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: 34

Answer

$c=\frac{4}{3}$ is the only critical number of $g$.

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 $f$. $$g(t)=|3t-4|$$ $D_g=R$ First, we need to deal with the absolute value. $$t\geq\frac{4}{3}\hspace{0.5cm}then\hspace{0.5cm}|3t-4|=3t-4 \\ t\lt\frac{4}{3}\hspace{0.5cm}then\hspace{0.5cm}|3t-4|=4-3t$$ 1) Now, we find $g'(t)$ - For $t\gt\frac{4}{3}$, $$g'(t)=(3t-4)'=3$$ - For $t\lt\frac{4}{3}$, $$g'(t)=(4-3t)'=-3$$ We see that there are no instances of $c$ where $g'(c)=0$ in this case. However, since there are 2 different values of $g'(t)$ for $t\gt\frac{4}{3}$ and $t\lt\frac{4}{3}$, $g'(\frac{4}{3})$ does not exist. 2) Examine whether $c$ lies in $D_g$ or not. We see that $c=\frac{4}{3}\in R$, so $c=\frac{4}{3}$ lies in $D_g$. We conclude that $c=\frac{4}{3}$ is a critical number of $g$.
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