Answer
Two horizontal tangents at $(1, \frac{1}{2})$ and at $(7, \frac{1}{14} ).$
Work Step by Step
Using the quotient rule: $fâ(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=(x-4); u'(x)=1$
$v(x)=(x^2-7); v'(x)=2x$
$f'(x)=\frac{(1)(x^2-7)-(x-4)(2x)}{(x^2-7)^2}=-\frac{x^2-8x+7}{(x^2-7)^2}$
$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 $(7, \frac{1}{14}).$
