Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.1 - Maximum and Minimum Values - 4.1 Exercises - Page 283: 29


$c=\frac{1}{3}$ is the only critical number of $f$.

Work Step by Step

How to find the critical numbers of a function according to definition 1) Find all numbers of $c$ satisfying either $f'(c)=0$ or $f'(c)$ does not exist. 2) See that all the $c$ we get lie in the domain of $f$ $(D_f)$ or not. - If $c$ lies in $D_f$, $c$ is a critical number of $f$. - If $c$ does not lie in $D_f$, $c$ is not a critical number of $f$. $$f(x)=4+\frac{1}{3}x-\frac{1}{2}x^2$$ $D_f=R$ First, we find $f'(x)$ $$f'(x)=0+\frac{1}{3}\times1-\frac{1}{2}\times2x$$ $$f'(x)=\frac{1}{3}-x$$ 1) Find $f'(c)=0$ $$f'(c)=0$$ $$\frac{1}{3}-c=0$$ $$c=\frac{1}{3}$$ There are no instances of $c$ where $f'(c)$ does not exist in this case. 2) Examine whether $c$ lies in $D_f$ or not. We see that $c=\frac{1}{3}\in R$, so $c=\frac{1}{3}$ lies in $D_f$. We conclude that $c=\frac{1}{3}$ is a critical number of $f$.
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.