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