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

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 $\pi/2$.
$$y=\cos^{-1}(x^2)$$ - 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 $\pi/2$.