Calculus 8th Edition

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

Chapter 3 - Applications of Differentiation - 3.1 Maximum and Minimum Values - 3.1 Execises - Page 212: 46


Absolute maximum: $f(3)=113$ Absolute minimum: $f(0) = 5$

Work Step by Step

Using closed interval method, as $f$ in continuous over the given range, let's first calculate the value of $f$ at critical numbers followed by values of $f$ at endpoints of the given range. For critical numbers, $f'(x)=0$ or $f'(x)$ should not exist. Let's differentiate $f(x)=5+54x-2x^3$ with respect to $x$ keeping $[0,4]$ as interval. $$\frac{df}{dx}=54-6x^2$$ $\frac{df}{dx}$ exists on the given interval. Thus, $\frac{df}{dx}=0$, $$54-6x^2=0\implies x^2=9.$$ $$x=-3,3$$ Thus, $f(3)=5+54(3)-2(3)^3=113$. Not considering $x=-3$ because not in given interval. At endpoints, $f(0)=5$ & $f(4)=93$. Thus, absolute maximum $=f(3)=113$ and absolute minimum $=f(0)=5$.
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.