Answer
Local maximum at $x=1$ and a local minimum at $x=3$.
Work Step by Step
Given the derivative of the function $f’(x)=6(x-1)(x-2)^2(x-3)(x-4)^4$, we can test the sign changes across the critical points (zeros) at $x=1,2,3,4$ and get $..(+)..(1)..(-)..(2)..(-)..(3)..(+)..(4)..(+)..$, thus the function has a local maximum at $x=1$ and a local minimum at $x=3$. Please note that the term containing $x-2$ and $x-4$ will always be positive and not give any sign changes of $f’$.
