Answer
Relative minimum: $x=1$
Work Step by Step
First derivative
\[
f^{\prime}(x)=(-4+x)^{3}+3 x(-4+x)^{2}=(x-4)^{2}(4 x-4)
\]
Critical points (zeros of first derivative or point where the first derivative does not exist)
$x=4$ and $x=1$
Second derivative
$f^{\prime \prime}(x)=(-4+4 x)(-4+x)2+2(-4+x)^{2}=(-8+5 x)(-4+x)2$
Second derivative
$f^{\prime \prime}(1)=36>0$
$f^{\prime \prime}(4)=0$
Values of the second derivative at critical points:
\[
\begin{array}{l}
f^{\prime \prime}(1)=36>0 \\
f^{\prime \prime}(4)=0
\end{array}
\]
Second derivative test
Relative minimum: $x=1$
