Answer
The curve, observing from left to right,
- rises on $(-\infty,0)$, concave down,
- reaches a local maximum at $(0,0)$, turns,
- at $(3,-162)$ changes concavity to "up" (inflection point),
- continues falling to a local minimum at $(4,-256)$,
- turns and contiunes rising on the interval $(4,\infty)$, concave up.
Work Step by Step
$y=f(x)=x^{4}(x-5)=x^{5}-5x^{4}$
$f'(x)=5x^{4}-20x^{3}=5x^{3}(x-4)$
$f''(x)=20x^{3}-60x^{2}=20x^{2}(x-3)$
Analyzing $f'(x)$, we see that it is defined everywhere,
and has zeros $ x=0, 4\qquad$... (critical points)
$\left[\begin{array}{lllllll}
y': & & ++ & | & -- & | & ++\\
& & & 0 & & 4 & \\
y: & & \nearrow & \max & \searrow & \min & \nearrow
\end{array}\right] \left\{\begin{array}{l}
f(0)=0\\
f(4)=-256
\end{array}\right.$
Analyzing $f''(x)$,
$f''(x)=0$ for $ x=0,3\qquad$... points of inflection.
$\left[\begin{array}{llllll}
y'': & -- & | & -- & | & ++\\
& & 0 & & 3 & \\
y: & \cap & & \cap & i.p. & \cup
\end{array}\right]\left\{\begin{array}{l}
f(0)=0\\
f(3)=-162
\end{array}\right.$
