Answer
Relative maximum: $x=-1$
Relative minimum: $x=-\frac{3}{5}$
Work Step by Step
First derivative
\[
f^{\prime}(x)=(1+x)^{2}3 x^{2}+(1+x) 2x^{3}=(1+x)\left(3 x^{2}+5 x^{3}\right)
\]
Critical points (zeros of first derivative or point where the first derivative does not exist)
\[
x=0, x=-\frac{3}{5} \text { and } x=-1
\]
Second derivative
\[
f^{\prime \prime}(x)=3 x^{2}+5 x^{3}+(1+x)\left(6 x+15 x^{2}\right)
\]
Values of the second derivative at critical points:
$f^{\prime \prime}(-1)=-2<0$
$f^{\prime \prime}\left(-\frac{3}{5}\right)=0.72>0$
\[
\begin{array}{l}
f^{\prime \prime}\left(-\frac{3}{5}\right)=0.72>0 \\
f^{\prime \prime}(0)=0
\end{array}
\]
Second derivative test
Relative minimum: $x=-\frac{3}{5}$
Relative maximum: $x=-1$
