How to find the monoticity of the following function

Question

I am trying to find where the following function is increasing and decreasing $$f(x)=\arcsin\left(\frac{1-x^2}{1+x^2}\right).$$ I found that at $x=0$ the function is not differentiable because the limit from above and below are not equal. Also, the point found is a global maxima. But with this information how can I find where it is increasing and decreasing? Taking the first derivative and studying where it is positive gives that it should increase for all $x$, even though it is not true.

What am I missing?

Answer 1

Note that

$$f'(x)=\frac{d}{dx} \sin^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) = -\frac{2 \sqrt{\frac{x^2}{(x^2 + 1)^2}}}{x}$$

thus since

• $f'(x)<0 \quad \forall x>0\implies \quad f(x)$ is increasing

• $f'(x)>0 \quad \forall x<0\implies \quad f(x)$ is decreasing

Answer 2

Hint:   $\sin(x)$ is strictly increasing on $[-\pi/2,\pi/2]$, so $f(x)$ has the same monotonicity as $\sin(f(x))=\dfrac{1-x^2}{1+x^2}=-1+\dfrac{2}{1+x^2}\,$, which should be easy to verify as even and decreasing for $x \ge 0$, so the only extremum is a maximum at $x=0$.