Answer
The equation of the tangent is $y=-8x+1.$
Work Step by Step
$f(x)=\dfrac{(x+1)-2+2}{x-1}=\dfrac{(x-1)+2}{x-1}=\dfrac{x-1}{x-1}+\dfrac{2}{x-1}$
$=1+\dfrac{2}{x-1}=1+2(x-1)^{-1}\rightarrow$
$f'(x)=0+(-1)(2)(x-1)^{-1-1}=-\dfrac{2}{(x-1)^2}$
$f'(\frac{1}{2})=-\dfrac{2}{(\frac{1}{2}-1)^2}=-8.$
Equation of tangent:
$(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$.
$(y+3)=-8(x-\frac{1}{2})\rightarrow y=-8x+1.$
