## University Calculus: Early Transcendentals (3rd Edition)

At $x=0$, $y$ has the absolute and local maximum value of $\pi/2$. There is no absolute or local minimum value of $y$.
$$y=\sin^{-1}(e^x)$$ - 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 $y$.