Answer
$f(x)$ has two critical numbers: {$-1, 1$}.
Work Step by Step
Using the quotient rule: $fâ(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=4x; u'(x)=4$
$v(x)=x^2+1; v'(x)=2x$
$f'(x)=\dfrac{4(x^2+1)-(2x)(4x)}{(x^2+1)^2}=\dfrac{4(1-x^2)}{(x^2+1)^2}.$
$f'(x)=0\to4(1-x^2)=0\to (1-x)(1+x)=0\to x=1$ or $x=-1.$
$f'(x)$ is defined for all values since the denominator is never $0.$
$f(x)$ has two critical numbers: {$-1, 1$}.
