Answer
Relative maximum: $x=-1$
Relative minimum: $x=0$
Work Step by Step
First derivative
\[
2 x^{-1 / 3}+2=f^{\prime}(x)
\]
Critical points (zeros of first derivative or points where the first derivative does not exist)
\[
x=0, x=-1
\]
Second derivative
\[
f^{\prime \prime}(x)=-\frac{2}{3} x^{-4 / 3}
\]
Values of the second derivative at critical points:
\[
f^{\prime \prime}(-1)=-\frac{2}{3}<0
\]
$f^{\prime \prime}(0)$ does not exist
Relative minimum: $x=0$
Relative maximum: $x=-1$
We used the first derivative test for $x=0$ and the second derivative test for $x=-1$.
