Explanation Given: The function is y = | 3 x − 9 | and the interval is [ 0 , 4 ] . Concept used: Critical points: For a function f ( x ) , critical points are the points at which the value of f ′ ( x ) is zero or f ′ ( x ) does not exists. Calculation: The given function is expressed as shown below: y = | 3 x − 9 | …..(1) Critical points are the points at which the value of f ′ ( x ) is zero or f ′ ( x ) does not exists. The given function is a modulus function and a modulus function is expressed as follows: y = { x x ≥ 0 − x x < 0 Similarly, the given function is expressed as follows: y = { 3 x − 9 x ≥ 3 − ( 3 x − 9 ) x < 3 For x ≥ 3 the derivative of the given function is calculated as shown below: y ′ = d ( 3 x − 9 ) d x = 3 This implies that in the interval [ 3 , ∞ ) the first derivative of the function is y ′ = 3 . For x ≤ 3 the derivative of the given function is calculated as shown below: y ′ = d ( − ( 3 x − 9 ) ) d x = d ( − 3 x + 9 ) d x = − 3 This implies that in the interval ( − ∞ , 3 ] the first derivative of the function is y ′ = − 3

4th Edition

ISBN: 9781319055905

Author: FRANZOSA

Publisher: VST

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Question

Chapter 4.2, Problem 66E

To determine

To find:

The critical points and the extreme values of the function y=|3x−9| on the interval [0,4].

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Chapter 4.2, Problem 65E

Chapter 4.2, Problem 67E

Students have asked these similar questions

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

Exercises 101–103 will help you prepare for the material covered in the next section. Use the graph of function f to solve each exercise. 5- 4- 3- 2- 1- -5-4 1 2 3 45 y = flx) 101. For what values of x is the function undefined? 102. Write the equation of the vertical asymptote, or the vertical line that the graph of f approaches but does not touch. 103. Write the equation of the horizontal asymptote, or the horizontal line that the graph of f approaches but does not touch.

In Exercises 27–58, find the critical points and the intervals on which the function is increasing or decreasing. Use the First Derivative Test to determine whether the critical point yields a local min or max (or neither).

Chapter 4 Solutions

EBK CALCULUS: EARLY TRANSCENDENTALS

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Prob. 13CRE Ch. 4 - Prob. 14CRE Ch. 4 - Prob. 15CRE Ch. 4 - Prob. 16CRE Ch. 4 - Prob. 17CRE Ch. 4 - Prob. 18CRE Ch. 4 - Prob. 19CRE Ch. 4 - Prob. 20CRE Ch. 4 - Prob. 21CRE Ch. 4 - Prob. 22CRE Ch. 4 - Prob. 23CRE Ch. 4 - Prob. 24CRE Ch. 4 - Prob. 25CRE Ch. 4 - Prob. 26CRE Ch. 4 - Prob. 27CRE Ch. 4 - Prob. 28CRE Ch. 4 - Prob. 29CRE Ch. 4 - Prob. 30CRE Ch. 4 - Prob. 31CRE Ch. 4 - Prob. 32CRE Ch. 4 - Prob. 33CRE Ch. 4 - Prob. 34CRE Ch. 4 - Prob. 35CRE Ch. 4 - Prob. 36CRE Ch. 4 - Prob. 37CRE Ch. 4 - Prob. 38CRE Ch. 4 - Prob. 39CRE Ch. 4 - Prob. 40CRE Ch. 4 - Prob. 41CRE Ch. 4 - Prob. 42CRE Ch. 4 - Prob. 43CRE Ch. 4 - Prob. 44CRE Ch. 4 - Prob. 45CRE Ch. 4 - Prob. 46CRE Ch. 4 - Prob. 47CRE Ch. 4 - Prob. 48CRE Ch. 4 - Prob. 49CRE Ch. 4 - Prob. 50CRE Ch. 4 - Prob. 51CRE Ch. 4 - Prob. 52CRE Ch. 4 - Prob. 53CRE Ch. 4 - Prob. 54CRE Ch. 4 - Prob. 55CRE Ch. 4 - Prob. 56CRE Ch. 4 - Prob. 57CRE Ch. 4 - Prob. 58CRE Ch. 4 - Prob. 59CRE Ch. 4 - Prob. 60CRE Ch. 4 - Prob. 61CRE Ch. 4 - Prob. 62CRE Ch. 4 - Prob. 63CRE Ch. 4 - Prob. 64CRE Ch. 4 - Prob. 65CRE Ch. 4 - Prob. 66CRE Ch. 4 - Prob. 67CRE Ch. 4 - Prob. 68CRE Ch. 4 - Prob. 69CRE Ch. 4 - Prob. 70CRE Ch. 4 - Prob. 71CRE Ch. 4 - Prob. 72CRE Ch. 4 - Prob. 73CRE Ch. 4 - Prob. 74CRE Ch. 4 - Prob. 75CRE Ch. 4 - Prob. 76CRE Ch. 4 - Prob. 77CRE Ch. 4 - Prob. 78CRE Ch. 4 - Prob. 79CRE Ch. 4 - Prob. 80CRE Ch. 4 - Prob. 81CRE Ch. 4 - Prob. 82CRE Ch. 4 - Prob. 83CRE Ch. 4 - Prob. 84CRE Ch. 4 - Prob. 85CRE Ch. 4 - Prob. 86CRE Ch. 4 - Prob. 87CRE Ch. 4 - Prob. 88CRE Ch. 4 - Prob. 89CRE

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The critical points and the extreme values of the function y = | 3 x − 9 | on the interval [ 0 , 4 ] .

Solution Summary: The author explains the critical points and extreme values of the function y=left|3x-9right| on the interval.

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The critical points and the extreme values of the function y=|3x−9| on the interval [0,4].

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Chapter 4 Solutions

The critical points and the extreme values of the function y = | 3 x − 9 | on the interval [ 0 , 4 ] .

Solution Summary: The author explains the critical points and extreme values of the function y=left|3x-9right| on the interval.

Concept explainers

Videos

To find: The critical points and the extreme values of the function y=|3x−9| on the interval [0,4].

Want to see the full answer?

Chapter 4 Solutions

To find:

The critical points and the extreme values of the function y=|3x−9| on the interval [0,4].