Answer
The curve, observing from left to right,
- rises on $(-\infty,0)$, concave down,
- at $(0,0)$ changes concavity to "up" (inflection point), continues rising
- at $(2.16)$ changes concavity to "down" (inflection point), continues rising
- reaches an absolute maximum at $(3,27)$, turns
- and contiunes to fall on the interval $(3,\infty)$, concave down.
Work Step by Step
$y=f(x)=x^{3}(4-x)=4x^{3}-x^{4}$
$f'(x)=-4x^{3}+12x^{2}=-4x^{2}(x-3)$
$f''(x)=-12x^{2}+24x=-12x(x-2)$
Analyzing $f'(x)$, we see that it is defined everywhere,
and has zeros $ x=0, 3\qquad$... (critical points)
$\left[\begin{array}{lllllll}
y': & & ++ & | & ++ & | & --\\
& & & 0 & & 3 & \\
y: & & \nearrow & & \nearrow & \max & \searrow
\end{array}\right] \left\{\begin{array}{l}
f(0)=0\\
f(3)=27
\end{array}\right.$
Analyzing $f''(x)$,
$f''(x)=0$ for $ x=0,2\qquad$... points of inflection.
$\left[\begin{array}{llllllll}
y'': & -- & | & ++ & | & -- & & \\
& & 0 & & 2 & & & \\
y: & \cap & i.p. & \cup & i.p. & \cap & &
\end{array}\right]\left\{\begin{array}{ll}
f(0)=0, & (0,0)\\
f(2)=16 & (2,16)
\end{array}\right.$
