Explanation Given information: The provided function is: y = ( x − 5 ) 2 ( x + 4 ) ( x − 2 ) Calculation: Consider the function: y = ( x − 5 ) 2 ( x + 4 ) ( x − 2 ) Differentiate above equation with respect to x to get: d y d x = d d x ( ( x − 5 ) 2 ( x + 4 ) ( x − 2 ) ) = 2 ( x − 5 ) ( x + 4 ) ( x − 2 ) + ( x − 5 ) 2 ( 1 ) ( x − 2 ) + ( x − 5 ) 2 ( x − 2 ) ( 1 ) = ( x − 5 ) [ 2 ( x + 4 ) ( x − 2 ) + ( x − 5 ) ( x − 2 ) + ( x − 5 ) ( x − 2 ) ] = ( x − 5 ) [ 2 x 3 + 8 x − 4 x − 8 + x 2 − 5 x − 2 x + 10 + x 2 − 5 x − 2 x + 10 ] Thus, d y d x = ( x − 5 ) ( 4 x 2 − 4 x − 26 ) 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-5 ) ^2* ( X+4 ) * ( X-2 ) . Step 4: Press “WINDOW” key and set Xmin = − 5 and Xmax = 6 . 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 ( X-5 ) * ( 4X^2-4X-26 ) . Step 4: Press “WINDOW” key and set Xmin = − 4 and Xmax = 4 . 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 and enter Y 1 = ( X-5 ) * ( 4X^2-4X-26 ) . Step 3: Press “WINDOW” key and set Xmin = − 5 and Xmax = 6 . Step 4: Press “GRAPH” key. Step 5: 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 ≈ − 2.10 , x ≈ 3

Finite Mathematics and Applied Calculus (MindTap Course List)

7th Edition

ISBN: 9781337274203

Author: Stefan Waner, Steven Costenoble

Publisher: Cengage Learning

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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 47E

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To calculate: The x− coordinates of the critical points, accurate to two decimal places for the function y=(x−5)2(x+4)(x−2) with domain [−5,6] and all relative and absolute maxima and minima by using graphing technology.

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Chapter 12.1, Problem 46E

Chapter 12.1, Problem 48E

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Exercises 121–140: (Refer to Examples 12–14.) Complete the following for the given f(x). (a) Find f(x + h). (b) Find the difference quotient of f and simplify. 121. f(x) = 3 122. f(x) = -5 123. f(x) = 2x + 1 124. f(x) = -3x + 4 %3D 125. f(x) = 4x + 3 126. f(x) = 5x – 6 127. f(x) = -6x² - x + 4 128. f(x) = x² + 4x 129. f(x) = 1 – x² 130. f(x) = 3x² 131. f(x) = 132. /(x) 3D글 = = 132. f(: 133. f(x) = 3x² + 1 134. f(x) = x² –- 2 135. f(x) = -x² + 2r 136. f(x) = -4xr² + 1 137. f(x) = 2x - x +1 138. f(x) = x² + 3x - 2 139. f(x) = x' 140. f(x) = 1 – x

In Exercises 57–62, find the zeros of ƒ and sketch its graph by plotting points. Use symmetry and increase/decrease information where appropriate. 57. f(x) — х? — 4 58. f(x) = 2x2 – 4 %3D %3D 59. f(x) — х3 — 4х 60. f(x) — х3 61. f(x) =2 – x3 62. f(x) = (x – A)¾i+ate Wind

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The x − coordinates of the critical points, accurate to two decimal places for the function y = ( x − 5 ) 2 ( x + 4 ) ( x − 2 ) with domain [ − 5 , 6 ] 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−5)2(x+4)(x−2) with domain [−5,6] and all relative and absolute maxima and minima by using graphing technology.

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