Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - Review - Exercises - Page 286: 1

Answer

$$ f(x)=x^{3}-9 x^{2}+24 x-2, \quad [0,5], $$ The local maximum value is $$ f(2)=18$$ and the local minimum value is $$f(4)=14 $$ The values of $f$ at the endpoints of the interval are $$ f(0)=-2 \text { and } f(5)=18 $$ Thus, $$ f(0)=-2 $$ is the absolute minimum value, $$ f(2)=f(5)=18 $$ is the absolute maximum value.

Work Step by Step

$$ f(x)=x^{3}-9 x^{2}+24 x-2, \quad [0,5], $$ Since $ f$ is continuous on [0,5],we can use the Closed Interval Method $$ f(x)=x^{3}-9 x^{2}+24 x-2 $$ $$ \begin{aligned} f^{\prime}(x) &=3 x^{2}-18 x+24 \\ & =3\left(x^{2}-6 x+8\right) \\ &=3(x-2)(x-4) \end{aligned} $$ Since $f^{\prime}(x)$ exists for all $x$ the only critical numbers of $f$ occur when $f^{\prime}(x)=0$ that is, $$ x=2 \text { or } x=4 . $$ For $0 \lt x \lt 2 $ $$ f^{\prime}(x) \gt 0$$ For $ 2 \lt x \lt 4 $ $$f^{\prime}(x) \lt 0$$ For $ 4 \lt x \lt 5 $ $$f^{\prime}(x) \gt 0$$ so $$f(2)=18$$ is a local maximum value and $$f(4)=14 $$ is a local minimum value. The values of $f$ at the endpoints of the interval are $$ f(0)=-2 \text { and } f(5)=18 $$ Thus, $$ f(0)=-2 $$ is the absolute minimum value, $$ f(2)=f(5)=18 $$ is the absolute maximum value.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.