Answer
$x=±1$
Work Step by Step
Using the quotient rule, we get
$f'(x)=\frac{\frac{d}{dx}(x)\times(x^{2}+1)-x\frac{d}{dx}(x^{2}+1)}{(x^{2}+1)^{2}}$
$=\frac{(x^{2}+1)-2x^{2}}{(x^{2}+1)^{2}}$
$=\frac{-x^{2}+1}{(x^{2}+1)^{2}}$
If $f'(x)$ does not exist or $f'(x)=0$, this implies $x$ is a critical point of the function. But, $f'(x)$ exists for all $x$ as the denominator does not vanish ($x^{2}+1$ is never equal to 0)
If $f'(x)=\frac{-x^{2}+1}{(x^{2}+1)^{2}}=0$
$\implies -x^{2}+1=0$
$\implies x^{2}=1$
Or $x= ±1$
Thus, we know that:
$x=±1$ are the critical points.
