Answer
Relative minimum: (9,-2187)
Work Step by Step
First derivative
\[
f^{\prime}(x)=-36 x^{2}+4 x^{3}=(-9+x)4 x^{2}
\]
Critical points (zeros of first derivative or point where the first derivative does not exist)
$x=9$ and $x=0$
Using the first derivative test to determine relative extrema:
$f^{\prime}(x)<0$ on the left and $f^{\prime}(x)<0$ on the right of $x=0,$ so $x=0$ isn't a relative extremum.
$f^{\prime}(x)<0$ on the left and $f^{\prime}(x)>0$ on the right of $x=0,$ so $x=9$ is a relative minimum.
$f(9)=-2187$
Second derivative
\[
f^{\prime \prime}(x)=-72 x+12 x^{2}=(-6+x)12 x
\]
Using second derivative test to determine relative extrema.
$f^{\prime}(0)=0$ and $f^{\prime \prime}(0)=0,$ so $x=\frac{4}{3}$ isn't a relative extremum
$f^{\prime}(9)=0$ and $f^{\prime \prime}(9)=324>0,$ so $x=9$ is a relative minimum.
