Answer
The curve, observing from left to right,
- falls on $(-\displaystyle \infty,-\frac{3}{2})$, concave up,
- reaches an absolute minimum at $(-\displaystyle \frac{3}{2},-\displaystyle \frac{27}{3})$, turns, rises,
- at $(-1,-1)$ changes concavity to "down" (inflection point), continues rising
- at $(0,0)$ changes concavity to "up" (inflection point),
- and contiunes rising on the interval $(0,\infty)$, concave up.
Work Step by Step
$y=f(x)=x^{3}(x+2)=x^{4}+2x^{3}$
$f'(x)=4x^{3}+6x^{2}=2x^{2}(2x+3)$
$f''(x)=12x^{2}+12x=12x(x+1)$
Analyzing $f'(x)$, we see that it is defined everywhere,
and has zeros $ x=0, -\displaystyle \frac{2}{3}\qquad$... (critical points)
$\left[\begin{array}{lllllll}
y': & & -- & | & ++ & | & ++\\
& & & -\frac{3}{2} & & 0 & \\
y: & & \searrow & \min & \nearrow & & \nearrow
\end{array}\right] \left\{\begin{array}{l}
f(-3/2)=-27/16\\
f(0)=0
\end{array}\right.$
Analyzing $f''(x)$,
$f''(x)=0$ for $ x=-1,0\qquad$... points of inflection.
$\left[\begin{array}{llllll}
y'': & ++ & | & -- & | & ++\\
& & -1 & & 0 & \\
y: & \cup & i.p. & \cap & i.p. & \cup
\end{array}\right]\left\{\begin{array}{ll}
f(-1)=-1, & (-1,-1)\\
f(0)=0 & (0,0)
\end{array}\right.$
