## Calculus 10th Edition

One Horizontal tangent at $(0, 0).$ One Horizontal tangent at $(2, 4).$
Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=x^2 u'(x)=2x$ $v(x)=x-1; v'(x)=1$ $f'(x)=\frac{(2x)(x-1)-1(x^2)}{(x-1)^2}=\frac{x(x-2)}{(x-1)^2}$ $f'(x)=0 \rightarrow (x-2)x=0 \rightarrow x=0$ or $x=2$ $f(0)=\frac{0^2}{0-1}=0 \rightarrow$ The point is $(0, 0)$. $f(2)=\frac{2^2}{2-1}=4 \rightarrow$ The point is $(2, 4).$