Answer
Over the specified interval, the function has an absolute maximum equal to $1$ and an absolute minimum equal to $-1.$
Work Step by Step
Using the quotient rule: $fâ(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=2x; u'(x)=2$
$v(x)=x^2+1; v'(x)=2x$
$f'(x)=\dfrac{2(x^2+1)-(2x)(2x)}{(x^2+1)^2}=\dfrac{2(1-x)(1+x)}{(x^2+1)^2}.$
$f'(x)=0\to x=1$ or $x=-1.$
Along with the interval's boundaries, both $x=1$ and $x=-1$ are possible candidates for absolute extrema.
$f(-2)=-\dfrac{4}{5}.$
$f(-1)=-1.$
$f(1)=1.$
$f(2)=\dfrac{4}{5}.$
Over the specified interval, the function has an absolute maximum equal to $1$ and an absolute minimum equal to $-1.$
