Answer
Absolute minimum is $1$ at $x=1,$
absolute maximum is $-8$ at $x=-32$
Work Step by Step
To find absolute extrema of a continuous function f on a closed interval:
1. Evaluate $f$ at all critical points and endpoints.
2. Take the largest and smallest of these values.
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$g$ is continuous on $[-32,1]$.
Critical points:
$g'(\theta)=0$
$\displaystyle \frac{3}{5}\theta^{-2/5}=0$
$g'(\theta)$ is undefined for
$\theta=0 , \quad f(0)=0.$
Endpoints:
$f(-32)=[(-32)^{1/5}]^{3}=(-2)^{3}=-8,$
$f(1)=(1^{1/5})^{3}=1^{3}=1$
Absolute minimum is $1$ at $x=1,$
absolute maximum is $-8$ at $x=-32$
