Explanation Given information: The provided function is: y = ( x + 3 ) 2 ( x − 2 ) 2 Calculation: Consider the function: y = ( x + 3 ) 2 ( x − 2 ) 2 Differentiate above equation with respect to x to get: d y d x = d d x ( ( x + 3 ) 2 ( x − 2 ) 2 ) = 2 ( x + 3 ) ( x − 2 ) 2 + ( x + 3 ) 2 2 ( x − 2 ) = 2 ( x + 3 ) ( x − 2 ) [ ( x − 2 ) + ( x + 3 ) ] = 2 ( x + 3 ) ( x − 2 ) ( 2 x + 1 ) Thus, d y d x = 2 ( x + 3 ) ( x − 2 ) ( 2 x + 1 ) Draw the graph of f ( x ) by using Ti-83 calculator and the steps are as follows: Step 1: Open Ti-83 calculator. Step 2: Press “Y=” key. Step 3: Press ( x + 3 ) ^ 2 * ( x − 2 ) ^ 2 . Step 4: Press “WINDOW” key and set Xmin = − 5 and Xmax = 5 . Step 5: Press “GRAPH” key. The graph of f ( x ) is shown below: Draw the graph of f ′ ( x ) by using Ti-83 calculator and the steps are as follows: Step 1: Open Ti-83 calculator. Step 2: Press “Y=” key. Step 3: Press 2 ( X+3 ) * ( X-2 ) * ( 2X+1 ) . Step 4: Press “WINDOW” key and set Xmin = − 5 and Xmax = 5 . Step 5: Press “GRAPH” key. The graph of f ′ ( x ) is shown below: Since, the stationary points are those points where the graph of f ′ ( x ) cuts the x − axis, thus the use the zoom in feature of the calculator to find the values of x where y = 0 and the steps are as follows: Step 1: Open Ti-83 calculator. Step 2: Press “Y=” key. Step 3: Press 2 ( X+3 ) * ( X-2 ) * ( 2X+1 ) . Step 4: Press “WINDOW” key and set Xmin = − 5 and Xmax = 5 . Step 5: Press “GRAPH” key. Step 6: Press “Zoom IN” key and move the cursor where Y=0 , and then press “ENTER” key. The results are illustrated below: And Also, Thus, the stationary points are x = − 3 , x = − 0.5 and x = 2

Bundle: Finite Mathematics and Applied Calculus, Loose-leaf Version, 7th + WebAssign Printed Access Card for Waner/Costenoble's Finite Math and Applied Calculus, 7th Edition, Single-Term

7th Edition

ISBN: 9781337604963

Author: Stefan Waner, Steven Costenoble

Publisher: Cengage Learning

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Domain

In simple mathematics terms, a domain is defined as the space on which all the possible inputs for the function will lie. It can also be expressed as a set of departures for the required function.

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Chapter 12.1, Problem 48E

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To calculate: The x− coordinates of the critical points, accurate to two decimal places for the function y=(x+3)2(x−2)2 with domain [−5,5] and all relative and absolute maxima and minima by using graphing technology.

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Chapter 12.1, Problem 47E

Chapter 12.1, Problem 49E

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The x − coordinates of the critical points, accurate to two decimal places for the function y = ( x + 3 ) 2 ( x − 2 ) 2 with domain [ − 5 , 5 ] and all relative and absolute maxima and minima by using graphing technology.

Solution Summary: The author calculates the x- coordinates of the critical points, accurate to two decimal places, using graphing technology.

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To calculate: The x− coordinates of the critical points, accurate to two decimal places for the function y=(x+3)2(x−2)2 with domain [−5,5] and all relative and absolute maxima and minima by using graphing technology.

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Chapter 12 Solutions