University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.1 - Extreme Values of Functions - Exercises - Page 216: 68

Answer

At $x=0$, $y$ has the absolute and local maximum value of $\pi/2$. There is no absolute or local minimum value of $y$.
1544105093

Work Step by Step

$$y=\sin^{-1}(e^x)$$ - Domain: $[-1,1]$ 1) Find all the critical points of the function: - Find $y'$: $$y'=\frac{(e^x)'}{\sqrt{1-(e^x)^2}}=\frac{e^x}{\sqrt{1-e^{2x}}}$$ - We have $y'=0$ when $e^x=0$. However, $e^x\gt0$ for all $x$, so there is no $x$ for which $y'=0$ here. - $y$ is undefined when $$\sqrt{1-e^{2x}}=0$$ $$1-e^{2x}=0$$ $$e^{2x}=1=e^0$$ $$2x=0$$ $$x=0$$ So $x=0$ is the only critical point of function $y$. 2) Evaluate $y$ at the critical points: - For $x=0$: $$y=\sin^{-1}(e^0)=\sin^{-1}(1)=\frac{\pi}{2}$$ Now take a look at the graph of $y$. From the graph, we see that at $x=0$, $y$ has the absolute and local maximum value of $\pi/2$. There is no absolute or local minimum value of $y$.
Small 1544105093
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.