Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 4 - Applications of the Derivative - 4.2 Extreme Values - Exercises - Page 181: 11

Answer

$t= 3,-1$

Work Step by Step

Given $$f(t)=t-4 \sqrt{t+1}$$ Since \begin{align*} f'(t)&= 1-4\frac{1}{2\sqrt{t+1}}\\ &=1-\frac{2}{\sqrt{t+1}}\\ &=\frac{\sqrt{t+1}-2}{\sqrt{t+1}} \end{align*} Then to find the critical points \begin{align*} f'(t)&=0\\ \frac{\sqrt{t+1}-2}{\sqrt{t+1}}&=0\\ \sqrt{t+1}-2&=0\\ t+1&=4 \end{align*} Hence $f(t)$ has a critical point at $t=3$ In addition, $t=-1$ is also a critical point because $f'(t)$ is undefined there (division by zero).
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