Answer
At $x=\pm1$, $y$ has the absolute and local minimum value of $0$ and at $x=0$, $y$ has the absolute and local maximum value of $\pi/2$.
Work Step by Step
$$y=\cos^{-1}(x^2)$$
- Domain: $[-1,1]$
1) Find all the critical points of the function:
- Find $y'$: $$y'=\frac{(x^2)'}{\sqrt{1-x^4}}=\frac{2x}{\sqrt{1-x^4}}$$
- We have $y'=0$ when $2x=0$ or $x=0$.
- $y$ is undefined when $$\sqrt{1-x^4}=0$$ $$1-x^4=0$$ $$x^4=1$$ $$x=\pm1$$
All these points $x=\{0,\pm1\}$ are critical points function $y$.
2) Evaluate $y$ at the critical points:
- For $x=-1$: $$y=\cos^{-1}((-1)^2)=\cos^{-1}(1)=0$$
- For $x=0$: $$y=\cos^{-1}(0^2)=\cos^{-1}(0)=\frac{\pi}{2}$$
- For $x=1$: $$y=\cos^{-1}(1^2)=\cos^{-1}(1)=0$$
Now take a look at the graph of $y$.
From the graph, we see that at $x=\pm1$, $y$ has the absolute and local minimum value of $0$ and at $x=0$, $y$ has the absolute and local maximum value of $\pi/2$.
