Answer
absolute minimum value = 0
absolute maximum value = 16
Work Step by Step
Given $f(x)=x^{\frac{4}{3}}$
The critical point occurs where the first derivative is zero:
so${\frac{df(x)}{dx}=\frac{x^{\frac{4}{3}}}{dx}}$
${\frac{df(x)}{dx}}=\frac{4}{3}x^{\frac{1}{3}}$
Critical point:
$f(0)=0$
Test end points:
$f(-1)=(-1)^{\frac{4}{3}}=-1$
$f(8)=8^{\frac{4}{3}}=16$
absolute minimum value = 0
absolute maximum value = 16
