Answer
The curve, observing from left to right,
- rises on $(-\infty,2)$, concave down,
- reaches a local maximum at $(2,512)$, turns,
- at $(4,324)$ changes concavity to "up" (inflection point),
- continues falling to a local minimum at $(10,0)$,
- turns and contiunes rising on the interval $(10,\infty)$, concave up.
Work Step by Step
$y=f(x)=x(\displaystyle \frac{x}{2}-5)^{4}$
$f'(x)=1\displaystyle \cdot(\frac{x}{2}-5)^{4}+x\cdot 4(\frac{x}{2}-5)^{3}\cdot\frac{1}{2}$
$=(\displaystyle \frac{x}{2}-5)^{3}[(\frac{x}{2}-5)+2x]$
$=(\displaystyle \frac{x}{2}-5)^{3}(\frac{5x}{2}-5)$
$f''(x)=3(\displaystyle \frac{x}{2}-5)^{2}\cdot\frac{1}{2}\cdot(\frac{5x}{2}-5)+(\frac{x}{2}-5)^{3}(\frac{5}{2})$
$=(\displaystyle \frac{x}{2}-5)^{2}[(\frac{15x}{4}-\frac{15}{2})+(\frac{5x}{4}-\frac{25}{2})]$
$=(\displaystyle \frac{x}{2}-5)^{2}(5x-20)$
$=5(\displaystyle \frac{x}{2}-5)^{2}(x-4)$
Analyzing $f'(x)$, we see that it is defined everywhere,
and has zeros $ x=2,10\qquad$... (critical points)
$\left[\begin{array}{lllllll}
y': & & ++ & | & -- & | & ++\\
& & & 2 & & 10 & \\
y: & & \nearrow & \max & \searrow & \min & \nearrow
\end{array}\right] \left\{\begin{array}{l}
f(2)=512\\
f(10)=0
\end{array}\right.$
Analyzing $f''(x)$,
$f''(x)=0$ for $ x=4,10\qquad$... points of inflection.
$\left[\begin{array}{llllll}
y'': & -- & | & ++ & | & ++\\
& & 4 & & 10 & \\
y: & \cap & i.p. & \cup & & \cup
\end{array}\right]\left\{\begin{array}{l}
f(4)=324\\
f(10)=
\end{array}\right.$
