Answer
absolute minimum value =0
absolute maximum value =27
Work Step by Step
Given $f(\theta)=3\theta^{\frac{2}{3}}$
Critical point occurs where first derivative is zero:
so${\frac{df(theta)}{d\theta}=\frac{3\theta^{\frac{2}{3}}}{d\theta}}$
${\frac{df(\theta)}{d\theta}}=\frac{2\times3}{3}\theta^{\frac{-1}{3}}=0$
$\theta=0$
so the critical point is:
$f(0)=0$
test end points:
$f(-1)=3(-27)^{\frac{2}{3}}=27$
$f(8)=3\times8^{\frac{2}{3}}=12$
absolute minimum value =0
absolute maximum value =27
