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 211: 34

Answer

Critical number: $t=\frac{4}{3}$.

Work Step by Step

The given function is: $g(t)=|3t-4|$ When $3t-4 \lt 0$ or $t\lt\frac{4}{3}$ then $g(t)=-(3t-4)=-3t+4$ When $3t-4 \gt 0$ or $t\gt \frac{4}{3}$ then $g(t)=3t-4$ When $3t-4 = 0$ or $t=\frac{4}{3}$ then $g(t)=0$ Notice that all the linear functions are differentiable on $\mathbb R$. The critical number $c$ of $g$ should satisfy the condition $g'(c)=0$. So when $3t-4 \lt 0$ or $t\lt\frac{4}{3}$ the first derivative of $g$ with respect to $t$ is $-3$. So when $3t-4 \gt 0$ or $t\gt\frac{4}{3}$ the first derivative of $g$ with respect to $t$ is $3$. When $3t-4 = 0$ or $t=\frac{4}{3}$ then $g'(t)=0$ Since $g'(t)=0$ for $t=\frac{4}{3}$ it follows that $g$ have critical number at $t=\frac{4}{3}$.
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