Answer
$ x=1,\ \ x=2$ and $ x= 3$
Work Step by Step
Given $$ g(x) =(x-1)^2(x-3)^3$$
Then to find the critical points
\begin{align*}
g'(x)&=0\\
(x-1)^{2} \cdot 2(x-3)(1)+2(x-1)(1) \cdot(x-3)^{2}&=0\\
2(x-3)(x-1)[(x-1)+(x-3)]&=0\\
4(x-3)(x-1)(x-2) &=0
\end{align*}
Hence the critical points are $ x=1,\ \ x=2$ and $ x= 3$
