Chapter 2 - Differentiation - 2.3 Exercises - Page 126: 74

One horizontal tangent at the point $(0,0)$.

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