Answer
General shape:
Work Step by Step
Step 2
$y'=(8x-5x^{2})(4-x)^{2}=-x(5x-8)(4-x)^{2}$
$y''=(8-10x)(4-x)^{2}+(8x-5x^{2})\cdot(-2)(4-x)$
$=2(4-x)[ (4-5x)(4-x) -(8x-5x^{2})]$
$=2(4-x)[ 16-4x-20x+5x^{2} -8x+5x^{2}]$
$=2(4-x)[ 10x^{2} -32x+16]$
$=4(4-x)(5x^{2}-16x+8)$
Step 3
$y'=0$ when $ x=0,\displaystyle \ x=\frac{8}{5} ,\ x=4 \qquad$... critical points.
Step 4
$\left[\begin{array}{ccccccc}
y': & -- & | & ++ & | & -- & | & --\\
& & 0 & & 8/5 & & 4 & \\
y: & \searrow & \min & \nearrow & \max & \searrow & & \searrow
\end{array}\right]$
Step 5
For concavity, $y''=0$ for $x=4,\ $and
$x=\displaystyle \frac{16\pm\sqrt{256-160}}{10}=\frac{16\pm 4\sqrt{6}}{10}=\frac{8\pm 2\sqrt{6}}{5}$
$\left[\begin{array}{ccccccc}
y': & ++ & | & -- & | & ++ & | & --\\
& & \frac{8-2\sqrt{6}}{5} & & \frac{8+2\sqrt{6}}{5} & & 4 & \\
y: & \cup & i.p. & \cap & i.p. & \cup & i.p. & \cap
\end{array}\right]$
