝑐𝑦 = (𝑇𝑜𝑡𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑠ℎ𝑒𝑒𝑡) × (𝑢𝑛𝑖𝑡 𝑐𝑜𝑠𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙) 𝑇𝑜𝑡𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑠ℎ𝑒𝑒𝑡 = (𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑛𝑘) × (𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑠ℎ𝑒𝑒𝑡) (2) 𝐶𝑐𝑦 = (𝑙 × 𝜋𝑑) × 150 Note that the length of the sheet is the length of the circumference formed when cutting the sphere in two identical parts (like the upper border of juice part of half-orange) Adding equations (1) and (2), the total cost will be: 𝐶𝑇𝑎𝑛𝑘 = 250 𝜋𝑑2 + 150 𝑙𝜋𝑑
(3)
Now, to express previous function in one variable we will use provided Volume Condition (150,000 liters), but first we have to write the volume of the Tank as per the next scheme:
𝑉𝑡𝑎𝑛𝑘 = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 + 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 2 𝑠𝑒𝑚𝑖𝑠𝑝ℎ𝑒𝑟𝑒𝑠 = = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 + 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑜𝑛𝑒 𝑠𝑝ℎ𝑒𝑟𝑒 𝑉𝑇𝑎𝑛𝑘 = (
𝜋𝑑2 𝑙 𝜋𝑑3 )+( ) = 150,000 𝑙𝑖𝑡𝑒𝑟𝑠 = 150 𝐾𝑖𝑙𝑜𝑙𝑖𝑡𝑒𝑟𝑠 4 6
(4)
Remember that we are working with the metric system, one cubic meter has a capacity of one thousand liters and 1000 𝑙𝑖𝑡𝑒𝑟𝑠 = 1 𝐾𝑖𝑙𝑜 𝑙𝑖𝑡𝑒𝑟 𝜋𝑑 2 𝑙 𝜋𝑑 3 ) + ( ) 4 6
(
= 150 𝐾𝑖𝑙𝑜 𝑙𝑖𝑡𝑒𝑟𝑠, Solving for 𝑙 ∶ 𝑙=
24(150) − 4𝜋𝑑3 6𝜋𝑑2
(5)
Now we can express the Total Cost of the material for the Tank in one variable as follow: 𝐶𝑇𝑎𝑛𝑘
24(150) − 4𝜋𝑑3 = 250 𝜋𝑑 + 150 ( ) 𝜋𝑑 6𝜋𝑑2 2
Simplifying: 𝐶𝑇𝑎𝑛𝑘
25 (3600 − 4 𝜋 𝑑3 ) = + 250 𝜋 𝑑2 𝑑
Differentiating and making it equal to zero to optimize: 152
(6)
Juan J. Prieto-Valdés
𝜕𝐶𝑇𝑎𝑛𝑘 300 (𝜋 𝑑 3 − 300) = =0 𝜕𝑥 𝑑2 Final Solution: 𝜋 𝑑3 − 300 = 0 3
𝑑= √
300 = ≅ 4.57 𝑚𝑒𝑡𝑒𝑟𝑠 3.1416
In this case the length of the cylindrical part of the Tank as per the equation (5), will be:
𝑙=
24(150) − 4𝜋(4.57)3 = 6.09802 … 6𝜋(4.57)2
First Conclusion: The optimal diameter will be 4.57 meter. Replacing optimal diameter value in equation (6), we can get the total cost of the special materials to store aggressive chemical: 𝐶𝑇𝑎𝑛𝑘 =
25 (3600 − 4 𝜋 (4.57)3 ) + 250 𝜋 (4.57)2 ≈ $29,535 (4.57)
For verification purposes, we can graph the Cost Function using Microsoft Mathematics (Figure 4). Observe that the result is the same: 𝑦 =
25 (3600−4 𝜋 𝑥 3 ) 𝑥
+ 250 𝜋 𝑥 2
Figure 4. Cost function versus the diameter of the Tank and its loupe magnification
Now let us work with the legs: As per structural engineer, the most convenient type of legs is represented in Figure 3. The tank lies on the legs through special solid pad located between circular and parabolic shapes to ensure a fixed position. Using this leg-form a high material economy can be reached without scarifying the strength of the legs/material. On the left of Figure 3, the expected shape in (𝑥, 𝑦) coordinate system while on the right the exact position of the border functions is provided to develop the equations, required to further calculate the area (in our case using integrals). The cost of the 5 legs will be: 𝐶𝑙 = 5(𝐴𝑟𝑒𝑎 𝑜𝑓 𝑜𝑛𝑒 𝑙𝑒𝑔) × (𝑢𝑛𝑖𝑡 𝑐𝑜𝑠𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙) 153
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Figure-3. Shape of the legs To calculate the area of the shape bordered by the parabola (open up) and the quadratic function (open down) we will 𝑑
use the following integral (due to even symmetry we will use 0 𝑎𝑛𝑑 2 as lower and upper limits of the integral (five legs 2 times the symmetric integral): 𝑑 2
(7)
𝐴𝑙 = 5 × 2 ∫ [𝑓𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑎 (𝑥) − 𝑓𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 (𝑥)]𝑑𝑥 0
𝑑
The equation of the circle with center at (ℎ, ℎ) = (0, ( 2 + 2)) , and radius 𝑟 =
𝑑 2
in the standard form can be
written as: 2
𝑑 𝑑 2 𝑥 + (𝑦 − ( + 2)) = ( ) , 2 2 2
at 𝑑 = 4.57 will be: 2
4.57 4.57 2 𝑥 + (𝑦 − ( + 2)) = ( ) 2 2 2
Simplifying: 𝑥 2 + (𝑦 − 4.285)2 = 5.22123 ≅ (2.28)2 , (8)
𝒙𝟐 + (𝒚 − 𝟒. 𝟐𝟖𝟓)𝟐 = 𝟓. 𝟐𝟐𝟏𝟐𝟑
The equation of the parabola open up can be written using the reflection of the parabola opens down with 𝑑
𝑑
𝑦𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = (0, 1.75), 𝑎𝑛𝑑 𝑥𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 ≡ (− 2 , 0) 𝑎𝑛𝑑 (+ 𝑠 , 0): 𝑑 𝑑 𝑦 = 𝑎 (𝑥 − ) (𝑥 + ), 2 2 𝑑 4.57 𝐹𝑜𝑟 𝑦 = 1.75 𝑚 𝑎𝑡 𝑥 = 0 , 𝑎𝑛𝑑 = , 𝑤𝑒 𝑔𝑒𝑡: 2 2 Finally, the equation of the parabola can be written as:
𝑎 = −0.192367
𝑦 = −0.192367 (𝑥 − 2.28)(𝑥 + 2.28) = 𝑦 = −0.240459 𝑥 2 + 1.75
(9)
Its reflection will be: 𝒚 = 𝟎. 𝟐𝟒𝟎𝟒𝟓𝟗 𝒙𝟐 + 𝟏. 𝟕𝟓 154
(10)
Juan J. Prieto-Valdés
For verification purposes on the right is exhibited the graph of the circle and both parabolas 𝑥 2 + (𝑦 − 4.285)2 = 5.22123 equation (8) y = −0.336642 𝑥 2 + 1.75 equation (9) y = 0.336642 𝑥 2 + 1.75 equation (10) Remember that we will not use the parabola opens down. For the inferior shape of the legs we will develop a more comfortable shape as recommended by the structural engineer. The equation of the quadratic function (as per expected shape in Figure-3) can be written using the (𝑥, 𝑦) intercepts: 𝑦𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = (0, 1.5); 𝑎𝑛𝑑 𝑥𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 = (0. 0), (0. 0), (−1.5, 0) 𝑎𝑛𝑑 (1.5, 0) On the first approximation, our equation can be written using 𝑎 = 1, in this case: 𝒚 = 𝒂𝒙𝟐 (𝒙 − 𝟏. 𝟓)(𝒙 + 𝟏. 𝟓) = −𝒙𝟒 + 𝟐. 𝟐𝟓𝒙𝟐
(11)
One more time, for verification purposes let’s see the graph of the borders of the legs together: y = 0.336642 𝑥 2 + 1.75 (𝑈𝑝𝑒𝑟 𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑎) 𝑦 = −𝑥 4 + 2.25𝑥 2 (𝐿𝑜𝑤𝑒𝑟 𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛) 𝑥 = 2.28 (𝑅𝑖𝑔ℎ𝑡 𝑤𝑎𝑙𝑙 𝑠𝑖𝑑𝑒) 𝑥 = −2.28 (𝐿𝑒𝑓𝑡 𝑤𝑎𝑙𝑙 𝑠𝑖𝑑𝑒)
Now we can substitute functions (10) and (11) into the integral (7) to calculate the total area of the shape of the legs: Right wall 2.28
𝐴𝑙 = 10 ∫
((0.336642 𝑥 2 + 1.75) − (−𝑥 4 + 2.25𝑥 2 )) 𝑑𝑥 =
0 2.28
= 10 ∫
(0.336642 𝑥 2 + 1.75 + 𝑥 4 − 2.25𝑥 2 ) 𝑑𝑥 = 8.75339 𝑚2
0
Finally, the cost of the five legs will be 5 × 8.75 × 200.00 = $8,750.00 The total cost of the material will be 𝑐𝑜𝑠𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑇𝑎𝑛𝑘 + 𝑐𝑜𝑠𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑒𝑔𝑠 ′ 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝐶𝑇𝑎𝑛𝑘+𝐿𝑒𝑔𝑠 = 29,535 + 8,750 = $38,285.00 Which is a good price for such type of Tank.
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PROJECT – 5. Profit Optimization (Transport Marine Company) This is a classic example, multiple different solutions can be found in public domain. The author of this book did not want to include a solutions from other authors. The student can review them on the internet or simply try to obtain their own solution with the assistance of his instructor. The "gas mileage'' of the ship depends of its weight and displacement speed. Suppose that you manage a company that delivers goods between two ports using small ships. You want to maximize profits by choosing the best shipping speed. Sailing faster increases the number of goods you can deliver per week (and thus, your revenue) but also increases your fuel costs because of the lower efficiency of the ship at higher speeds. Equivalently, increasing the number of transporting goods also increases your revenue but reduces fuel efficiency. This project requires to create a model to calculate the optimal speed and number of transporting goods to maximize the profit. Create fictitious data to prepare a Business Calculus Project. Optimize the ship speed and number of goods to maximize the Profit. Be creative by considering different situations. You can search on internet some information about gas consumption of ships to approximate your database to reality. Recommendation: Use the following parameters to build your math model. s = speed of the ship (miles/hour) d = distance of the delivery route (miles) w = weight of the ship plus its cargo (lbs.) f = fuel consumption of the ship, as a function of s and w (gallons/miles) R = revenue earned per delivery (US $) C= cost per gallon of diesel fuel ($/gallon) + cost per cargo handling ($/lbs.) + salary of workers ($/hours) P= average profit earned per traveling. Will be a function of s, d, r, f, w and e ($/hours)
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PROJECT – 6. Optimization of fictitious city evacuation. This project has to be developed by yourself with the assistance of your instructor. Optimize the evacuation process of the represented population in the Figure. The evacuation happens through three different exits, two of them have two parallel paths, and one has three paths. The maximum speed in each rout is indicated. Determine the number of elements to follow each path to diminish the traffic jam. How the elements have to be
directed to each exit and from which fictitious marked zones to faster reach safe territory? This problem has some similarity with Torricelli exercise to drain a water tank. By applying Pascal/Torricelli formulation and Flow/Work equations, this problem can be solved. The atmospheric pressure and gravitation effect has to be diminished. Such effect will be given by the awareness/enthusiasm of the evacuated people. Use some awareness coefficient when creating your math model.
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CHAPTER 5. INTRODUCTORY STATISTICS WITH CALCULUS. 5.1. Continuous Random Variable, Probability, and Histograms. Random Variable: A random variable is a function X that assigns to each possible outcome in an experiment a real number. If X may assume any value in some given interval I (the interval may be bounded or unbounded), it is called a continuous random variable. If it can assume only a number of separated values, it is called a discrete random variable. Examples: 1. Roll a die and take X to be the number on the uppermost face. Then X is a discrete random variable with possible values 1, 2, 3, 4, 5, and 6. 2. Open the business stock website and take X to be the closing price of Specific Company Stock. Then X can take any positive real value, so we can think of X as a continuous random variable. Probability: In Example-1, if the die is fairly, the probability will be 1/6: one outcome at time from 6 equally probable outcomes. If X is a random variable, the probability that X takes on a value in a certain range has to be determined by it appearance or success. The probability of each number appearance will be 1/6. In Example 2, suppose that 50% of the time the closing price X price fluctuates between $20 and $25. We can say that 0.5 is the probability between 20 and 25. We can write this statement mathematically as follows: P(20≤ X ≤25) = 0.5 Histogram: A bar chart representing the Probability of X in selected ranges is called the histogram.
5.2. Probability Density Functions: Uniform, Exponential, and Normal. The histograms are a convenient way to visualize the probability distribution associated with a continuous random variable X. Using a subdivisions of some specific unit, the probability P(c ≤ X ≤d) is given by the area under the histogram between X=c and X=d. However, it is not easy to calculate probabilities for ranges of X that are not a whole number of units. The following example shows the main idea of histogram construction. EXAMPLE-1: A survey finds the following probability distribution for the number of houses sold in some city in one year: Number of houses sold (𝒙)
0-1
1-2
2-3
3-4
4-5
5-6
6-7
Probability 𝒇(𝒙)
0.18
0.30
0.20
0.15
0.10
0.05
0.02
Using Excel DATA ANALYSIS TOOL, we can get a quick view of all statistical characteristics of our sample of study: Follow the next directions: 1.
158
Go to the DATA table and click on the DATA ANALYSIS tool:
Juan J. Prieto-Valdés
2.
3.
If we want a quick overview of the SCORE variable, we can use the DESCRIPTIVE STATISTICS tool. From the list of tools, choose DESCRIPTIVE STATISTICS:
Check SUMMARY STATISTICS:
4.
Get the results:
Some of these variables can be required for specific operations, For example, if creating a regression, you would need some of these parameter to be fairly close to the real result, or for the analysis of your data:
5.
Using Excel, we can also create the histogram (next Figure). The shape of the probability function can be simulate manually, as a possible probability function, something like the displayed curve (right side on the next figure). We will call this function a Probability density function. The Domain of 𝑓(𝑥) has to be [0, ∞), because these are the possible values of 𝑋 (random variable). Note that we use 𝑥 to refer possible values of 𝑋 .
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Histogram
Histogram and possibkle probability function
If we would like to calculate the probability of selling between 0 and 3 houses, we will write: 𝑃(0 ≤ 𝑋 ≤ 3) = 0.18 + 0.30 + 0.2 = 0.68 Which from the point of view of calculus can be written as: 3
𝑃(0 ≤ 𝑋 ≤ 3) = ∫ 𝑓(𝑥) 𝑑𝑥 = 0.68 0
Next figure shows the area (in black color) which represent the probability of selling between 0 and 3 houses.
Similar calculation can be done for any other interval, including not whole numbers by supposing that we can sell half house, which would be correctly for other type of random variables, for example the number of McDonalds that people can eat at time; one can eat 1, another 2, 3, or 3 and half. 3.5
𝑃(1 ≤ 𝑋 ≤ 3.5) = ∫ 𝑓(𝑥) 𝑑𝑥 1
Although we haven’t define a formula for 𝑓(𝑥) we know based in our knowledge of introductory statistics that total probability of all events will be equal to one. +∞
𝑃(0 ≤ 𝑋 ≤ +∞) = ∫ 𝑓(𝑥) 𝑑𝑥 = 1 0
Now, Using Wolfram Alpha, we can create a regression to approach the shape of f(x). However, such regression do not has statistical meaning because we don’t know the real physical meaning of the coefficients. We will study later appropriate statistical functions. The next regression serves only for illustration purposes about the total area.
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If considering the previous four-degree polynomial as a possible approximation, we can see that the total probability will be approximately equal to one. 7
𝑦 = ∫ ( − 0.00132576 𝑥 4 + 0.024596 𝑥 3 − 0.153428 𝑥 2 + 0.317293 𝑥 + 0.0644464)𝑑𝑥 = 0.99 0
This was calculated only for illustration purposes, the function does not provides statistical information despite that the shape of the probability function may be like this one, but we can observe that the function is defined in an 𝑏
interval (𝑎, 𝑏) where the probability function (area under the curve) is equal to one: 𝑓(𝑥) ≥ 0, and ∫𝑎 𝑓(𝑥)𝑑𝑥 = 1.
5.3. Uniform Density Function Uniform density function is a constant function: 𝑓(𝑥) = 𝑘. In such case 𝑏
𝑏
∫ 𝑓(𝑥)𝑑𝑥 = 1 = ∫ 𝑘𝑑𝑥 = 𝑘(𝑏 − 𝑎) 𝑎
𝑎
Thus, we have: 1 𝑏−𝑎 In other words, the uniform density function on the interval [𝑎, 𝑏] is given by: 𝑘=
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𝑓(𝑥) =
1 𝑏−𝑎
EXAMPLE-2: An example of uniform probability is rolling one dice. The probability of getting any number from 1 1
1
1
to 6 is the same: 1𝑜𝑢𝑡𝑐𝑜𝑚𝑒 𝑓𝑟𝑜𝑚 6, 𝑘 = 6 = 1.666, while on the histogram 𝑘 = 6−0 = 6, as the [𝑎, 𝑏] exist from 0 to 6.
1
2
3
4
5
6
EXAMPLE-3: Suppose a random number N is taken from 670 to 850 in uniform distribution (for example: ID number of the participants in a group conference of 180 people). Find the probability number N is greater than the 790? The interval number in probability distribution is = [670, 850]. Thus, the Density of Probability will be: =
1 850−670
=
1 180
We will calculate the probability 𝑃(790 < 𝑥 ≤ 850) of successful events considering first the interval of probability of successful events: = [790, 850] = 60 60
1
So, the probability ratio will be = 180 = 3 . 1 3
The probability of N>790 is = = 0.33 … 1 )will 180
We can get the same result using calculus; the integral of the probability uniform function 𝑓(𝑥) = ( 850
∫ 790
be:
1 𝑥 850 850 790 1 𝑑𝑥 = │ = − = = 0.3333 180 180 180 180 3 790
5.4. Exponential Density Functions A continuous random variable X is said to have an Exponential(λ) distribution if it has probability density function with λ > 0 rate of the distribution. 𝑓𝑋 (𝑥) = {
𝜆𝑒 −𝜆𝑥 𝑓𝑜𝑟 𝑥 > 0 0 𝑓𝑜𝑟 𝑥 ≤ 0
The exponential distribution is usually used to model the time until something happens in the process. The mean of the Exponential (λ) distribution is calculated using integration: ∞
∫ 𝜆𝑒 0
∞ −𝜆𝑥
𝑑𝑥 = 𝜆 ∫ 𝑒 0
𝜆(−𝑥)
∞ ∞ {𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑑𝑥 , 𝑢 = 𝜆(−𝑥)} = − ∫ 𝑒 𝑢 𝑑𝑢 = −𝑒 𝑢 = −𝑒 𝜆(−𝑥) ] = 1 0 0
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EXAMPLE-4: Suppose that the amount of mosquitoes is falling 2.5% per day after implementing an ecological fumigation system in a small island on Caribbean Sea (2.5% is calculated from remaining amount each day; it is the combined result of reproduction and mortality). What is the probability of extinguishing the mosquitoes in the 𝑇 next days? To answer this question we consider 100% the initial amount of mosquitoes. Using decay representation of the population of mosquitoes (failing continuously at a rate of 2.5% per day), the population left after T days is given by the following equation: 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑙𝑒𝑓𝑡 = 100𝑒 −0.025𝑇 𝑇ℎ𝑒 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑡ℎ𝑎𝑡 𝑓𝑎𝑖𝑙 = [𝑇𝑜𝑡𝑎𝑙 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 − 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑙𝑒𝑓𝑡] = 100 − 100𝑒 −0.025𝑇 = 100(1 − 𝑒 −0.025𝑇 ) Thus, the percentage that will have failed by the time, which will be the probability that we are looking for: 100(1 − 𝑒 −0.025𝑇 ) = 1 − 𝑒 −0.025𝑇 100 Now, let X be the number of days a randomly chosen will take to extinguish the mosquitoes, 𝑃=
𝑃(0 ≤ 𝑋 ≤ 𝑇) = 1 − 𝑒 −0.025𝑇 𝑏
This result also can be obtained applying calculus by calculating the probability integral ∫𝑎 𝜆𝑒 −𝜆𝑥 𝑑𝑥: 𝑇
𝑇 ∫ 0.025 𝑒 −0.025𝑇 𝑑𝑥 = 𝑒 −0.025𝑇 │ = 1 − 𝑒 −0.025𝑇 0 0 Thus, we use 𝑓(𝑥) = 0.025 𝑒 −0.025𝑇 as a probability density function to model previous situation. 𝑇
𝑃(0 ≤ 𝑋 ≤ 𝑇) = ∫ 0.025 𝑒 −0.025𝑇 𝑑𝑇 0
Question: Find the probability of mosquitoes extinction during 5 and 6 days from the first day of experiment: 6
6
𝑃(5 ≤ 𝑋 ≤ 6) = ∫ 0.025 𝑒 −0.025𝑇 𝑑𝑇 = ∫ 5
5
{𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑢 = − which give us new bound: upper bound 𝑢 = −
6 40
=
𝑇 , 40
3 20
1 −𝑇 𝑒 40 𝑑𝑇, 40
𝑎𝑛𝑑 𝑑𝑢 = −
1 𝑑𝑇} 40
and lower bound 𝑢 = −
5 50
= −
1 8
Our integral becomes: −
−∫ −
3 20
1 8
3 − 20 1 − 8
𝑒 𝑢 𝑑𝑢 = − 𝑒 𝑢 │
=−
1 3 𝑒 20
+
1 1 𝑒8
= 0.0218
Mathematically, 2% of mosquitoes will be extinguished during the 5 and 6 days after beginning the experiment. The probability of extinction after 6 days can be expressed as: ∞
𝐿
𝑃(6 ≤ 𝑋 ≤ ∞) = ∫ 0.025 𝑒 −0.025𝑇 𝑑𝑇 = lim ∫
𝐿→∞ 6
6 6
1 −𝑇 𝑒 40 𝑑𝑇 40
3
Using same procedure, the new lower bound will be: 𝑢 = − 40 = − 20 1 −∞ −𝑢 1 1 1 ∫ 𝑒 𝑑𝑢 = − 𝑒 𝑢 │∞ 3 = − ∞ + 3 = 3 = 0.86 3 − 40 − 𝑒 20 𝑒 20 𝑒 20 20 Means that 86% of mosquitoes will be extinguished after 6 days, while from 0 to 6 days will be only 14 %: 𝑃(6 ≤ 𝑋 ≤ ∞) =
−
−∫ 0
164
3 20
3
𝑒 𝑢 𝑑𝑢 = 1 − 1/𝑒 20 ≈ 0.13929.
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The full extinction can be estimated from the graph or evaluating the function: 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑙𝑒𝑓𝑡 = 100𝑒 −0.025𝑇 Let’s consider that full extinction can be acknowledged when only one mosquito remain (not possible reproduction). We will not use zero (0) mosquitoes because the distribution equation gets undefined when the 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 = 0. Solving for 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 = 1 = 100 𝑒 −0.025𝑇 (Use software to solve the equation, or simple checking the graph) We get 𝑇 = 184.2 𝑑𝑎𝑦𝑠 to full extinction Now, considering the Probability of having mosquitoes after ∞
185 days as: 𝑃(185 ≤ 𝑋 ≤ ∞) = ∫185 0.025 𝑒 −0.025𝑇 𝑑𝑇 and using the same previous procedure ∞
𝑃(185 ≤ 𝑋 ≤ ∞) = − ∫ 𝑒 𝑢 𝑑𝑢 = − 𝑒 𝑢 │∞185 = 185 40
= −𝑒 ∞ +
−
1 185 𝑒 40
=
1 𝑒 185/40
40
= 0.0098
The probability of finding a mosquito is 0.9%; practically NO MORE mosquitoes. The probability distribution 𝑃(𝑋) = 0.025 𝑒 −0.025𝑇 is shown in the next graph. Note that after 185 days the remaining area under the function is practically zero, or 0.9%, one mosquito.
5.5. Normal Distribution A normal (or Gaussian or Gauss or Laplace–Gauss) distribution in a variate 𝑋 with mean 𝜇 and variance 𝜎 2 is a statistic distribution with probability density function 𝑃(𝑋) =
1 𝜎√2𝜋
𝑒
−
(𝑥−𝜇)2 2𝜎 2
Properties of the Normal Distribution Curve: 1. 2. 3. 4. 5. 6. 7. 8.
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It’s Domain: (−∞, +∞) It is “bell-shaped” with the peak occurring at x = μ. It is symmetric about the vertical line x = μ. It is concave down in the range μ − σ ≤ x ≤ μ + σ . It is concave up outside that range, with inflection points at 𝑥 = 𝜇 – 𝜎 and 𝑥 = 𝜇 + 𝜎. 68.2% of the values lie within these two values (μ - σ) and ( μ + σ). 95.4% of the values lie within 2 standard deviations of the mean, between (μ −2σ) and (μ +2σ). The Normal Density Function is expressed as: (𝑥−𝜇)2 1 − 𝑃(𝑋) = 𝑒 2𝜎2 𝜎√2𝜋
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The normal density function applies in many situations that involve measurement and testing, it is wide used in quality control; practically all process of life can be described using normal distribution. For instance, average weight of NBA players, Average female shoe size, Population blood Pressure, birthweight of newborn babies, repeated imprecise measurements of determinate characteristics of a material, etc. The universality of normal distribution make it convenient in assessing the results of standardized tests (previous examples are classically sited by the majority of authors, but you can represent any other data or process). In order to use the normal density function to compute probabilities, we need to calculate integrals of the form of 𝑏
∫𝑎 𝑓(𝑥)𝑑𝑥. However, the antiderivative of the normal density function cannot be expressed in terms of any commonly used functions, must be Gaussian distribution. Our previously four degree polynomial used to simulate the shape of the probability of selling houses in EXAMPLE-1 does not provide statistical description, it was use only to simulate the shape and to verify the value of the area under this curve. The traditional statistical way to calculate normal probabilities is using tables. Commonly published tables (named zscore Tables) are for the standard normal distribution, with mean 𝜇 = 0 and standard deviation 𝜎 = 1. If 𝑋 is a normal variable with mean 𝜇 and standard deviation 𝜎, the standard normal variable Z, named z-score, will be expressed as 𝑍 =
𝑋−𝜇 𝜎
, which represent the distance from the mean to the random variable value in the
normalized density curve. In this book, we will not use z-score Tables, as we are interested in calculus-statistics link and technology utilization. In addition, Tables are obsoletes, the can be generated using excel, for example, assuming that we have two z-sores equal to: 𝑍𝑙𝑜𝑤𝑒𝑟 = −0.96 𝑎𝑛𝑑 𝑍𝑢𝑝𝑝𝑒𝑟 = 0.96 In such situation, the total probability as displayed on the right Excel screen shot will be 66.29%. No integral calculation is required to find the area under the density curve between corresponding random variables: 𝑋𝑙𝑜𝑤𝑒𝑟 𝑎𝑛𝑑 𝑋𝑢𝑝𝑝𝑒𝑟 . Example-5: Elba Co. produces 24-volt batteries for special applications. If the battery voltage differs in 1% of its nomination, it is rejected. Assuming that battery voltage is a normal random variable with a mean 24 and standard deviation 0.25 volts, find the percentage of batteries rejected (0.24 is one percent of 24). For the battery to be accepted is required (24 − 0.24) ≤ 𝑋 ≤ (24 + 0.24). Thus, the probability that the batteries will be accepted can be represented as 𝑃(23.76 ≤ 𝑋 ≤ 24.24), where 𝑋 is a normal random variable, with 𝜇 = 24 𝑎𝑛𝑑 𝜎 = 0.25. 24.24
𝑃(23.76 ≤ 𝑋 ≤ 24.24) = ∫ 23.76
1 0.25√2(𝜋)
𝑒
−
(𝑥−24)2 2(0.25)2
𝑑𝑥
The step by step mannual solution method of the previous integral is rapidly becoming obsolete as the technology of spreadsheets, handheld computers, and programmable calculators can solve integration problems quickly and accurately. Using Wolfram Alfa application, we can get fast solution (on the right): 24.24
∫ 23.76
1 0.25√2(𝜋)
𝑒
−
(𝑥−24)2 2(0.25)2
𝑑𝑥 = 0.6629
In other words, 66.29% of the batteries will be accepted. Follow the traditional statics method we can calculate the z score for 𝑥 = 24.24 𝑎𝑛𝑑 𝑥 = 23.76 168
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𝑋𝑢𝑝𝑝𝑒𝑟 − 𝜇 24.24 − 24 𝑋𝑙𝑜𝑤𝑒𝑟 − 𝜇 23.76 − 24 = = −0.96 𝑎𝑛𝑑 𝑍𝑢𝑝𝑒𝑟 = = = 0.96 𝜎 0.25 𝜎 0.25 The next graph illustrate the result. Note that shaded area represent 66.29 % as per previously Excel calculation using the command 𝑍𝑙𝑜𝑤𝑒𝑟 =
= NORMSDIST(𝑑𝑎𝑡𝑎 1) − NORMSDIST(𝑑𝑎𝑡𝑎 2):
Example-6: Follow previous example, using the same assumptions; find the percentage of batteries rejected if the battery voltage differs in 2% of its nomination. For the battery to be accepted is required (24 − 0.48) ≤ 𝑋 ≤ (24 + 0.48) as 2% of 24 is 0.48. Thus, the probability that the batteries will be accepted can be represented as 𝑃(23.52 ≤ 𝑋 ≤ 24.48), where 𝑋 is a normal random variable 𝜇 = 24 𝑎𝑛𝑑 𝜎 = 0.48. 24.48
𝑃(23.52 ≤ 𝑋 ≤ 24.48) = ∫ 23.52
1 0.25√2(𝜋)
𝑒
−
(𝑥−24)2 2(0.25)2
𝑑𝑥 = 0.945
In other words, 94.5% of the batteries will be accepted.
5.6. Mean and Median. Let’s review from elemental statistics to calculus application follow the next example: Example 7: Roll one die. Let’s represent the set of possible outcomes as 𝑋 = {1, 2, 3, 4, 5, 6}. So 𝑋 will be the value on the die. 1
𝑋(𝑥) = 𝑥. If the die is fair, then the probability model has 𝑃{𝑥} = 6 for each outcome 𝑥 and the expected value will be 1 + 2 + 3 + 4 + 5 + 6 21 7 = = 6 6 2 Based on elemental statistical theory the expected value also can be calculate by adding the products of the (𝑥𝑖 ) by its probability 𝑃(𝑥𝑖 ): 𝜇=
𝐸(𝑋) = 1 · 𝑃{1} + 2 · 𝑃{2} + 3 · 𝑃{3} + 4 · 𝑃{4} + 5 · 𝑃{5} + 6 · 𝑃{6} = 1 1 1 1 1 1 21 7 𝐸(𝑋) = 1 ( ) + 2 ( ) + 3 ( ) + 4 ( ) + 5 ( ) + 6 ( ) = = 6 6 6 6 6 6 6 2 For discrete case: 𝐸(𝑋) = ∑ (𝑥)(𝑃{𝑥}) = 𝜇𝑥 𝑎𝑙𝑙 𝑥
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For continuous case: If 𝑋 is a continuous random variable with probability density function 𝑓(𝑥) defined on an interval with endpoints a and b (can be infinite number of values), then the mean or expected value of 𝑋 is: ∞
𝐸(𝑋) = ∫ 𝑥 ∙ 𝑓(𝑥)𝑑𝑥 = 𝜇𝑥 −∞
Example 8: Revisiting Example-4. The probability of mosquitoes’ extinction was 𝑇
defined as 𝑃(0 ≤ 𝑋 ≤ 𝑇) = ∫0 0.025 𝑒 −0.025𝑇 𝑑𝑇 Therefore, the Expected Life span of mosquitoes will be: 𝑏
∞
𝐸(𝑋) = ∫ 𝑥𝑓(𝑥 )𝑑𝑥 = ∫ 0.025 𝑥 𝑒 −0.025𝑥 𝑑𝑥 𝑎
0
This integral can be solved by-parts, by substitution method, or using technology (wolfram Alpha solution is shown on the right Figure). For our level in calculus, next table shows the solution by substitution: ∞
−∞
∫ 𝜆 𝑥 𝑒 −𝜆𝑥 𝑑𝑥 = ∫ 0
0
Use substitution:
𝑢 𝑢 𝑑𝑢 1 −∞ 𝑢 𝜆 (− ) 𝑒 (− ) = ∫ 𝑢𝑒 𝑑𝑢 𝜆 𝜆 𝜆 0
𝑢 = −𝜆𝑥, 𝑑𝑢 = −𝜆𝑑𝑥, 𝑢 𝜆
𝑥 = − , 𝑑𝑥 = −
1 −∞ 𝑢 1 −∞ ∫ 𝑢𝑒 𝑑𝑢 = ∫ (𝑢𝑒 𝑢 + 𝑒 𝑢 − 𝑒 𝑢 )𝑑𝑢 𝜆 0 𝜆 0
Add and subtract 𝑒 𝑢
−∞ −∞ −∞ 1 𝑑 1 −∞ (𝑢𝑒 𝑢 )𝑑𝑢 − ∫ 𝑒 𝑢 𝑑𝑢) = (𝑢𝑒 𝑢 ] − ∫ 𝑒 𝑢 𝑑𝑢) (∫ 0 𝜆 0 𝑑𝑢 𝜆 0 0
Apply product rule in inverse way:
1 1 −∞ (( lim 𝑢𝑒 𝑢 − (0)𝑒 0 ) − 𝑒 𝑢 ] ) = (0 − 0 − (𝑒 −∞ − 𝑒 0 )) = 0 𝜆 𝑢→−∞ 𝜆
𝑢→−∞
1 1 (0 − 0 − (0 − 1)) = 𝜆 𝜆
𝑜𝑟
1 = 40 0.025
𝑑 (𝑢𝑒 𝑢 ) 𝑑𝑢
𝑏
∞
𝐸(𝑋) = ∫ 𝑥𝑓(𝑥 )𝑑𝑥 = ∫ 0.025 𝑥 𝑒 −0.025𝑥 𝑑𝑥 = 𝑎
Mathematically speaking 𝜇 = 40, which in our example represent the expected life span of mosquitoes. If this is TRUE, the Normal Density distribution function 𝜆 𝑥 𝑒 −𝜆𝑥 has a maximum in 𝑥 = 40. The graph of 𝑓(𝑥) = 0.025 𝑥 𝑒 −0.025𝑥 is shown on the right matching exactly our result.
0
= 𝑢 ∙ 𝑒𝑢 + 𝑒𝑢
lim 𝑢𝑒 𝑢 𝑢 {𝑎𝑝𝑝𝑙𝑦 𝐿′ 𝐻𝑜𝑠𝑝𝑖𝑡𝑎𝑙} = lim 𝑢→−∞ 1 𝑒𝑢 = lim − 𝑒 𝑢 = 0 𝑢→−∞
In our example:
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𝑑𝑢 𝜆
1 = 40 0.025
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In concordance with previous solution in general form for the integral equal to 𝜆, the first derivative test (looking for maximum of the function) should have the same value (Maximum position): 𝑑 (𝜆 𝑥 𝑒 −𝜆𝑥 ) = 𝜆 (−𝑥𝜆 𝑒 −𝜆𝑥 + 𝑒 − 𝜆 𝑥 ) = 𝜆𝑒 −𝜆𝑥 (−𝜆𝑥 + 1) 𝑑𝑥 𝑑𝑓 1 𝐴𝑠 𝑝𝑒𝑟 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑡𝑒𝑠𝑡 = 0, − 𝜆𝑥 + 1 = 0, 𝑥 = 𝑑𝑥 𝜆 1 𝜆
In our Examples 4 and 8:
=
1 0.025
= 40
The value of the area under the graph from 0 to ∞ gets the same value as the mean (𝜇𝑥 = 𝑋𝑚𝑎𝑥𝑖𝑚𝑢𝑚 ) on the distribution graph. This is an interesting property of data.
5.7. Variance and Standard Deviation Let 𝑥𝑖 be a random variable and 𝜇 the mean; using traditional notation, the standard deviation 𝜎 2 can be written as: 𝜎2 = Let’s analyze the numerator: 𝑁
∑(𝑥𝑖2 𝑖=1
𝑁 2
− 2𝑥𝑖 𝜇 + 𝜇 ) =
𝑁
∑ 𝑥𝑖2 𝑖=1
2 ∑𝑁 𝑖=1(𝑥𝑖 − 𝜇) 𝑁 𝑁
𝑁 2
− ∑ 2𝑥𝑖 𝜇 + ∑ 𝜇 = 𝑖=1
𝑖=1
∑ 𝑥𝑖2 𝑖=1
𝑁
𝑁 2
− 2𝜇 ∑ 𝑥𝑖 + 𝜇 ∑ 1 𝑖=1
𝑖=1
Dividing by 𝑁, we get: 2
2 𝟐 ∑𝑁 ∑𝑵 2𝜇 ∑𝑁 𝜇 2 ∑𝑁 𝟐𝝁 ∑𝑵 𝝁𝟐 ∑𝑵 𝜇2 𝑁 (∑𝑁 𝑖=1 𝑥𝑖 𝑖=1 𝑥𝑖 𝑖=1 1 𝒊=𝟏 𝒙𝒊 𝒊=𝟏 𝒙𝒊 𝒊=𝟏 𝟏 𝑖=1 𝑥𝑖 ) − + = − + = − 2𝜇𝜇 + = 𝑁 𝑁 𝑁 𝑵 𝑵 𝑵 𝑁 𝑁 2
(∑𝑁 𝑖=1 𝑥𝑖 ) 𝜎 = − 𝜇2 𝑁 2
Later we will use this formula; note that ∑𝑁 ∑𝑁 𝑁 𝑖=1 𝑥𝑖 𝑖=1 1 = 𝜇, 𝑎𝑛𝑑 = = 1. 𝑁 𝑁 𝑁 In previous formula 𝜎 2 =
2 ∑𝑁 𝑖=1(𝑥𝑖 −𝜇)
𝑁
, 𝑁 represent how frequently each number occurs, so it is equivalent to the
probability ( 𝑓(𝑥) ) of occurrence of each 𝑥 continuous variable, which is the probability of each outcome; so we can write: 2 𝜎 2 =∑𝑁 𝑖=1(𝑥𝑖 − 𝜇) 𝑓(𝑥)
Now let’s incorporating calculus. To facilitate the analysis we will use more advance notation: Let 𝑋 be a continuous random variable with density function 𝑓(𝑥) defined on the interval (𝑎, 𝑏), and let 𝜇 = 𝐸(𝑋) be the mean of 𝑋, which is the Expected Value 𝐸(𝑋). Then the variance of 𝑋 is given by 𝑏
𝑉𝑎𝑟(𝑋) = 𝐸((𝑋 − 𝜇)2 ) = ∫ (𝑥 − 𝜇)2 𝑓(𝑥)𝑑𝑥 𝑎
Now Let’s review some statistical properties of variance and deviation: By definition, the standard deviation of X is the square root of the variance 𝜎 = √𝑉𝑎𝑟(𝑋) Which also can be written as 𝜎 = 𝐸((𝑋 – 𝜇)2 ). 174
𝑜𝑟 𝜎 2 = 𝑉𝑎𝑟(𝑋),
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1. Applying algebra to the definition: 𝜎 = 𝐸((𝑋 – 𝜇)2 ) = 𝐸(𝑋 2 − 2𝑋𝜇 + 𝜇2 ) {𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝜇 = 𝑐𝑜𝑛𝑠𝑡, 𝑎𝑛𝑑 𝐸(𝑋) = 𝜇},
𝑟𝑒𝑠𝑢𝑙𝑡𝑠:
𝜎 = 𝐸((𝑋 – 𝜇)2 ) = 𝐸(𝑋 2 ) − 2𝜇(𝐸(𝑋)) + 𝜇2 = 𝐸(𝑋 2 ) − 2𝜇𝜇 + 𝜇2 = 𝐸(𝑋 2 ) − 𝜇2 Note that, In order to calculate the variance and standard deviation, we first need to calculate the Mean, Where 𝑉𝑎𝑟(𝑋) is the expected value of the function (𝑥 − 𝜇)2 , which measures the square of the distance of 𝑋 from its mean 𝜇. We will use this formula for 𝑉𝑎𝑟(𝑋) calculations. It is often easier for hand calculation which is also a theorem in statistics: 𝑽𝒂𝒓(𝑿) = 𝑬(𝑿𝟐 )– [𝑬(𝑿)]𝟐 = 𝑬(𝑿𝟐 ) − µ𝟐 .
2. For 𝑌 = 𝑎𝑋 + 𝑏, where 𝑎 and 𝑏 are constants, we can write: 𝑉𝑎𝑟(𝑌) = 𝑉𝑎𝑟(𝑎𝑋 + 𝑏) = 𝑎 ∙ 𝐸(𝑋) + 𝑏 = 𝑎𝜇𝑋 + 𝑏, because 𝑎 = 𝑐𝑜𝑛𝑠𝑡 𝑎𝑛𝑑 𝐸(𝑋) = 𝜇, the mean of 𝐸(𝑎𝑋 + 𝑏) = (𝑎𝜇𝑋 + 𝑏) Using previous property and to simplify the notation use 𝜇𝑋 = 𝜇: 𝑉𝑎𝑟(𝑎𝑋 + 𝑏) = 𝐸([(𝑎𝑋 + 𝑏) − (𝑎𝜇𝑋 + 𝑏)]2 ) = 𝐸((𝑎𝑋 + 𝑏)2 − 2(𝑎𝑋 + 𝑏)(𝑎𝜇𝑋 + 𝑏) + ( 𝑎𝜇𝑋 + 𝑏)2 ) = = 𝐸(𝑎2 𝑋 2 + 2𝑎𝑏𝑋 + 𝑏 2 − 2𝑎2 𝑋𝜇 − 2𝑎𝑏𝑋 − 2𝑏 2 − 𝑎𝑏𝜇 + 𝑎2 𝜇2 + 2𝑎𝑏𝜇 + 𝑏 2 ) = 𝐸[𝑎2 (𝑋 − 𝜇𝑋 )2 ] = 𝑎2 𝐸(𝑋 − 𝜇𝑋 )2 = 𝑎2 (𝜎𝑋 )2 𝑽𝒂𝒓(𝒂𝑿 + 𝒃) = 𝒂𝟐 (𝝈𝑿 )𝟐 3. If 𝑋 and 𝑌 are independent variables, then considering 𝑊 = 𝑋 + 𝑌 2
𝑉𝑎𝑟 (𝑊) = 𝐸 ((𝑊 – 𝜇𝑊 ) ) 𝑜𝑟 𝑉𝑎𝑟(𝑊) = 𝐸(𝑊 2 )– [𝐸(𝑊)]2 , 𝑤ℎ𝑒𝑟𝑒
𝑉𝑎𝑟(𝑋 + 𝑌 ) = 𝐸(𝑋 + 𝑌)2 − [𝐸(𝑋 + 𝑌)]2 𝑜𝑟 𝑉𝑎𝑟(𝑋 + 𝑌 ) = 𝐸(𝑋 + 𝑌)2 − (𝜇𝑋 + 𝜇𝑌 )2 Simplifying: 𝐸(𝑋 2 + 2𝑋𝑌 + 𝑌 2 ) − (𝜇𝑋2 + 2𝜇𝑋 𝜇𝑌 + 𝜇𝑌2 ) = (𝐸𝑋 2 + 2𝐸𝑋𝑌 + 𝐸𝑌 2 ) − (𝜇𝑋2 + 2𝜇𝑋 𝜇𝑌 + 𝜇𝑌2 ); 𝑉𝑎𝑟(𝑋 + 𝑌 ) = (𝐸𝑋 2 − 𝜇𝑋2 ) + (𝐸𝑌 2 − 𝜇𝑌2 ) + (2𝐸𝑋𝑌 − 2𝜇𝑋 𝜇𝑌 ) 𝑽𝒂𝒓(𝑿 + 𝒀 ) = 𝑽𝒂𝒓(𝑿) + 𝑽𝒂𝒓(𝒀 ) + 𝟐𝑪𝒐𝒗(𝑿𝒀) Similarly can be calculated for 𝑊 = 𝑋 − 𝑌, resulting: 𝑽𝒂𝒓(𝑿 − 𝒀 ) = 𝑽𝒂𝒓(𝑿) + 𝑽𝒂𝒓(𝒀 ) − 𝟐𝑪𝒐𝒗(𝑿𝒀) The term 𝐶𝑜𝑣(𝑋𝑌) = (𝐸𝑋𝑌 − 𝜇𝑋 𝜇𝑌 ) can be defined as the expected value of the product of the deviations of two random variables from their respective means. It is called the covariance. If two variables are independent, then their covariance is zero. When the variables are independent the expected values of X and Y will be also independently and the expected value of the random variables will be its mean: 𝐸𝑋𝑌 = 𝐸𝑋𝐸𝑌, 𝐸𝑋 = 𝜇𝑋 , 𝐸𝑌 = 𝜇𝑌 Resulting that for independent variables the covariance will be zero: 𝐶𝑜𝑣(𝑋, 𝑌) = 𝐸𝑋𝑌 − 𝜇𝑋 𝜇𝑌 − 𝜇𝑋 𝜇𝑌 + 𝜇𝑋 𝜇𝑌 = 𝐸𝑋𝐸𝑌 − 𝜇𝑋 𝜇𝑌 − 𝜇𝑋 𝜇𝑌 + 𝜇𝑋 𝜇𝑌 = 𝐶𝑜𝑣(𝑋, 𝑌) = 𝜇𝑋 𝜇𝑌 − 𝜇𝑋 𝜇𝑌 − 𝜇𝑋 𝜇𝑌 + 𝜇𝑋 𝜇𝑌 = 0 For independent variables 𝑽𝒂𝒓(𝑿 + 𝒀 ) = 𝑽𝒂𝒓(𝑿) + 𝑽𝒂𝒓(𝒀 ) 176
and
𝑽𝒂𝒓(𝑿 − 𝒀 ) = 𝑽𝒂𝒓(𝑿) + 𝑽𝒂𝒓(𝒀 ) while 𝑪𝒐𝒗(𝑿𝒀) = 𝟎
Juan J. Prieto-Valdés
Important facts: a.
Make sure that the variables are independent or that it's reasonable to assume independence, before combining variances. b. Even when we subtract two random variables, we still add their variances; subtracting two variables increases the overall variability in the outcomes. c. We can find the standard deviation of the combined distributions by taking the square root of the combined variances. 4. Variance-Covariance Matrix. When working with multiple variables, variance and covariance are often displayed together in a variance-covariance matrix. The variances appear along the diagonal and covariance appear in the off-diagonal elements, as shown below for 𝑋𝑖 = 𝑋1 , 𝑋2 , … 𝑋𝑛 . ∑ 𝑋12 /𝑁 𝑉 = ∑ 𝑋2 𝑋1 /𝑁 ⋮ [∑ 𝑋𝑛 𝑋1 /𝑁 Where:
∑ 𝑋1 𝑋2 /𝑁
…
∑ 𝑋1 𝑋𝑛 /𝑁
∑ 𝑋22 /𝑁
⋯
∑ 𝑋2 𝑋𝑛 /𝑁
⋮
⋱
⋮
∑ 𝑋𝑛 𝑋2 /𝑁
⋯
∑ 𝑋𝑛2 /𝑁 ]
V is a (𝑛 × 𝑛) variance-covariance matrix N is the number of scores in each of the n sets X i is a deviation score the ith data set X i = xi − μi ∑ 𝑋𝑖2 /𝑁 is the variance of elements from the ith data set ∑ 𝑋𝑖 𝑋𝑖 /𝑁 is the covariance for elements from the ith to nth data sets
Variance-covariance matrix are extremely important when performing optimization of the variables with respect to a selected component of the system in order to take a convenient decision. Further, we will discuss a couple of examples. 5. Covariance and Correlation: Consider two random variables X and Y. Here, we define the covariance between X and Y, written Cov(X,Y). The covariance gives some information about how X and Y are statistically related. Using previous notation, let us discuss the properties and applications of covariance. 𝐶𝑜𝑣(𝑋, 𝑌) = 𝐸𝑋𝑌 − 𝜇𝑋 𝜇𝑌 Adding and subtracting the product of the mean: 𝐶𝑜𝑣(𝑋, 𝑌) = 𝐸𝑋𝑌 − 𝜇𝑋 𝜇𝑌 − 𝜇𝑋 𝜇𝑌 + 𝜇𝑋 𝜇𝑌 = 𝐸(𝑋𝑌 − 𝑋𝜇𝑌 − 𝜇𝑋 𝑌 + 𝜇𝑋 𝜇𝑌 ) = 𝐸(𝑋 − 𝜇𝑋 )𝐸(𝑌 − 𝜇𝑌 ) = 𝐸[(𝑋 − 𝜇𝑋 )(𝑌 − 𝜇𝑌 )] The covariance between 𝑋 and 𝑌 indicates how the values of 𝑋 𝑎𝑛𝑑 𝑌 move relative to each other. If large values of 𝑋 tend to happen with large values of 𝑌, then (𝑋 − 𝜇𝑋 )(𝑌 − 𝜇𝑌 ) is positive on average. In this case, the covariance is positive and we say 𝑋 𝑎𝑛𝑑 𝑌 are positively correlated. On the other hand, if 𝑋 tends to be small when 𝑌 is large, then (𝑋 − 𝜇𝑋 )(𝑌 − 𝜇𝑌 ) is negative on average. In this case, the covariance is negative and we say 𝑋 𝑎𝑛𝑑 𝑌 are negatively correlated. The level of correlation can be evaluated using the following definition: 𝐶𝑜𝑟𝑟(𝑋, 𝑌) =
𝐶𝑜𝑣 (𝑋, 𝑌) 𝜎𝑋 𝜎𝑌
The high correlation value occurs when the relation between variables 𝑋 and 𝑌 is strong. 177
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Example 9 As we saw before, the covariance can take positive or negative values. The values are interpreted as follows: •
Positive covariance: Indicates that two variables tend to move in the same direction.
•
Negative covariance: Reveals that two variables tend to move in inverse directions.
In finance, the covariance concept can be used in stock market analysis and in portfolio theory among other applications. Portfolio theory is a model on how risk-averse investors can construct portfolios to maximize expected return based on a given level of market risk. For example, by choosing assets that do not exhibit a high positive covariance with each other, the unsystematic risk can be partially eliminated or diminished. Covariance can be used to predict one stock-value tendency following another. For example, if oil price increase how it could affect electrical automobile market or gasoline sales. Let’s apply previous developed theory to solve one real problem. All calculations will be performed using Microsoft Excel. From https://finance.yahoo.com/ you can copy the historic price (public information) of any company stock. For this example let us copy the stock price daily-variation of Crude Oil per barrels (CL=F), a transport company Frontline Ltd. (FRO), and ecological corporation Tesla, Inc. (TSLA). All prices are in real time US dollars. For the analysis, we use price-data from August 2019 to March 2020. The calculation for such a small period decreases the confidence, but our main objective in this example is to discuss the method not a final investment proposal. To calculate the covariance and correlation between the monthly prices variations of the named stock we will use the following Excel commands. Follow the next instructions; we can perform all math/stat operations previously described in seconds. 𝑖𝑛 𝑐𝑒𝑙𝑙 𝐶26 = 𝐶𝑂𝑉𝐴𝑅𝐼𝐴𝑁𝐶𝐸. 𝑆(𝐵3: 𝐵22, 𝐶3: 𝐶22) 𝑖𝑛 𝑐𝑒𝑙𝑙 𝐷26 = 𝐶𝑂𝑉𝐴𝑅𝐼𝐴𝑁𝐶𝐸. 𝑆(𝐵3: 𝐵22, 𝐷3: 𝐷22) 𝑖𝑛 𝑐𝑒𝑙𝑙 𝐶30 = 𝐶𝑂𝑅𝑅𝐸𝐿(𝐵3: 𝐵22, 𝐶3: 𝐶22) 𝑖𝑛 𝑐𝑒𝑙𝑙 𝐷30 = 𝐶𝑂𝑅𝑅𝐸𝐿(𝐵3: 𝐵22, 𝐷3: 𝐷22) Now look to the screenshot on the right. The covariance between the daily-variation of Crude Oil (CL=F) price per barrels and the Frontline Ltd (FRO) stock price is -3.11039. Negative value because the more expensive a barrel of oil, the less the FRO stocks are worth, which is a logic result. Frontline Ltd is one of the world biggest crude oil transportation companies, but the certainty of such correlation is only about 22.7%. Other factors also influence the studied correlated process. By the other hands, when the oil prices increases, ecological companies becomes very successful. Note that the covariance between Crude Oil and Tesla Inc. has high positive value (631.5) with a 22.3% of correlation. If oil prices increase, do not invest in transportation companies but it is advisable to invest in ecological companies. It is important to clarify that this example do not considers possible companies strategies, for example when abruptly occurs price change, the company can apply some dividend schemes. Such strategies varies the conditions of the analysis, requiring the incorporation of new variables in the calculation. For more accurate evaluation is required advanced portfolio theory, which we are not covering in this business calculus course. 178
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5.8. Variance in different types of distributions: 1.
Case Study Uniform Distribution
From elemental statistics, we know that variance can be written as 𝜎 2 =
∑(𝑥−𝜇)2 𝑁
, or based on previous property
(Topic 5.7), the variance can be written as 𝑉𝑎𝑟(𝑋) = 𝐸(𝑋 2 )– [𝐸(𝑋)]2 . Now let’s re-visit Topic 5.3, to calculate the variance of the uniform distribution on the interval (𝑎, 𝑏), which 1
probability the is given by 𝑓(𝑥) = 𝑏−𝑎. therefore, the Expected Value will be 𝑏
𝐸(𝑋) = ∫ 𝑥 𝑎
(𝑎 + 𝑏) 1 1 1 𝑏 2 − 𝑎2 𝑏 𝑑𝑥 = ∙ 𝑥2 │ = = . 𝑎 2(𝑏 − 𝑎) 𝑏−𝑎 2 𝑏−𝑎 2
2
While 𝐸(𝑋 ) will be: 𝑏
𝐸(𝑋 2 ) = ∫ 𝑥 2 𝑎
(𝑏 2 + 𝑎𝑏 + 𝑎2 ) 1 1 1 𝑏 3 − 𝑎3 𝑏 𝑑𝑥 = ∙ 𝑥3 │ = = . 𝑎 3(𝑏 − 𝑎) 𝑏−𝑎 3 𝑏−𝑎 3
Thus, subtracting these two expressions, we get: 2
𝑉𝑎𝑟(𝑋) = 𝐸(𝑋
2 )–
(𝑎 + 𝑏) (𝑏 2 + 𝑎𝑏 + 𝑎2 ) 𝑏 2 + 𝑎𝑏 + 𝑎2 𝑏 2 + 2𝑎𝑏 + 𝑎2 [𝐸(𝑋)] = −[ ] = − 3 2 3 4 2
4𝑏 2 + 4𝑎𝑏 + 4𝑎2 − 3𝑏 2 − 6𝑎𝑏 − 3𝑎2 𝑏 2 − 2𝑎𝑏 + 𝑎2 (𝑎 − 𝑏)2 = = 12 12 12 Under such conditions, the variance of the uniform distribution can be written as: 𝑉𝑎𝑟(𝑋) =
𝝈=
2.
𝒃−𝒂 √𝟏𝟐
Case Study Exponential Distribution
Follow previous properties, the variance can be written as 𝑉𝑎𝑟(𝑋) = 𝐸(𝑋 2 )– [𝐸(𝑋)]2 . From Example-8 Topic 5.6 we know that ∞
𝐸(𝑋) = ∫ 𝜆 𝑥 𝑒
−𝜆𝑥
𝑑𝑥 = 𝟏/𝝀
0
Using similar procedure as shown in Example-8 we can solve the integral in 𝑋 2 . ∞
𝐸(𝑋
2)
=∫ 𝜆 0
∞
𝑥2
𝑒
−𝜆𝑥
1 𝑑𝑥 = {𝑐ℎ𝑎𝑛𝑔𝑖𝑛𝑔 𝜆𝑥 = 𝑢} = 2 ∫ 𝑢2 𝑒−𝑢 𝑑𝑢 𝜆
0
Applying similar procedure as in Example-8 1 ∞ 2 [−2𝑒 −𝑢 − 2𝑢𝑒 −𝑢 − 𝑢2 𝑒 −𝑢 ] = 2 2 0 𝜆 𝜆 So by property-3, 2 1 2 1 𝑉𝑎𝑟(𝑋) = 𝐸(𝑋 2 )– [𝐸(𝑋)]2 = 2 − ( ) = 2 and 𝛔𝐱 = 𝟏/𝝀 𝜆 𝜆
180
𝜆
Juan J. Prieto-Valdés
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3.
Case Study Normal Distribution
As we learned before; The Normal Density Function is expressed as: 𝑃(𝑋) =
1 𝜎√2𝜋
𝑒
−
(𝑥−𝜇)2 2𝜎2
The Mean: Let’s find maximum of the Normal Distribution by making the first derivative equals to zero. Solve by substitution or use Wolfram Alpha for fast view: 𝑑 1 ( 𝑑𝑥 𝜎√2𝜋
(𝑥−𝜇)2 − 𝑒 2𝜎2 )
=−
(𝑥 – 𝜇)𝑒
−
(𝑥 – 𝜇)2 2 𝜎2
√(2 𝜋) 𝜎 3
Applying zero product rule: 𝒙 − 𝝁 = 𝟎, 𝒔𝒐:
=0
𝒙𝑴𝒂𝒙 = 𝝁, 𝒂𝒏𝒅 𝑬(𝑿) = 𝝁
The Variance: Now let’s locate the points of inflexion (check Topic 5.5) by making the second derivative equals to zero. Solve by substitution or use Wolfram Alpha for fast view: −
(𝑥 – 𝜇)2
𝑑 (𝑥 – 𝜇)𝑒 2 𝜎2 𝑒 ( )= 3 𝑑𝑥 √(2 𝜋) 𝜎
−
(𝑥 − 𝜇)2 2 𝜎 2 (−𝜇2
+ 𝜎 2 − 𝑥 2 + 2 𝜇 𝑥)
√(2 𝜋) 𝜎 5
=0
Applying zero product rule: (−𝜇2 + 𝜎 2 − 𝑥 2 + 2 𝜇 𝑥) = −(𝑥 2 − 2𝜇𝑥 − 𝜇2 − 𝜎 2) = 𝜎 2 − (𝑥 − 𝜇)2 = 0 𝑺𝒐: 𝝈𝟐 = (𝒙 − 𝝁)𝟐 . 𝒕𝒉𝒖𝒔,
𝑽𝒂𝒓(𝑿) = 𝝈𝟐 , 𝒂𝒏𝒅 𝑽𝒂𝒓(𝑿) = 𝑬((𝑿 − 𝝁)𝟐 )
We have deduced all basic statistical parameter using calculus; by calculating the 1st and 2nd derivative of the normal function. Next: Chapter quiz and some practicing exercises available at Professor Prieto-Valdes Web Page, as well as further topics in statistics based calculus and physics interpretations.
Homework Chapter 5 Homework and practicing exercises at Prieto-Valdes web-page
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Juan J. Prieto-Valdés
PROJECT – 7. Portfolio Optimization The term “portfolio” refers to any combination of financial assets such as stocks, boons, etc. A portfolio is designed according to the investor's risk tolerance, period, and investment objectives. The monetary value of each asset may influence the risk and reward ratio. In the present project, we will optimize the volume of investment when having 3 different stock options. Based on previously exposed statistical theory, we will optimize using Microsoft Excel. 1. Create a spreadsheet including asset prices for the chosen frequency. In our case, asset prices were sampled in monthly intervals for a period of one year. Public data from https://finance.yahoo.com Let’s consider three of the publicly recognized successful companies during the last year. Our portfolio will be composed of: Amazon.com, Inc. (AMZN); Tesla, Inc. (TSLA); and Netflix, Inc. (NFLX) stocks. Larger number of investments and period are preferably. However, for the simplification we use only three stocks and price variation during one year (COVID19 first pandemic year). 2. Our main interest is the optimization of the return (maximizing our partial profit), so we need to work with returns. We transform the monthly prices into monthly returns using the natural logarithm of the monthly relationship: 𝑃𝑖 𝑃𝑅𝐼𝐶𝐸𝑡 𝑀𝑜𝑛𝑡𝑙𝑦 𝑅𝑒𝑡𝑢𝑛𝑠 = 𝐿𝑁 ( ) 𝑜𝑟 𝑅 = 𝐿𝑁 ( ). 𝑃𝑖−1 𝑃𝑅𝐼𝐶𝐸𝑡−1 Where 𝑃𝐼 − current price and 𝑃𝑖−1 − is the price on the month before. 𝑖 − represents the time 𝑡.
3.
In this exercise, we will focus on the Return (𝑅) optimization relative to standard deviation, not on other characteristics of the investment. Let’s assume a risk-free rate of 0.05% per month, and the expected returns (EXP. RET.) over the year equal to the average return of each stocks (next screen shots): for AMZN (cell L2) Expected Return = 𝐴𝑉𝐸𝑅𝐴𝐺𝐸(𝐹3, 𝐹14) = −0.01812 , for TSLA (cell M2) Expected Return = 𝐴𝑉𝐸𝑅𝐴𝐺𝐸(𝐺3: 𝐺14) = −0.11935, and for NFLX (cell N2) Expected Return = 𝐴𝑉𝐸𝑅𝐴𝐺𝐸(𝐻3: 𝐻14) = −0.01656. The reflecting COVID19 pandemic period denotes negative average returns. Now, we will distribute the weight in our portfolio equivalently (33.33%) for each stock to further perform the real weight optimization; these values are located in cells L6, M6, and N6. Now we can construct the Variance-Covariance Matrix using the following Excel commands: 183
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Structure of the Matrix (Check topic 5.7-4) 𝑊1 𝑉𝑎𝑟1 𝑊2 [𝐶𝑜𝑣2−1 𝑊3 𝐶𝑜𝑣3−1
𝐶𝑜𝑣1−2 𝑉𝑎𝑟2 𝐶𝑜𝑣3−2
𝐶𝑜𝑣1−3 𝐶𝑜𝑣2−3 ] 𝑉𝑎𝑟3
𝐶𝑜𝑛𝑡𝑆 1
𝐶𝑜𝑛𝑡𝑆 2
𝐶𝑜𝑛𝑡𝑆 3
Excel commands/formulas (See the screenshot of the excel sheet below)
After completing the matrix, we need to calculate the Contribution of each stock (1, 2, and 3) by multiplying each stock Weigh by its Variance/Covariance and adding them to complete the last row (𝐶𝑜𝑛𝑡𝑆# ) of the matrix. These operations can be perform directly using the following commands (locate the command in a corresponding cell as per the next screenshot): In cell L10 perform = 𝐿6 ∗ 𝑆𝑈𝑀𝑃𝑅𝑂𝐷𝑈𝐶𝑇($𝐾$7: $𝐾$9, 𝐿7: 𝐿9), In cell M10 perform = 𝑀6 ∗ 𝑆𝑈𝑀𝑃𝑅𝑂𝐷𝑈𝐶𝑇($𝐾$7: $𝐾$9, 𝑀7: 𝑀9), and In cell N10 perform = 𝑁6 ∗ 𝑆𝑈𝑀𝑃𝑅𝑂𝐷𝑈𝐶𝑇($𝐾$7: $𝐾$9, 𝑁7: 𝑁9). 4. The last part to optimize the portfolio requires the following operations: Sum of all weights. It is equal to one. Our total money invested is 100%: In cell K1: Σ Weight = 𝑆𝑈𝑀(𝐾7: 𝐾9) Portfolio Expected Return Which is the Sum of each Expected Return by its weight: In cell K13: P. E. R. = 𝑆𝑈𝑀𝑃𝑅𝑂𝐷𝑈𝐶𝑇(𝐿2: 𝑁2, 𝐿6: 𝑁6) Portfolio Standard Deviation, which is the square rood of the sum of all contributions to variance: In cell K14: Portfolio S. D. = = SUM(L10: N10)^(1/2) Expected Sharped Ratio is the excess return over the standard deviation: In cell K15: 𝐸. 𝑆. 𝑅. −(𝐾13 − 𝐽2)/𝐾14 5. Next we will do ACTUAL OPTIMIZATION: Maximize the Expected Sharp Ratio by stablishing the optimum weight of each investment (stocks 1, 2, and 3). Up to this point our Excel work sheet looks as follow:
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Juan J. Prieto-Valdés
To optimize the amount of each investment we will use the SOLVER Add-in to Excel. The parameters to be satisfied are:
Set the Object: Maximize cell K15 (Expected Sharp Ratio) By Changing the variables 𝐿6, 𝑀6, 𝑎𝑛𝑑 𝑁6 (each investment/stock weight) Fixing a group of constrains for the three selected stocks. Weights have to be positive and greater than one dollar while the total weight is equal to one (100%).
Click on Solve and get the Math/Stat/Technology recommendation for your investment. Next, ratify the Solver Optimization by keeping your solution or simple save the scenario. The screenshot of the SOLVER control panel on the next image.
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The resulting Excel sheet is displayed in the next screenshot. Follow our optimization, the most convenient investment (March 22, 2021) will be amazo.com, which is an expected result according to the currently public information. amazon.com is probably one the most successful company during the first pandemic year. Underline that we have observed the stock’s price variation for the period from April 2020 to March 2021.
Application of previously developed template: Now using our Excel Template, let’s analyze three other randomly selected stocks from the communication field 𝐴𝑇&𝑇 (𝐴𝑇𝑇),
𝐶𝑜𝑚𝑐𝑎𝑠𝑡 (𝐶𝑀𝐶𝑆𝐴),
𝑎𝑛𝑑 𝑉𝑒𝑟𝑖𝑧𝑜𝑛 (𝑉𝑍).
Perform exactly the same operations: Fill out columns A, B, C, and D. And follow Excel directions. We can see on the next screenshot that the investments should be directed 60% in AT&T, 40% in Verizon to get maximum profit. The optimization was performed using only three stocks to facilitate the math/stat operations. In addition, we analyzed only one-year period under the effect of pandemic emergencies. In real practice, the variancecovariance matrix should be built from 5, 10, and more company components, during five or more years. Other elements of Portfolio Theory can be incorporate to approach better and more confident result. This is a simple case example in a limited period analysis to show the calculation method. Do not invest based on this example.
The present project offers a real vision of portfolio optimization to distribute the investment in the most convenient weight to maximize the return (profit). Hope this project opens up new points of view on the application of business calculus with statistics. There are many software for business application in progress which can be reviewed during lectures by students request. But now we are working on the fundaments of Calculus and Statistics. 186
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REVIEW CHAPTER. R-1. SOME ELEMENTARY ALGEBRA CONCEPTS In the next pages you will find some simple numeric examples and basic concepts indispensables to understand calculus. I detected that some student who were not working with mathematics during the last 5 years do not remember well these concepts. If this is your case, you need to recover your level to succeed in Business Calculus.
R-1.1. Common Factors When some particular factor appears in every term, we say we are observing a common factor. For example: In 12𝑥 3 + 6𝑥 3 + 4𝑥 3 , the factor 𝑥 3 appears in each and every term, 𝑥 3 is a common factor. In 32𝑥 2 + 3𝑥 2 + 4𝑥, the factor 𝑥 appears in each and every term, x is a common factor. In 12(𝑥 − 5)3 + 5(𝑥 − 5)2 + (𝑥 − 5), the factor (𝑥 − 5) is a common factor because it appears in each term.
R-1.2. Exponents The following Table introduces the properties of exponents. Note that (𝑛, 𝑚; 𝑎 𝑎𝑛𝑑 𝑏 ≠ 0) 𝑎0 = 1
Zero Exponent Property Negative Exponent Property
𝑎−𝑛 =
1 , 𝑎𝑛
𝑎𝑚 =
1 𝑎−𝑚
Product of Powers Property
𝑎𝑛 ⋅ 𝑎𝑚 = 𝑎𝑛+𝑚
Quotient of Powers Property
𝑎𝑛 = 𝑎𝑛−𝑚 𝑎𝑚
Power of a Product Property
𝑎𝑛 ⋅ 𝑏 𝑛 = (𝑎𝑏)𝑛
Power of a Quotient Property
𝑎𝑛 𝑎 𝑛 = ( ) 𝑏𝑛 𝑏
Power of a Power Property
(𝑎𝑛 )𝑚 = 𝑎𝑛⋅𝑚
Rational Exponent Property
𝑎
𝑏⁄ 𝑐
𝑐
= √𝑎𝑏
R-1.3. Roots; Products, Addition, and Subtraction Rules for Square Roots The product and quotient rules for n degree roots are: 𝑛
𝑎𝑛 𝑏 𝑛 = (𝑎𝑏)𝑛 𝑎 𝑛
𝑛
𝑛
and √𝑎𝑏 = √𝑎 × √𝑏 𝑎𝑛
𝑛
𝑎
(𝑏 ) = 𝑏𝑛 and √𝑏 =
𝑛
√𝑎 √𝑏
𝑛
𝑛
𝑛
𝑛
The addition and/or subtraction can be represented as follow: 𝑎 √𝑚𝑥 + 𝑏 √𝑚𝑥 = (𝑎 + 𝑏) √𝑚𝑥 Also remember that we must identify 𝑎 188
1⁄ 𝑛
𝑛
𝑤𝑖𝑡ℎ √𝑎.
Juan J. Prieto-Valdés
R-1.4. Rationalizing the Denominators We cannot divide by a number written inside the radical symbol; for example,
𝑎 √𝑏
, however we can solve the problem
by rationalizing the denominator (multiply the numerator and the denominator by the same denominator or by the smallest number that produces the square root of a perfect square in the denominator). 𝑎 √𝑏
×
√𝑏 √𝑏
=
𝑎√𝑏 48 √2 48√2 48√2 𝑡ℎ𝑒 𝑛𝑒𝑥𝑡 𝑖𝑠 𝑎 𝑛𝑢𝑚𝑒𝑟𝑖𝑐 𝑒𝑥𝑎𝑚𝑝𝑙𝑒: × = = = 12√2 𝑏 4 √8 √2 √16 𝑥+2
Now you have to remember how to rationalize an algebra fraction using the same principle, for example 3
√𝑥−5
3
3
√
:
3
𝑥 + 2 √(𝑥 − 5)2 √(𝑥 + 2)(𝑥 − 5)2 × = 𝑥 − 5 3√(𝑥 − 5)2 𝑥−5
R-1.5. Polynomials, Addition/Subtraction, Multiplication, Special Products, and Factoring A polynomial is a single term or sum of two or more terms containing variables in the numerator with whole number exponents. A polynomial in 𝑥 of degree 𝑎𝑛 is an algebraic expression of the form 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 +𝑎𝑛−3 𝑥 𝑛−3 + ⋯ + 𝑎0 Where 𝑎𝑛 , 𝑎𝑛−1 , 𝑎𝑛−3 , … , 𝑎0 are real numbers and 𝑎𝑛 ≠ 0
R-1.6. The Product of Two Binomials (Foil Method) In elementary algebra, FOIL is a mnemonic for the standard method of multiplying two binomials—hence the method may be referred to as the FOIL method. The word FOIL is an acronym for the four terms of the product: First (“first” terms of each binomial are multiplied together) Outer (“outside” terms are multiplied—that is, the first term of the first binomial and the second term of the second) Inner (“inside” terms are multiplied—second term of the first binomial and first term of the second) Last (“last” terms of each binomial are multiplied) F O I L The general form is
(𝑥 + 𝑎)(𝑥 + 𝑏) = 𝑥 2 + (𝑏𝑥 + 𝑎𝑥) + 𝑎𝑏 First + (outer + inner) + last
R-1.7. Special Product Rules Some products occur so frequently in algebra that it is advantageous to be able to recognize them. You have to remember the following special product rules: square of binomial (1) and (2), and difference of square (3): (𝑥 + 𝑎)(𝑥 + 𝑎) = 𝑥 2 + 2𝑎𝑥 + 𝑎2
(1)
(𝑥 − 𝑎)(𝑥 − 𝑎) = 𝑥 2 − 2𝑎𝑥 + 𝑎2
(2)
(𝑥 + 𝑎)(𝑥 − 𝑎) = 𝑥 2 − 𝑎2
(3)
Other important special product rules are Sum or Difference of Cubes. (𝑥 3 + 𝑦 3 ) = (𝑥 + 𝑦)(𝑥 2 − 𝑥𝑦 + 𝑦 2 )
(4)
(𝑥 3 − 𝑦 3 ) = (𝑥 − 𝑦)(𝑥 2 + 𝑥𝑦 + 𝑦 2 )
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R-1.8. Greatest Common Factor (GCF) and Factoring by Grouping In some cases, there is not a GCF for all terms in a polynomial; however, you can create a common factor by grouping, for example, factor: 𝑥 3 + 3𝑥 2 + 2𝑥 + 6 Step 1: Group the first and the last terms together: 𝑥 3 + 3𝑥 2 + 2𝑥 + 6 = (𝑥 3 + 3𝑥 2 ) + (2𝑥 + 6) Step 2: Factor out a GCF from each separate binomial: (𝑥 3 + 3𝑥 2 ) + (2𝑥 + 6) = 𝑥 2 (𝑥 + 3) + 2(𝑥 + 3) Step 3: Factor out the common binomial: 𝑥 2 (𝑥 + 3) + 2(𝑥 + 3) = (𝑥 + 3)(𝑥 2 + 2)
R-1.9. Factoring Trinomials in the Form 𝒙𝟐 + 𝒃𝒙 + 𝒄 and 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 where 𝒂 ≠ 𝟏 We search two binomials that when you multiply them you get the trinomial, for example: 𝑥 2 − 5𝑥 − 14 Step 1: Set up a product of two binomials. It will look like this: (binomial)(binomial) Step 2: Find the factors that go in the first positions; to get the x squared: (𝑥+ )(𝑥+ )
Step 3: Find the factors that go in the last positions. They have to be factors of the constant term, for example: 𝑥 2 − 5𝑥 − 14 = (𝑥 + 2)(𝑥 − 7). If the constant 𝒄 is positive, your factors are going to both have the same sign depending on 𝒃’s sign. If 𝒄 is negative, your factors are going to have opposite signs. In the case of 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 we need to find factors of c, but also a. Step 1: Set up a product of binomials. It will look like this ( )( ) Step 2: Use trial and error to find the factors as needed. You can apply the FOIL method as follow: 1. Multiply the leading coefficient, 𝑎, and the constant, 𝑐 . 2. Find the factors of 𝑎𝑐 whose sum is 𝑏 and rewrite the middle term, 𝑏𝑥, as a sum or difference as follow: The following illustration shows the path of the operation:
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 =
(𝐴𝑥 + 𝐵)(𝐶𝑥 + 𝐷)
𝐵𝐶𝑥 𝐴𝐷𝑥 2 𝐴𝐶𝑥 + (𝐴𝐷 + 𝐵𝐶)𝑥 + 𝐵𝐷 Where 𝐚, 𝐛, and 𝐜 are the coefficients, and; 𝑨𝑪 = 𝒂,
𝐴𝑫 + 𝑩𝑪 = 𝒃, and 𝑩𝑫 = 𝒄. 𝑨 and 𝑪 are = 1 , when 𝒂 = 1
The following numerical example illustrates the operations of previous diagram; try to establish full correlation and you will improve your factoring abilities:
12𝑥 2 + 6𝑥 − 2 =
(3𝑥 + 2)(4𝑥 − 1)
12𝑥 2 𝟑 × 𝟒 = 𝒂,
+8𝑥 −2𝑥 + (8 − 2)𝑥
−2
(2 × 4) + (𝟐 × (−𝟏)) = 𝒃, and (2 × (−1)) = 𝒄.
Further will be discussed more advanced algebra method for factoring polynomials of n degree using factor theorem. 190
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R-1.10. Rational Expressions, Multiplication and Division: A rational expression is a fraction in which the numerator and/or the denominator are polynomials. To multiply 2 fractions, both the numerators and the denominators multiply together, and common factors may be cancelled before multiplying, for example; given Equation:
𝑥 2 +3𝑥+2 𝑥−2
×
𝑥 2 −4 𝑥+1
First, factor all the expressions, then cancel common factors (factor on the top can cancel with factor on the bottom): (𝑥+1)(𝑥+2) (𝑥−2)
×
(𝑥−2)(𝑥+2) (𝑥+1)
=
(𝑥+1)(𝑥+2)(𝑥−2)(𝑥+2) (𝑥−2)(𝑥+1)
= (𝑥 + 2)2
Division works the same way with rational expressions, multiply by the reciprocal of the divisor: 𝑥 2 −25 5𝑥−𝑥
÷ 2
𝑥 2 +2𝑥−15 5−𝑥
=
𝑥 2 −25 5𝑥−𝑥 2
×
5−𝑥
=
𝑥 2 +2𝑥−15
(𝑥−5)(𝑥+5) 𝑥(5−𝑥)
5−𝑥
× (𝑥+5)(𝑥−3) =
(𝑥−5) 𝑥(𝑥−3)
.
R-1.11. Addition/subtraction. Finding the Low Common Denominator (LCD) The product of all the denominators is always a common denominator, but not necessarily the LCD (the final answer may have to be reduced). For example: 𝑥−5 4𝑥 𝑥−5 4𝑥 + 2 = + 2 (𝑥 + 2)(𝑥 − 2) (𝑥 − 2)(𝑥 − 2) 𝑥 − 4 𝑥 − 4𝑥 + 4 Note that the LCD have to contains both denominators (𝑥 + 2)(𝑥 − 2)(𝑥 − 2) (𝑥−5)(𝑥−2)
4𝑥(𝑥+2)
+ (𝑥−2)(𝑥−2)(𝑥+2) = (𝑥+2)(𝑥−2)(𝑥−2)
𝑥 2 −7𝑥+10
4𝑥 2 +8𝑥
+ (𝑥−2)(𝑥−2)(𝑥+2) = (𝑥+2)(𝑥−2)(𝑥−2)
Now that they are over the same denominator, you can add the numerators and simplify if possible: 5𝑥 2 − 𝑥 + 10 = (𝑥 − 2)(𝑥 − 2)(𝑥 + 2)
R-1.12. Compound Rational Expressions A compound fraction is a fraction whose numerator or denominator contains one or more fractions. They can be simplified by using the methods we have already learned for simplifying rational expressions. You may use one or more of the following two methods. Method 1: Multiply by the reciprocal of the denominator in both parts (numerator and denominator), to create a single fraction using previous described division method: 4𝑥 + 1 1 4𝑥 1 ( 𝑥 ) 4+𝑥 +𝑥 4𝑥 + 1 𝑥2 4𝑥 + 1 𝑥2 4𝑥 2 + 𝑥 𝑥 = 2 = = ( ) × = × = ( ) 2 3𝑥 2 3𝑥 2 + 2 𝑥 3𝑥 2 + 2 𝑥 3𝑥 2 + 2 3𝑥 2 + 2 3+ 2 + ( ) 2 2 2 𝑥 𝑥 𝑥 𝑥 Method 2: Determine the LCD of all the fractions in the compound fraction, them, multiply both the numerator and the denominator of the compound fraction by the LCD. If possible, simplify the resulting rational expression. 1 1 𝑥2 2 4+ 𝑥 =( 𝑥 ) ( 1 ) = (4𝑥 + 𝑥 ) 2 2 𝑥2 3𝑥 2 + 2 3+ 2 3+ 2 𝑥 𝑥 1 Obviously, the second method is notably faster. 4+
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R-2. ALGEBRA REVIEW R-2.1. Lines, Slope, and Graphing Linear Functions. Given two points 𝑷𝟏 and 𝑷𝟐 , the slop m of the line passing through these to points can be written as follow: ∆𝒚
𝒚 −𝒚
𝒎 = ∆𝒙 = 𝒙𝟐 −𝒙𝟏 𝟐
𝟏
If considering 𝑷𝟏 (𝟎, 𝒃) and 𝑷𝟐 (𝒙, 𝒚), after substituting in previous equation, we get: 𝑚 =
𝒚−𝒃 𝒙−𝟎
. Solving for 𝑦 wen get the
equation of the line in standard form: 𝒚 = 𝒎𝒙 + 𝒃 While the equation of the line in general form can be written as follow: 𝒂𝒙 + 𝒃𝒚 + 𝒄 = 𝟎 𝑎 𝑏
𝑐 𝑏
Where 𝑚 = and 𝑏 =
R-2.2. Solving Linear Equations A linear equation will have a variable raised to a power of one. A linear equation in one variable can be written in the form 𝑎𝑥 + 𝑏 = 0 were 𝑎 and 𝑏 are real numbers and 𝑎 ≠ 0. When the equation has at least one real solution it is called a Conditional Equation. In not conditional, linear equations can be classified depending on the final answer: Identity: An equation is true for all values of 𝑥: – 4(𝑥 + 7) = – 4𝑥 – 28 Inconsistent Equation: An equation that is NOT true for any real number. – 4(𝑥 + 7) = – 4𝑥 – 21 A rational equation involves a variable in the bottom of a fraction. For rational equations you must be sure your answer does not make a denominator go to zero. This is because any number divided by zero is undefined.
Simple linear equation containing the variable in both sides. 9𝑥 − 20 = 4𝑥 + 5
𝑦 = 9𝑥 − 20
Subtract 4𝑥 from both sides to get all the variables on one side.
𝑦 = 4𝑥 + 5
5𝑥 − 20 = 5 Them, add 20 in both sides, and divide by 5 both sides, so we are done
5𝑥 = 25,
5 5
𝑥=
25 5
,
so 𝑥 = 5
. Graphic consideration: the equation on the left represent a line (𝑦1 = 9𝑥 − 20) and on the right another line (𝑦2 = 4𝑥 + 5) . If graphing both lines, they meet (intercept) at 𝑥 = 5, when 𝑦1 = 𝑦2 . Graphing linear functions will be reviewed in topics 3.1
Linear equation containing the variable in both sides (distribution is required). 2(𝑥 − 2) + 11 = 𝑥 − 4(2𝑥 + 5) Distribute the 2 and then -4. 192
Juan J. Prieto-Valdés
2𝑥 − 4 + 11 = 𝑥 − 8𝑥 − 20 To get all the variables to one side, ad 7𝑥 to both sides, and them, subtract 7 from both sides to get all the numbers on the right side. 9𝑥 − 4 + 11 = −20 Simplify 9𝑥 = −27 Therefore, divide both sides by 9 to get the final answer 9
𝑥= 9
−27 9
equivalently; 𝑥 = −3
Graphic solution also can be found.
R.2.3. Rational Equations Simple rational equation with the variable on denominator in binomial form. 11 −3 −7= 𝑥+3 𝑥+3 First multiply both sides of the equation by the Lower Common Denominator (LCD). 𝑥 + 3 11 𝑥+3 𝑥 + 3 −3 ( ) −( )7 = ( ) 1 𝑥+3 1 1 𝑥+3 Notice that all the fractions are gone after multiplying, so now solve a resulting simple lineal equation applying previously learned procedures. 11 7𝑥 + 21 −3 −( )=( ) 1 1 1 11 − 7𝑥 − 21 = −3 −7𝑥 = 7, dividing by −7, we get: 𝑥 = 1
Rational equation with the variable on denominator (factorable trinomial). 11 7 28 − = 2 𝑥 + 6 𝑥 − 2 𝑥 + 4𝑥 − 12 First, we need to find the LCD. If any of the denominators can be factored, we have to do that first. The polynomial can be factored into (𝑥 + 6)(𝑥 – 2). Now we can see that the LCD is (𝑥 + 6)(𝑥 – 2). [(𝑥 + 6)(𝑥 – 2)] [
11 7 28 − = ] 𝑥 + 6 𝑥 − 2 (𝑥 + 6)(𝑥 – 2)
Multiply both sides by the LCD and then reduce each fraction. This will result in a simple linear equation: 11(𝑥 − 2) 7(𝑥 + 6) 28 − = 1 1 1 Solve the equation: 11𝑥 − 22 − 7𝑥 − 42 = 28, so; 𝑥 = 23
Rational equation with the variable on denominator (factorable binomial). 11 7 −28 − = 2 𝑥+2 𝑥−2 𝑥 −4 𝑥 2 − 4 can be factored into (𝑥 + 2)(𝑥 – 2). The LCD is (𝑥 + 2)(𝑥 – 2). 193
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Multiply both sides by this LCD, [(𝑥 + 2)(𝑥 – 2)] [
11 7 −28 − = ] 𝑥 + 2 𝑥 − 2 (𝑥 + 2)(𝑥 – 2)
then reduce each fraction by simplifying, this will result in a simple lineal equation. 11(𝑥 – 2) 7(𝑥 + 2) −28 − = 1 1 1 11𝑥 − 22 − 7𝑥 − 14 = −28 Its solution is = 2 ; however, this solution has to be rejected because when 𝑥 = 2 the denominator becomes equal to zero, and division by zero is not allowed (undefined). This example has NO SOLUTION.
R-2.4.
Complex Numbers. Complex numbers are combination of real and imaginary number parts: 𝑎 + 𝑏 𝑖, WHERE 𝑎 AND 𝑏 ARE REAL NUMBERS (COEFFICIENTS). 𝑖 = √ −1 and 𝑖 2 = −1 . For any real number 𝑐 > 0; √−𝑐 = 𝑖 √𝑐 .
Complex numbers can be added, subtracted multiplied and divided. Always we apply algebra rules to perform this operation, we have to combine imaginary part and real part separately; they are two different like terms.
Multiplying (power) 𝒊 number 𝑖 2 = −1
𝑖4 = 1
𝑖 11 = (𝑖 2 )5 𝑖 = −𝑖
𝑖 3 = −𝑖
𝑖5 = 𝑖
𝑖 112 = (𝑖 2 )56 = 1
Operations with complex numbers Multiplication Apply algebra considering the number-letter as a variable: (2 + 2𝑖)(3 − 5𝑖) = 6 − 10𝑖 + 6𝑖 − 10𝑖 2 Because 𝑖 2 = −1, 10 becomes positive. Then, collect like terms resulting: 16 − 4𝑖
Addition / Subtraction Apply algebra (distribute if needed), clear parenthesis, and collect like terms (2 + 2𝑖) + (3 + 5𝑖) = 5 + 7𝑖 (5 + 6𝑖) − (3 − 2𝑖) = 2 + 8𝑖
Division We have to clear imaginary (complex) numbers from the denominator as we cannot divide by imaginary number. Perform this operation by multiplying both, numerator and denominator by a conjugated expression of the denominator. Then multiply horizontally. Remember that 𝑖 2 = −1, so you substitute 𝑖 2 𝑏𝑦 − 1 . (5 + 6𝑖) (5 + 6𝑖) (2 − 3𝑖) 10 − 15𝑖 + 12𝑖 − 18𝑖 2 28 − 3𝑖 28 3 = × = = = − 𝑖 (2 + 3𝑖) (2 + 3𝑖) (2 − 3𝑖) 4 − 9𝑖 2 13 13 13
Graphic interpretation of the 𝒊 number Find 𝒙 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕𝒔 of the graph 𝑦 = 𝑥2 − 4 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 exist when 𝑦 = 0 , so 𝑥 2 − 4 = 0 or 𝑥 2 = 4 194
Find 𝒙 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕𝒔 of the graph 𝑦 = 𝑥2 + 4 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 exist when 𝑦 = 0 , so 𝑥 2 + 4 = 0 or 𝑥 2 = −4
Juan J. Prieto-Valdés
√𝑥 2 = ±√4, so 𝑥 = −2 𝑜𝑟 𝑥 = 2 There are two real value solutions. The graph intercepts the axis 𝑥 in two places.
√𝑥 2 = ±√−4, so 𝑥 = −2𝑖 𝑜𝑟 𝑥 = 2𝑖 Imaginary solution because there are not real intercepts with axis 𝑥
R-2.5. Quadratic Equation. Quadratic equation has the following form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0:, where 𝑎, 𝑏, and 𝑐 are real numbers (coefficients). We solve a quadratic equation using at least 5 different methods. 1. Factoring and applying ZERO PRODUCT RULE: When coefficient 𝑐 = 0:
3𝑥 2 − 4𝑥 = 0
Factor, apply zero product rule to solve When 𝑎, 𝑏, 𝑎𝑛𝑑 𝑐 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠 𝑒𝑥𝑖𝑠𝑡 :Factor, apply zero product rule to solve
→
4
𝑥(3𝑥 − 4) = 0 → 𝑥 = 0 𝑜𝑟 𝑥 = 3
𝑥 2 − 4𝑥 − 12 = 0 → (𝑥 − 6)(𝑥 + 2) = 0 → 𝑥 = −2 𝑜𝑟 𝑥 = 6
1. By using SQUARE ROOT PROPERTY: when 𝒙𝟐 = 𝒏𝒖𝒎𝒆𝒓𝒊𝒄 𝒗𝒂𝒍𝒖𝒆 , 𝒙 = ±√𝒏𝒖𝒎𝒆𝒓𝒊𝒄 𝒗𝒂𝒍𝒖𝒆 𝑥 2 − 16 = 0 → 𝑥 2 = 16
When coefficient 𝑏 = 0
√𝑥 2 = ±√16 → 𝑥 = −4 𝑜𝑟 𝑥 = 4
Apply square root property to solve
(𝑥 − 6)2 − 25 = 0 √(𝑥 − 6)2 = ±√25
→
(𝑥 − 6)2 = 25
→ 𝑥 − 6 = ±5 → 𝑥 = 6 ± 5
When the variable is in binomial form: Apply square root property to solve and get 𝑥 = 1 𝑜𝑟 𝑥 = 11
2. By COMPLETING THE SQUARE and using square root property. In this case we have to construct a Perfect Square Trinomial and make it equal to square of Binomial: 𝑏 2
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = (𝑥 + 2) . Where 𝑎 = 1 . Example 1: 𝑥 2 + 6𝑥 − 40 = 0 𝑥 2 + 6𝑥+
= 40+__
𝑥 2 + 6𝑥 + 9 = 40 + 9 (𝑥 + 3)2 = 49 𝑥 = −10 𝑜𝑟 𝑥 = 4 Example 2: 3𝑥 2 + 15𝑥 − 150 = 0 3 2 15 150 𝑥 + 𝑥− =0 3 3 3
Note that the coefficient of the term 𝑥 2 is equal to one (𝑎 = 1). In this case, move 40 to the right, and complete the left side of the equation by adding required number to create a perfect 6 2
square trinomial. In this case add (2) = 9. Then, transfer the trinomial into a square of binomial and solve applying square root property method. If the coefficient in 𝑥 2 is different than 1 (𝑎 = 3), we have to divide the equation by the value of 𝑎 coefficient, to make a new coefficient. The new numbers are 𝑎 = 1, 𝑏 =
15 3
and 𝑐 =
150 . 3
195
Business Calculus Notebook 𝑥2 +
15 25 150 25 𝑥+ = + 3 4 3 4
2
5 150(4) 25(3) 225 (𝑥 + ( )) = + =( ) 2 3(4) 4(3) 4 2
5 225 √(𝑥 + ( )) = √( ) 2 4 5
15
2
2
𝑥=− ±
Then move
150 3
to the right and add
𝟐𝟓 𝟒
in both parts to complete
the square, use the following operation:
( By adding
15 3
NEW b 2 2 𝟐𝟓 𝟒
2
15
2
5 2
) = ( 2 ) = ((3)(2)) = (2) =
𝟐𝟓 𝟒
.
to both part of the equation, we create a Perfect
Square Polynomial, and we write it in the form of Square of Binomial. Then, solve by applying Square Root Property: 𝑥 = −10 𝑜𝑟 𝑥 = 5 −𝒃±√𝒃𝟐 −𝟒𝒂𝒄
3. By using QUADRATIC FORMULA: 𝒙𝟏,𝟐 = , where a, b, and c are the coefficients (real 𝟐𝒂 2 numbers) in the quadratic equation 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0 . The next Table shows the Quadratic Formula demonstration: We will solve the quadratic equation applying the method described 𝑎 2 𝑏 𝑐 0 above in point 3 (by completing the square). To facilitate this operation, 𝑥 + 𝑥 + = 𝑎 𝑎 𝑎 𝑎 it is convenient to have leading coefficient 𝑎 = 1 (see point 3). So, we divide all elements by 𝑎, constructing a new equivalent equation
𝑥2 +
Them, move to the right the constant element; and add in both sides the equivalent value to the square of the half of the new b coefficient, 𝑏 𝑎
𝑥2 +
2
which is: ( 2 ) ; resulting the following equation: Now, on the left side we have the configuration of a Perfect Square Trinomial which is equal to the square of binomial, while on the right side we multiply and divide the last term by 4𝑎 to create a common denominator, them perform the addition of the two right elements: The new equation can be solved applying the method described above in point 2 (Square Root Property): Solving for 𝑥 we obtain the quadratic formula:
𝑥2 +
𝑏 𝑐 𝑥 + =0 𝑎 𝑎
𝑏 𝑥+ 𝑎
=
−
𝑐 𝑎
𝑏 𝑏 2 𝑏 2 𝑐 𝑥 +( ) =( ) − 𝑎 2𝑎 2𝑎 𝑎
(𝑥 +
𝑏 2 𝑏 2 (4𝑎)𝑐 ) =( ) − (4𝑎)𝑎 2𝑎 2𝑎
𝑏 2 𝑏 2 − 4𝑎𝑐 (𝑥 + ) = 2𝑎 4𝑎2 √(𝑥 +
𝑥+
𝑏 2 𝑏 2 − 4𝑎𝑐 ) = ±√ 2𝑎 4𝑎2 𝑏 ±√𝑏 2 − 4𝑎𝑐 = 2𝑎 2𝑎
4. Graphic method: Graphing the function 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 ; the solution will be the interception of the graph with axis 𝑥. 𝑥 2 − 4𝑥 − 12 = 0 → (𝑥 − 6)(𝑥 + 2) = 0 The solutions will be at 𝑥 = −2 𝑜𝑟 𝑥 = 6 Graph of 𝑦 = 𝑥 2 − 4𝑥 − 12 on the right Type equation here.
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Juan J. Prieto-Valdés
R-2.6. Radical Equations. A "radical" equation is an equation in which at least one variable expression is stuck inside a radical, usually a square root. To solve an equation with a radical expression you have to isolate the radical on one side from all other terms and then power (rice to power) both sides of the equation to the level (index) of the radical; with other words; square both sides if you are dealing with square roots. Let’s check some examples: 3
1. Simple radical value: √𝑥 = 2 or
𝑥
1⁄ 3
=2
Solve by raising both sides to the third power 3
3
( √𝑥 ) = 23 because the index of the radical is 3. (𝑥
1⁄ 3 3)
= 23
Resulting: 𝑥 = 8
2. Simple Radical Equation: √𝑥 + 8 − 2 = 𝑥 Isolate the radical. The index is 2. Therefore, raise both sides to the second power. 2
(√𝑥 + 8) = (𝑥 + 2)2
√𝑥 + 8 = 𝑥 + 2
Graphic interpretation (2 functions) 𝑦 = √𝑥 + 8 𝑦 =𝑥+2
(𝑥 + 8) = 𝑥 2 + 4𝑥 + 4 after collecting like elements, we get: 𝑥 2 + 3𝑥 − 4 = 0 Factor it and apply zero product rule to solve: (𝑥 + 4)(𝑥 − 1) = 0 so: 𝑥 = −4 𝑜𝑟 𝑥 = 1 Finally, we have to recheck if the solutions are correct: for 𝑥 = −4, doesn’t work because √−4 + 8 − 2 ≠ −4 For 𝑥 = 1, the equation √1 + 8 − 2 = 1 is correct. (Reject 𝑥 = −4) 𝑥 = 1 is the only solution
3. Two Radical Equation (generating first degree equation): √𝑥 + 10 + √𝑥 − 6 = 4 Graphic interpretation (2 functions): Isolate one of the radicals: √𝑥 + 10 = 4 − √𝑥 − 6
𝑦 = √𝑥 + 10 + √𝑥 − 6
The index is 2. Therefore, raise both sides to the second power:
𝑦=4
2
2
(√𝑥 + 10) = (4 − √𝑥 − 6 )
𝑥 + 10 = 16 − 8√𝑥 − 6 + (√𝑥 − 6 )
2
A simple radical equation can be obtained. 𝑥 + 10 = 16 − 8√𝑥 − 6 + 𝑥 − 6 Then, isolate the new radical √𝑥 − 6 and raise it one more time to the second power (to clear the radical). 0 = 8√𝑥 − 6, The resulting linear equation has a simple solution 𝑥 − 6 = 0, so:
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Business Calculus Notebook
4. Two-Radical Equation (generating a second degree equation) √𝑥 − 1 − √3𝑥 + 1 = −2 Isolate one of the radicals: √𝑥 − 1 = √3𝑥 + 1 − 2 The index is 2. Therefore, raise both sides to the second power; resulting a simple radical equation. 2
2
(√𝑥 − 1) = (√3𝑥 + 1 − 2) 2
𝑥 − 1 = (√3𝑥 + 1 ) − 4√3𝑥 + 1 + 4 𝑥 − 1 = 3𝑥 + 1 − 4√3𝑥 + 1 + 4 −2𝑥 − 6 = −4√3𝑥 + 1 Apply previously described method (raise it one more time to the second power to clear the radical) 2
(𝑥 + 3)2 = (2√3𝑥 + 1 )
𝑥 2 + 6𝑥 + 9 = 4(3𝑥 + 1) Multiply and collecting like elements: 𝑥 2 + 6𝑥 + 9 = 12𝑥 + 4 The resulting quadratic equation can be factored: 𝑥 2 − 6𝑥 + 5 = (𝑥 − 1)(𝑥 − 5) = 0 The solutions are: 𝑥 = 1 𝑜𝑟 𝑥 = 5; Both solutions work 5. Two-Radical Equation (generating a second-degree equation): 1 + √𝑥 + 4 = √3𝑥 + 1 2
2
First raise to the second power both sides of the equation: (1 + √𝑥 + 4) = (√3𝑥 + 1) 2
Then, multiply and simplify: 1 + 2√𝑥 + 4 + (√𝑥 + 4) = 3𝑥 + 1 2√𝑥 + 4 = 2𝑥 − 4 Them divide by 2 and power one more time to the second degree. Multiply, collect like terms and get new equation. 2
(√𝑥 + 4) = (𝑥 − 2)2 𝑥 + 4 = 𝑥 2 − 4𝑥 + 4, Solved by factoring: 𝑥 2 − 5𝑥 = 0 𝑥 2 − 5𝑥 = 𝑥(𝑥 − 5) = 0 so 𝑥 = 0 𝑜𝑟 5 After checking, 𝑥 = 5 and 𝑥 = 0 both work
R-2.7. Other Types of Equations Solving an algebra equation sometimes means to find the interceptions of the graph with the axis 𝑥. To solve an equation, we can create a fictitious function and analyze its graph, or we make it equal to zero (𝑦 = 0) to later apply zero-product rule. The function interpretation is useful to understand the problem. Let’s consider that our function is equal to zero: 𝑦 = 𝐴𝑙𝑔𝑒𝑏𝑟𝑎 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 = 0. If the equation is 3rd degree, for example (𝑦 = 𝑎3 𝑥 3 + 𝑎2 𝑥 2 + 𝑎1 𝑥 + 𝑎0 = 0)
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Juan J. Prieto-Valdés
Where 𝑎1 , 𝑎2 , etc… are real coefficients. This equation could have 3, 2, or 1 real solution, (how many times it cuts the axis 𝑥). On the graphs are displayed 𝑥1, 𝑥2, 𝑎𝑛𝑑 𝑥3 interceptions. Equations can be of any degree; however, in this topic, to facilitate the solution process, we will use a second-degree interpretation, factorable by substituting the variable or by creating a common term (factor that could be considered of a second or first degree). The following are some descriptive examples: 1. Third degree equation, factorable by grouping: 4𝑥 3 + 16𝑥 2 = 16𝑥 + 64 Create the function and make it equal to zero, then, factor by grouping. 4𝑥 3 + 16𝑥 2 − 16𝑥 − 64 = 0 To factor, regroup in two and two terms, then factor using a convenient Common Factor to produce the same binomial combination. 4𝑥 2 (𝑥 + 4) − 16(𝑥 + 4) = 0 equivalently to: (𝑥 + 4)(4𝑥 2 − 16) = 0. Then factor: (𝑥 + 4)(2𝑥 + 4)(2𝑥 − 4) = 0 Finally, apply zero-product rule to solve the equation. 𝑥 = −4 𝑜𝑟 𝑥 = −2 𝑜𝑟 𝑥 = 2
2. Four-degree equation, factorable using variable replacement: 𝑥 4 − 29𝑥 2 + 100 = 0 Replace the variable to convert 4th degree equation in quadratic form. ( 𝑢 = 𝑥 2 𝑜𝑟 𝑢2 = 𝑥 4 ) 𝑢2 − 29u + 100 = 0, which can ben factor as the product of two binomials: (𝑢 − 25) (𝑢 − 4) = 0 Apply zero product rule and solve: 𝑢 = 25 𝑜𝑟 𝑢 = 4 After having resulted 𝑢. Substitute these values into variable replacement equation (𝑢 = 𝑥 2 ) and get final solution using square root property 25 = 𝑥 2 𝑎𝑛𝑑 4 = 𝑥 2 𝑥 = −5 𝑜𝑟 𝑥 = 5 𝑜𝑟 𝑥 = −2 𝑜𝑟 𝑥 = 2 The equation also can be solved using direct factorization (no variable replacement): 𝑥 4 − 29𝑥 2 + 100 = (𝑥 − 5)(𝑥 + 5)(𝑥 − 2)(𝑥 + 2) = 0 Where: 𝑥 = −2 𝑜𝑟 𝑥 = 2 𝑜𝑟 𝑥 = −5 𝑜𝑟 𝑥 = 5
3. Binomial second degree, factorable using variable replacement: (3𝑥 + 3)2 − 4(3𝑥 + 3) − 12 = 0 Replace the variable to convert 4th degree equation in quadratic form: (𝑢 = 3𝑥 + 3) .
(u)2 − 4(u) − 12 = 0 Factor the new equation and solve for 𝑢
(𝑢 + 2)(𝑢 − 6) = 0
so
𝑢 = −2 𝑜𝑟 𝑢 = 6
From replacement equation we get two simple lineal equations: −2 = 3x + 3 and 6 = 3x + 3 So: 𝑥 = −
5 3
𝑜𝑟 𝑥 = 1
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Business Calculus Notebook
4. Binomial Four-degree equation, factorable using variable replacement: (𝑥 2 − 9) − 8(𝑥 2 − 9) + 12 = 0 Convert 4th degree equation in quadratic form replacing 𝑢 = 𝑥 2 − 9 . u2 − 8u + 12 = 0 Factor new equation and solve applying zero product rule 𝑢 = 2 𝑜𝑟 𝑢 = 6 After having resulted 𝑢; substitute these values into variable replacement equation (𝑢 = 𝑥 2 − 9) and get final solution using square root property. 2 = 𝑥 2 − 9 and 6 = 𝑥 2 − 9 𝑥 = √11 𝑜𝑟 𝑥 = −√11 𝑜𝑟 𝑥 = √15 𝑜𝑟 𝑥 = −√15 2
1
5. Rational equation factorable as quadratic in form: 𝑥 3 + 5𝑥 3 − 6 = 0 1
2
1
Convert radical equation in quadratic form: (𝑥 3 ) + 5 (𝑥 3 ) − 6 = 0 1
Replace 𝑢 = 𝑥 3 and get 𝑢2 + 5𝑢 − 6 = 0, solve by factoring (𝑢 + 6)(𝑢 − 1) = 0 𝑢 = −6 𝑜𝑟 𝑢 = 1 After having resulted 𝑢 ; substitute these values into variable replacement equation 1
(𝑢 = (𝑥 3 )) and get final solution using the root property. 1
1
−6 = 𝑥 3 and 1 = 𝑥 3
or
3
3
−6 = √𝑥 and 1 = √𝑥
or 𝑥 = −216 and
𝑥=1
R-2.8. Absolute Value Equations & Inequalities. Absolute value equations. To solve an absolute value equation, isolate the absolute value on one side of the equal sign, and establish (create) two cases as follow: for |𝑓(𝑥)| = 𝑎, create: Case 1:
𝑓(𝑥) = 𝑎
Case 2: 𝑓(𝑥) = −𝑎
CHECK your answers for the two "derived" equations. Also try to use graphic interpretation by plotting: 𝑦1 = 𝑓(𝑥), 𝑦2 = −𝑓(𝑥), and 𝑦3 = 𝑎, all them in the 𝑥, 𝑦 coordinate system. Check the second example below. Inequalities. To solve an inequality, we can use numeric and/or graphic interpretations: Numeric solution of an inequality is a number (or group of numbers) which when substituted for the variable makes the inequality a true statement. Graphically: 𝑔(𝑥) > 𝑓(𝑥), means, when 𝑔(𝑥) is/goes over 𝑓(𝑥); check the graph and note that the set of numbers (𝑥0 , ∞) satisfy the inequality. The figure on the right side illustrates general case, however, in this topic we will work only with linear equations. The following examples illustrate the solution procedures of Absolute value equations, and inequalities: 200
Juan J. Prieto-Valdés
1. Simple absolute value equations and its graphic interpretation 9|𝑥 − 3| = 36 → |𝑥 − 3| = 4 +(𝑥 − 3) = 4 and −(𝑥 − 3) = 4 𝑥−3=4
and
−𝑥+3=4
𝑥=7
and
𝑥 = −1
In one-line case solution: −4 = 𝑥 − 3 = 4 +3 … + 3 … + 3
Isolate the absolute value part of the equation; in this case |𝑥 − 3|. Consider that it content can be positive as well as negative by creating two new linear equations. Solve them isolating the variable. Note that the absolute value equation can be solve in one line creating a compound equation, later by adding +3 in all sides +3 for this specific case).
−1 = 𝑥 = 7 |2(𝑥 + 1) + 8| = 16 −[2(𝑥 + 1) + 8] = 16 and [2(𝑥 + 1) + 8] = 16 −2𝑥 − 10 = 16 and 2𝑥 + 10 = 16 𝑥 = −13 and 𝑥 = 3
Isolate the absolute value part of the equation; in this case |2𝑥 + 10|. Consider that it content can be negative (on the left) as well as positive (right) by creating two new linear equations. Solve each isolating the variable. We have three equations (underlined): 𝑦1 = −2𝑥 − 10 𝑦2 = 2𝑥 + 10 𝑦3 = 16 This example can be considered as a system of three linear equations. Graphing them, we can see the solution at the interceptions between 𝑦1 𝑎𝑛𝑑 𝑦3 𝐴𝑁𝐷 𝑦2 𝑎𝑛𝑑 𝑌3 .
2. Simple monomial/number |𝑥| < 2 𝑥 can be positive and negative: 𝑥 < 2
AND −𝑥 < 2
Note that −𝑥 < 2 is equivalent to 𝑥 > −2
multiplying by −1 both sides of the second inequality we get – (1)(−𝑥) < −(1)(2) and swish the inequality symbol 𝑥 > −2
Solution set: (−2 , 2) |𝑥| > 2 𝑥 can be positive and negative; in both cases greater than 2. 𝑥>2
OR −𝑥 > 2 (equivalent to 𝑥 < −2)
We have to remember the multiplication/division property of inequalities: When multiplying or dividing by a negative number, we have to change the direction of the inequality. In this case the answer (solution set) will be: (−∞, −2) 𝑜𝑟 (2, ∞)
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3. Simple Binomial |𝑥 − 6| > 2
Note when we use AND and OR, depending of the type of solution: this is a resulting interception set (in this case the solutions are in both sets) or two independent set are solutions (in this case the solution can be ascribed to one OR another set).
𝑥 − 6 > 2 AND −𝑥 + 6 > 2 𝑥>8
or −𝑥 > −4 (equivalent to 𝑥 < 4) are solutions
Previous example can be solved in one line −2 > 𝑥 − 6 > 2 +6
+6
+6
4>𝑥>8 Add 6 in each section and we get the result automatically. On the infinity side use parenthesis because we never touch the infinite. Always use parenthesis to represent greater than or less than. Solution in set notation can be written as:
When using greater than or equal the corresponding symbol will be bracket as displayed in a following supplementary example/solution: 4 ≥ 𝑥 ≥ 8 can be represented using set notation as (−∞, 4] 𝑜𝑟 [8, ∞).
(−∞, 4) 𝑜𝑟 (8, ∞).
R.2.9. Factoring Before Calculus. Do not copy the solution try to solve by yourself:) Factoring out the common greatest factor and simplify: 1.
(𝑥 2 − 3)2 (𝑥 + 5) − 3(𝑥 + 5)(𝑥 2 − 3)3 = (𝑥 2 − 3)2 (𝑥 + 5)(1 − 3(𝑥 2 − 3)) = (𝑥 2 − 3)2 (𝑥 + 5)(−3𝑥 2 + 10)
2.
(𝑥 2 − 5)2 (𝑥 + 5)−2 + (𝑥 + 5)2 (𝑥 2 − 5) = (𝑥 (𝑥 + 1)(𝑥 2 − 5))(𝑥 + 5)−2
1
3.
1
(𝑥 2 − 1)2 (𝑥 + 2) − (𝑥 + 2)2 (𝑥 2 − 1)3 (𝑥 2 − 1)
1
= −(𝑥 + 2)(𝑥2 − 1)(𝑥 (𝑥 (𝑥 + 2) − 1) − 3)
= (−1)(𝑥 − 1)(𝑥 + 1)(𝑥 + 2)(𝑥 3 + 2 𝑥 2 − 𝑥 − 3) 4.
3
5
(𝑥 3 − 2𝑥)2 (𝑥 + 7)−2 − (𝑥 + 7)2 (𝑥 3 − 2𝑥) =
𝑥 (−2 + 𝑥 2 )(−2 𝑥 + 𝑥 3 − (7 + 𝑥)4 ) 3
(7 + 𝑥)2
5
3
1
((2𝑥)2 − 3)−2 (𝑥 − 2)2 − (𝑥 − 2)((2𝑥)2 − 3)3 (4 𝑥 2 −
3 3)−2
6
9
= √(𝑥 − 2) − (𝑥 − 2)(4 𝑥2 − 3)4 1
(𝑥 3 −
2 2)3 (𝑥
+
3 5)−2
− (𝑥 +
5 5)2
(𝑥 3 −
1 2)2
(𝑥 3 − 2) ((𝑥 3 − 2)6 + (𝑥 + 5)4 ) − (𝑥 +
202
3 5)2
1
= (𝑥 3 − 2)6 + (𝑥 + 5)4
Juan J. Prieto-Valdés
Simplify complex fractions: 7.
𝑥 2 − 16 141 − 𝑥 2 5 = 𝑥−4 5 𝑥 − 20
25 −
8.
3−𝑥 ( − √𝑥 − 1) 3−𝑥 𝑥−3 2 (𝑥 − 3)(𝑥 − 2) √𝑥 − 1 =( − √𝑥 − 1) ( )= 𝑥−1 √𝑥 − 1 √𝑥 − 1 √𝑥 − 1 3−𝑥
9.
2𝑥 − 2𝑥 √𝑥 − 2 = −2𝑥 (√(𝑥 − 2) − 1) (𝑥 − 2) √𝑥 − 2 (𝑥 − 2)2
10.
(𝑥 + 2)2 𝑥 − 2 − 2 𝑥 (𝑥 − 6) 𝑥 (𝑥 − 6) 𝑥2 − 4 = − = − 2 2 𝑥 −4 (𝑥 − 2) (𝑥 + 2) (𝑥 − 2)2 (𝑥 + 2) 2
11.
𝑥 2 − 16 25 − 𝑥 + 4 (𝑥 − 9)(𝑥 2 + 25 𝑥 − 104) 141 − 𝑥 2 − 5) = ( )( 𝑥−4 5 𝑥 − 20 5 (𝑥 − 4)2
R-2-10. The Distance, Midpoint Formulas, and Circles. The Distance Formula is a variant of the Pythagorean Theorem (𝑐 2 = 𝑎2 + 𝑏 2 ) that you used before in geometry to find the length of the hypotenuse of a right triangle. Given the two points 𝑃1 (𝑥1 , 𝑦1 ) and 𝑃2 (𝑥2 , 𝑦2 ), the distance between these points is given by the formula: 𝑑2 𝑃1 −𝑃2 = (𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 or 𝑑𝑃1 −𝑃2 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 Where 𝑎 = (𝑥2 − 𝑥1 ) and 𝑏 = (𝑦2 − 𝑦1 ). Whichever one you call "first POINT" or "second" is up to you; the distance will be the same because the distance is an absolute (scalar) value, not depending of direction. Example: given two points located at (-2,1) and (3,6) find the distance between them. Applying the distance formula, we get: 𝑑 = √(3 + 2)2 + (6 − 1)2 = 5 √2
Middle Point: Middle Point of the segment line is located at: 𝑥𝑀 = In example-1, the middle point is located at: 𝑥𝑀 =
(3− 2) 2
(𝑥2 + 𝑥1 ) 2
= 0.5 and 𝑦𝑀 =
and 𝑦𝑀 = (6+ 1) 2
(𝑦2 + 𝑦1 ) 2
.
= 3.5 ;so (𝑥𝑀 , 𝑦𝑀 ) ≡ (0.5 , 3.5)
Circles: To write the equation of circle first we need to know-- “what is a circle?” If you cannot answer to this question, you never will be able to write the equation of the circle.
Circle is a line, equidistant to one fixed point called center. The fixed distance between any (𝑥. 𝑦) point of the line and the center (ℎ, 𝑘) is called the radius. We can rewrite the equation of the circle using the distance between two points: 203
Business Calculus Notebook
𝑅 = √(𝑥 − ℎ)2 + (𝑦 − 𝑘)2
or (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑅 2 ,
which is the equation of the circle in standard form. Example 1: Write the equation of the circle with center at (2, 4) and radios equal to 4. By substituting in a previously developed formula we get: (𝑥 − 2)2 + (𝑦 − 4)2 = 16. Example 2: Graph the circle given its equation in general form: 𝑥 2 + 𝑦 2 + 6𝑥 + 10 𝑦 − 15 = 0 We can provide x values and calculate y, but the work will be tedious and very consuming time. Notable easier will be by completing the square to create a two-binomial expression, with another word, re-write the equation in a standard form by completing the square. To complete 6 2 2
10 2 2
the square, we can use: ( ) = 9 and ( ) = 25 Resulting: 𝑥 2 + 6𝑥 + 9 + 𝑦 2 + 10 𝑦 + 25 = 15 + 9 + 25 or in binomial form: (𝑥 + 3)2 + (𝑦 + 5)2 = 49. As we can see, the center is located at (ℎ, 𝑘) = (−3, −5), and the radius 𝑅 = 7 units.
R-2.11. Functions and Their Graph. A function from 𝑋 into 𝑌 is a relation that associates with each element of X exactly one element of Y. The idea of a function machine is often helpful for conveying the notion of a function. The machine applies the rule that assigns to each member of a set called the domain (x) exactly one member of a set called the range (y). You can produce ONE juice from different fruits (one, two, or more flavors), but you cannot produce different types of juices (one, two, or more different juices) from the same (one) fruit. The first case is an example of a function (only one 𝒚). An equation can serve as a function if it assigns each 𝑥 value (one or more) exactly only one 𝑦 value.
Examples of lineal, quadratic and rational functions: LINEAR Functions (1st degree) one x value for each y value
y 23 x 3
=>
f ( x) 23 x 3
f (1) 20(1)2 80(1) 60 f (3) 20(3) 80(3) 60
f (3) 23 (3) 3 5
The Domain and Range will be:
x x any reals}
RATIONAL Functions
f ( x) 20 x 2 80 x
f ( x) 23 x 3
Domain: 204
QUADRATIC Functions (2nd degree function) two x values for each y value
2
{0 x 4} and {0 y 80}
f ( x) f (2)
D:
2x 1 x2
2x 1 undefined ! (2) 2
{x x reals, x 2}
Juan J. Prieto-Valdés
R-2.12 Graphing Technics and Transformation of Functions. Translating a graph vertically (up / down) Given Original Simple Function: 𝑦 = 𝑥 2. When 𝑦 = 0, we get: 𝑥 = 0. One solution, one interception with axis 𝑥 at (0,0) Same function adding two units: 𝑦 = 𝑥 2 + 2. In this case, you have to move the graph two units up, as we are subtracting two units to the 𝑦 value ( 𝑦 − 2 = 𝑥 2 ). When 𝑦 = 0 we get complex solutions (no real solution because no interception with axis 𝑥). At 𝑥 2 + 2 = 0, we get (𝑥 2 = −2) two solutions: 𝑥 = ±√2 𝑖 Same function subtracting two units: 𝑦 = 𝑥 2 − 2 . Move two unit down as we are adding two units to the 𝑦 ( 𝑦 + 2 = 𝑥 2 ). When 𝑦 = 0 (𝑥 2 − 2 = 0), we get two real solutions, means two interceptions with axis 𝑥 at: 𝑥 =
𝑦 = 𝑥2 + 2 2
𝑦 = 𝑥 −2
𝑦=𝑥
2
√2 𝑜𝑟 𝑥 = −√2
Translating a graph horizontally (to the right or to the left) 2
𝑦 = 𝑥 , adding/subtracting two (2) units to the 𝑥: When subtracting: 𝑦 = (𝑥 − 2)2 , you have to move the graph two units to the right. If adding, 𝑦 = (𝑥 + 2)2 , you have to move the graph two units to the left. 𝑦 = (𝑥 + 2)2 𝑦 = 𝑥2 𝑦 = (𝑥 − 2)2
Stretch or Compress 2
In 𝑦 = 𝑎𝑥 , if a>1, stretch. If a 0) to satisfy the condition of the existence of conjugates. For real; solutions (𝑏 = 0); in this case previous trinomial becomes a perfect square trinomial with two identical solutions coming from the square of binomial (𝑥 − 𝑎)2 = (𝑥 − 𝑎)(𝑥 − 𝑎). Example: Factor 𝒇(𝒙) = 2 𝑥 5 + 17 𝑥 4 + 49 𝑥 3 + 38 𝑥 2 − 46 𝑥 − 60 Factors of 2 are: ± 1; ±2. Factors of 60 are: ± 1; ±2; ±3; ±4; ±5; ±6; 𝑒𝑡𝑐 3 2
The possible root are: ± 1; ± ; ±2; ±3; 𝑒𝑡𝑐 (Let’s try with the first four root, if needed later we can 3
recalculate other roots): 𝑓(1) = 0; (𝑂𝐾 𝑖𝑡 𝑤𝑜𝑟𝑘𝑠); 𝑓(−1) ≠ 0; 𝑓(2) = 0 (𝑂𝐾); 𝑓(−2) ≠ 0; 𝑓 (− 2) = 0 (𝑂𝐾) We found 3 roots. if needed we can calculate more. Use divisor {1} Use divisor {−2}
1/ -2/
2 2
17 49 38 -46 -60 2 19 68 106
60
19 68 106 60
0
(𝑥 − 1) (𝑥 + 2)
-4 -30 -76 -60 Use divisor {-3/2 }
3
−2
2
15 38 6
−2 − 2
36 2
12 20
30 −
0
(2𝑥 + 3)
60 2
0
(2𝑥 2 + 12𝑥 + 20)
So: (2𝑥 2 + 12𝑥 + 20) can be factored using the product rule for conjugates (Example #7 above) Finally, we can write: 𝑓(𝑥) = 2(𝑥 2 + 6𝑥 + 10)(𝑥 − 1)(𝑥 + 2)(2𝑥 + 3) = 2(𝑥 + 3 + 𝑖)(𝑥 + 3 − 𝑖)(𝑥 − 1)(𝑥 + 2)(2𝑥 + 3) 2
The solutions are: 𝑥 = −3 − 𝑖. 𝑥 = −3 + 𝑖, 𝑥 = 1, 𝑥 = −2, 𝑎𝑛𝑑 𝑥 = − 3 .
Steps for Finding the Real Zeros of a Polynomial Function. When graphing a Polynomial Function is convenient to follow the following steps: 1. Determine the end behavior of the graph as per leading term (coefficient) value 2. Determine the y-intercept by evaluating f(0) 3. Determine real zeros, evaluating (solving) the equation at y=0, apply Factor and Remainder Theorems if needed, also if needed use long division or synthetic division as you feel more comfortable, and 4. Determine their Multiplicity 5. Plot all intercepts with x and y axis and sketch the graph character as per zeros multiplicity that you have determined before. 6. Use Symmetry characteristics and some testing points (if needed) to construct the graph. 216
Juan J. Prieto-Valdés
R-2.19. Property and Graph of Rational Functions. A function 𝑓(𝑥) is called a rational function if and only if it can be written in the form of 𝑃(𝑥)
𝑅(𝑥) = 𝑄(𝑥) , where 𝑃 and 𝑄 are polynomials in 𝑥 and 𝑄 never equal to zero. The Domain of 𝑓(𝑥) is the set of all points 𝑥 for which the denominator 𝑄(𝑥) is not zero, assuming 𝑃 and 𝑄 have no common factors. The values of 𝑥 that make the denominator equal to zero, are “prohibit”, as the rational function gets undefined value when division by zero is performing, the rational function goes to − ∞ or to + ∞ as 𝑥 1
1
approaches this value from positive or negative side, similar to the functions 𝑥 or 𝑥 2 (on the right). These are two standard simple cases (division by 𝑥 𝑎𝑛𝑑 𝑏𝑦 𝑥 2 ), is not allowed when 𝑥 approaches 0 (𝑧𝑒𝑟𝑜); under this condition both graph approach 𝑦 𝑎𝑥𝑖𝑠 but never touch it. In this case, axis 𝑦 is the vertical asymptote. If any 𝑥 − 𝑣𝑎𝑙𝑢𝑒 makes the denominator equal to zero, in this position will be located the Vertical Asymptote.
Descriptive examples: Function
Denominator analysis
𝑓(𝑥) =
5 𝑥−2
𝑓(𝑥) =
𝑥2
𝑥 − 4𝑥 − 12
𝑥 2 − 4𝑥 − 12 =
𝑥−2=0
(𝑥 − 6) (𝑥 + 2) = 0
𝑥≠2
𝑥 ≠ 6 𝑜𝑟 𝑥 ≠ −2
Domain
(−∞, 2) ∪ (2, ∞)
(−∞, −2) ∪ (−2, 6) ∪ (6, ∞)
Vertical asymptote
𝑥=2
𝑥 = 6 𝑎𝑛𝑑 𝑥 = −2
Horizontal asymptote
𝑦ℎ.𝑎. = 0
𝑦ℎ.𝑎. = 0
2𝑥 2 𝑥2 + 4 𝑥 2 + 4 = 0 produces imaginary solution (𝑥 2 + 4 ≠ 0) 𝑓(𝑥) =
𝑥 = 2𝑖 𝑜𝑟 𝑥 = −2𝑖 All Real numbers {R} 𝑦ℎ.𝑎. = 2 𝑎
The position of the Horizontal Asymptote can be found per the next rule:
If 𝑛 = 𝑚
the Horizontal Asymptote is located at 𝑦ℎ.𝑎. = 𝑏
If 𝑛 > 𝑚
there is NO Horizontal Asymptote (will be slant asymptote)
𝑎𝑥 𝑛 + ⋯ 𝑏𝑥 𝑚 + ⋯ 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 exist when 𝑦 = 0 (the numerator of the rational function is equal to 0.
If 𝑛 < 𝑚
the Horizontal Asymptote is located at 𝑦ℎ.𝑎. = 0
𝑓(𝑥) =
𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 , it exists when 𝑥 = 0.
No 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡
𝑦=−
5 2
𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡. In 𝑥 = 0
𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡. In 𝑥 = 0
𝑦=0
𝑦=0
Graph
217
Business Calculus Notebook
When graphing rational function, pay attention to the following steps: 1. 2. 3. 4. 5.
Domain, Vertical, Horizontal and/or Slant Asymptotes, X and Y Intercepts, Increasing and Decreasing Intervals, and Evaluate some testing points. Graph: 𝑦
=
Quick example. Determine the following characteristics:
4 𝑥 2 +1 𝑥 2 −4𝑥−12
Domain: (make the denominator equal to zero to analyze undefined situations) 𝑥 2 − 4𝑥 − 12 = (𝑥 − 6) (𝑥 + 2) = 0. So, the Domain will be: (∞, −2) ∪ (6. ∞); and correspondently, in 𝑥 = −2 𝑎𝑛𝑑 6 are located the Vertical Asymptotes the Horizontal 4𝑥 2 =4; 1𝑥 2
Asymptote is located at:
at 𝑦 = 4 1
𝒀 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕, is located at 𝑓(𝑥 = 0) = − 12 𝑿 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕𝒔, are located at 𝑓(𝑥) = 0, 4𝑥 2 + 1 = (2𝑥 − 1)(2𝑥 + 1) = 0 1 1 𝑆𝑜 𝑥 = − 𝑎𝑛𝑑 𝑥 = 2 2
R-2.21. Polynomial and Rational Inequalities Example 1: Solve the following polynomial inequality 𝑥 3 − 6𝑥 2 ≥ 9 𝑥 − 54: Using Wolfram Alpha, we can visualize the interceptions between two functions (upper figure): 𝑓1 (𝑥) = 𝑥 3 − 6 𝑥 2 𝑎𝑛𝑑 𝑓2 (𝑥) = 9𝑥 − 54 The interceptions between these two functions are the borders of the sectors were the statement expressed by the given inequality is true. To provide effective solution we will apply more traditional method by creating a unique function: 𝑓(𝑥) = 𝑥 3 − 6 𝑥 2 − 9 𝑥 + 54 ≥ 0 Factor by grouping and solve the equation: 𝑓(𝑥) = 𝑥 3 − 6 𝑥 2 − 9 𝑥 + 54 = (𝑥 − 6)(𝑥 − 3)(𝑥 + 3) = 0 𝑥 = −3, 𝑥 = 3, 𝑎𝑛𝑑 𝑥 = 6 Evidently, there are three points where the function 𝑓(𝑥) can change its sign from positive to negative and vice versa. In this case we have four intervals where the left part of the inequality may by greater or lesser than the right part of the given inequality. Also, by testing these intervals, we can find the solution: [−3, 3] ∪ [6, ∞) as displayed in the graph. Intercepts Testing Points Plugin the values of 𝒙 Statement 218
Interval-1
-3
𝑥 = −5 (−5)3
Interval-2
3
𝑥=0 2
− 6(−5)
(0)3
Interval-3
6 Interval-4
𝑥=4 2
− 6(0)
(4)3
𝑥 = 10 2
≥ 9 (−5) − 54
≥ 9(0) − 54
− 6(4) ≥ 9 (4) − 54
False
True: [−3, 3]
False
(10)3 − 6 (10)2 True: [6, ∞)
≥ 9 (10) − 54
Juan J. Prieto-Valdés
To graph the solution, we consider the segments were the inequality statement has true value. (𝑓(𝑥) ≥ 0) in highlighted regions of the graph. Note that for set representation we have to use parenthesis if we do not touch the point (greater or less than) and bracket when touching the point (les tan or equal OR greater than or equal): Example 2: Solve the following rational inequality: 1 𝑥 2 −4𝑥−12
>
1 6−𝑥
Firstly, create the function, arrange and make it equal to zero to analyze 1
𝑦=
[−3, 3] ∪ [6, ∞)
1 − 4𝑥 − 12 1 y= 6−𝑥 𝑥2
1
𝑓(𝑥) = (𝑥 2 −4𝑥−12) + (𝑥−6) = 0. Factor and create a common denominator (multiply as needed) 1 (x−6)(x+2)
1
+ (𝑥−6) = 0;
1 (x−6)(x+2)
1(𝑥+2)
+ (𝑥−6)(𝑥+2) = 0
Them, simplify and solve the resulting equation: 𝑥 + 3 = 0 We get: 𝑥 = −3. As we can verify, only a number greater than -3 satisfy given inequality. So, the solution set will be: (−3, ∞) After 𝑥 = −3 one part of the first rational function goes over the second one, as illustrated on the right side.
Another solution can be represented by graphing the algebraic solution: 𝑦 = 𝑥 + 3. In this case, the graph will show that
after 𝑥 = −3 the resulting function is positive (goes over the 𝑎𝑥𝑖𝑠 𝑥).
Previously included review is recommended to update yourself before working with business calculus. Sincerely yours Prof. Prieto-Valdes
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Tentative course content / schedule of regular course Date
Topic No.
Topic Title
1.1.
The Idea of a Limit
1.2.
Formal Limit Definition
1.3.
Algebraic Properties of Limits (formal notation)
1.4.
Continuity. and One-Side Limits:
1.5.
Limits at the Infinity (Analyzing also One Side Limit)
1.6.
Example of Limit Calculation Using Technology
2.1.
Definition of Derivative
Test-1 2.2.
Techniques of differentiation.
2.3.
Product and Quotient Rules
2.4.
The Chain Rule
2.5.
Implicit Differentiation.
2.6.
High Order Derivatives (second, third, …):
2.7.
Real Case Examples (Marginal Analysis and Approximations)
2.8.
Increasing and Decreasing Functions (critical points)
2.9.
Rolle's Theorem and The Mean Value Theorem
2.10.
Concavity and Point of Inflexion
Test-2 3.1.
Remembering Exponential and Logarithmic Functions
3.2.
Differentiation of Exponential Functions
3.3.
Differentiation of Logarithmic Functions
3.4.
L’ Hospital Rule / Revisiting Limits
3.5.
Optimization and Business Applications Examples.
Test-3 4.1.
Indefinite Integration and Simple Differential Equation
4.2.
Integration by substitution
4.3.
Riemann Integration / Riemann Sum.
4.4.
Fundamental Theorem of Calculus and Definite Integral
4.5.
Application of Integrals (Average Function Value).
4.6.
Application of Integrals (Area Under and Between Curves).
Test-4 220
Juan J. Prieto-Valdés
221