Let $f\left(x\right)=(x-a{)}^m(x-b{)}^n$ when $m,n\in \mathbf{I}$ and $m, n>1$ Then

 

$f^{\text{'}}\left(x\right)=m(x-a{)}^{m-1}(x-b{)}^n+n(x-a{)}^m$

$(x-b{)}^{n-1}$

$=(x-a{)}^{m-1}(x-b{)}^{n-1}\left[\right(m+n)x-(mb+na\left)\right].\text{ Thus }{f}^{\text{'}}\left(x\right)=$

$0\Rightarrow x=a,x=b,x=\frac{mb+na}{m+n}$ So f has exactly three critical points.