Answer
Over the specified interval, the function has an absolute maximum equal to $\dfrac{1}{4}$ and an absolute minimum equal to $0$.
Work Step by Step
Using the quotient rule: $g'(t)=(\frac{u(t)}{v(t)})'=\frac{u'(t)v(t)-v'(t)u(t)}{(v(t))^2}$.
$u(t)=t^2 ;u'(t)=2t$.
$v(t)=t^2+3 ;v'(t)=2t$
$g'(t)=\dfrac{(2t)(t^2+3)-(2t)(t^2)}{(t^2+3)^2}=\dfrac{6t}{(t^2+3)^2}.$
$g'(t)=0\to 6t=0\to t=0$
$g'(t)$ is defined for all values of $t.$
Along with the interval endpoints, $t=0$ is a possible value at which the function might attain an absolute extremum.
$g(-1)=\dfrac{1}{4}.$
$g(0)=0.$
$g(1)=\dfrac{1}{4}$
Over the specified interval, the function has an absolute maximum equal to $\dfrac{1}{4}$ and an absolute minimum equal to $0$.
