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: 42

Answer

$t=\pm\frac{2\sqrt2}{3}$ and $t=\pm1$ are critical numbers of $h$.

Work Step by Step

How to find the critical numbers of a function $f$ according to the 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 $f$. $$h(t)=3t-\arcsin t$$ The domain of $h$ here is, in fact, the domain of $\arcsin t$. Therefore, $$D_h=[-1, 1].$$ 1) Now, we find $h'(t)$ $$h'(t)=3-\frac{1}{\sqrt{1-t^2}}\hspace{2cm}(\frac{d(\arcsin t)}{dt}=\frac{1}{\sqrt{1-t^2}}).$$ We find $h'(t)=0$, $$h'(t)=0$$ $$\frac{1}{\sqrt{1-t^2}}=3$$ $$\sqrt{1-t^2}=\frac{1}{3}$$ $$1-t^2=\frac{1}{9}$$ $$t^2=\frac{8}{9}$$ $$t=\pm\frac{2\sqrt2}{3}$$ These values belong to $D_h$, so they are critical numbers of $h$. Also, looking at $$h'(t)=3-\frac{1}{\sqrt{1-t^2}}$$ we find that for $t=\pm1$, $h'(t)$ does not exist. Since they also belong to $D_h$, $t=\pm1$ are critical numbers of $h$. We conclude that $t=\pm\frac{2\sqrt2}{3}$ and $t=\pm1$ are critical numbers of $h$.
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