Answer
Local minimum at $x=2.$
No local maxima.
Inflection points at $x=1$ and $x=\displaystyle \frac{5}{3}$
Work Step by Step
$y'=(x-1)^{2}(x-2)$
Critical points: $x=1,2$
$\left[\begin{array}{llllll}
y': & -- & | & -- & | & ++\\
& & 1 & & 2 & \\
y: & \searrow & & \searrow & \min & \nearrow
\end{array}\right]$
$y $has a local maximum at $x=2.$
$y''=2(x-1)(x-2)+(x-1)^{2}(1)$
$=(x-1)[2(x-2)+(x-1)$
$=(x-1)(3x-5)$
$y''=0$ for $x=1$ and $x=\displaystyle \frac{5}{3}$
$\left[\begin{array}{llllll}
y'': & ++ & | & -- & | & ++\\
& & 1 & & 5/3 & \\
y: & \cup & inf & \cap & inf & \cup
\end{array}\right]$
