Answer
absolute minimum value =-8
absolute maximum value =1
Work Step by Step
Given $f(\theta)=\theta^{\frac{3}{5}}$
Critical point occurs where first derivative is zero:
so${\frac{df(theta)}{dx}=\frac{\theta^{\frac{3}{5}}}{d\theta}}$
${\frac{df(\theta)}{dx}}=\frac{3}{5}\theta^{\frac{-2}{5}}$
so the critical point is:
$f(0)=0$
test end points:
$f(-1)=(-32)^{\frac{3}{5}}=-8$
$f(8)=1^{\frac{3}{5}}=1$
absolute minimum value =-8
absolute maximum value =1
