Chapter 2 - Differentiation - 2.3 Exercises - Page 126: 73

There is one horizontal tangent at the point $(1, 1)$.

Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=(2x-1); u'(x)=2$ $v(x)=x^2; v'(x)=2x$ $f'(x)=\frac{(2)(x^2)-(2x)(2x-1)}{x^4}=\frac{2-2x}{x^3}$ $f'(x)=0 \rightarrow (2-2x)=0 \rightarrow x=1$ $f(1)=\frac{2(1)-1}{1^2}=1.$ The point is $(1, 1)$.