Calculus 10th Edition

Two horizontal tangents at $(1, \frac{1}{2})$ and at $(7, \frac{1}{14} ).$
Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=(x-4); u'(x)=1$ $v(x)=(x^2-7); v'(x)=2x$ $f'(x)=\frac{(1)(x^2-7)-(x-4)(2x)}{(x^2-7)^2}=-\frac{x^2-8x+7}{(x^2-7)^2}$ $f'(x)=0 \rightarrow-(x^2-8x+7)=0\rightarrow x=1$ or $x=7$ $f(1)=\frac{1-4}{1^2-7}=\frac{1}{2}\rightarrow$ The point is $(1, \frac{1}{2}).$ $f(7)=\frac{7-4}{7^2-7}=\frac{1}{14}\rightarrow$ The point is $(7, \frac{1}{14}).$