Graph of $\cos^{-1}\left(\sqrt{1-x^2}\right)$

Question

Plot the Graph of $f(x)=\cos^{-1}\left(\sqrt{1-x^2}\right)$

Sol: We see that $$f'(x)=\frac{d}{dx}\cos^{-1}\left(\sqrt{1-x^2}\right)=\frac{-1}{\sqrt{1-(1-x^2)}}\frac{d}{dx}\sqrt{1-x^2}$$ $\implies$ $$f'(x)=\frac{x}{|x|\sqrt{1-x^2}}$$ Now its evident that $$f'(0^+)=1,f'(0^-)=-1$$ But $f'(0)$ is not defined and $f(0)=0$. Do we say this is removable discontinuity of $f'(x)$? How will this information be conveyed in graph?

Answer 1

Hint:

Let $y=\cos^{-1}\sqrt{1-x^2}$

$\implies0\le y\le\pi$

Now for real $x^2\le1,\sqrt{1-x^2}\ge0\implies\le y\le\dfrac\pi2$

$\cos y=\sqrt{1-x^2}\implies x^2=\cdots=\sin^2y\implies\sin y=|x|$ as $\sin y\ge0$

So, $f'(0^-)=?$

Answer 2

Here $f(-x)=f(x)$, $f(0)=0$,$f(\pm 1)=\pi/2$; $f'(0^+)>0$, $f'(0^-)<0$, $f'(\pm 1)=\infty.$ So the curve is symmetric, passing through $x=0$ which is non-differentiable at $x=0$ (left/right derivatives are finite but unequal) and at $x=\pm 1$ (vertical tangent).ts domain is $[-1,1]$ and range is $[0, \pi/2]$. Hence, it looks like: