Answer
Relative maximum: $\frac{4}{3}$
Work Step by Step
First derivative
\[
f^{\prime}(x)=8-6 x
\]
Critical points (zeros of first derivative or point where first derivative does not exist)
\[
x=\frac{4}{3}
\]
Using first derivative test, determine relative extrema
$f^{\prime}(x)>0$ on the left and $f^{\prime}(x)<0$ on the right of $x=\frac{4}{3},$ so $x=\frac{4}{3}$ is a relative maximum.
Second derivative.
\[
-6=f^{\prime \prime}(x)
\]
$f^{\prime}\left(\frac{4}{3}\right)=0$ and $f^{\prime \prime}\left(\frac{4}{3}\right)<0,$ so $x=\frac{4}{3}$ is a
relative maximum.