Answer
Local maximum at $x=2$ and a local minimum at $x=4$.
Inflection points $x=1, \frac{5\pm\sqrt {3}}{2}$.
Work Step by Step
Step 1. Given $y'=(x-1)^2(x-2)(x-4)$, as the squared term will always be positive, we only need to test $y'$ sign changes across points $x=2,4$ as: $..(+)..(2)..(-)..(4)..(+)$. Thus, there is a local maximum at $x=2$ and a local minimum at $x=4$.
Step 2. Find the second derivative as $y''=2(x-1)(x-2)(x-4)+(x-1)^2(x-4)+(x-1)^2(x-2)=2(x-1)(x^2-6x+8)+(x-1)^2(2x-6)=2(x-1)(x^2-6x+8+x^2-4x+3)=2(x-1)(2x^2-10x+11)$
Step 3. The zeros for $y''$ can be found as $x=1$ and $x=\frac{10\pm\sqrt {10^2-88}}{4}=\frac{5\pm\sqrt {3}}{2}$ or we can let $x_1=\frac{5-\sqrt {3}}{2}, x_2=\frac{5+\sqrt {3}}{2}$
Step 4. Test the signs of $y''$ across its zeros; we have $..(-)..(1)..(+)..(x_1)..(-)..(x_2)..(+)$; thus the function is concave down on $(-\infty,1),(x_1,x_2)$ and concave up on $(1,x_1),(x_2,\infty)$ and the inflection points are $x=1, \frac{5\pm\sqrt {3}}{2}$.
