Answer
Relative maximum: $x=-\frac{2}{5}$
Relative minimum: $x=0$
Work Step by Step
First derivative
\[
f^{\prime}(x)=3 (1+x)^{2}x^{2}+2 (1+x)^{3}x=(1+x)^{2}\left(2 x+5 x^{2}\right)
\]
Critical points (zeros of the first derivative or the point where the first derivative does not exist).
\[
x=0, x=-\frac{2}{5} \text { and } x=-1
\]
Second derivative
\[
f^{\prime \prime}(x)=2(1+x)\left(2x+5 x^{2}\right)+(1+x)^{2}(2+10 x)
\]
Values of the second derivative at critical points:
$f^{\prime \prime}(-1)=0$
\[
f^{\prime \prime}\left(-\frac{2}{5}\right)=-0.72<0
\]
\[
f^{\prime \prime}(0)=2>0
\]
Second derivative test
Relative maximum: $x=-\frac{2}{5}$
Relative minimum: $x=0$
