Calculus 10th Edition

Published by Brooks Cole

Chapter 2 - Differentiation - Review Exercises: 43

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 $m$. $(y+3)=-8(x-\frac{1}{2})\rightarrow y=-8x+1.$

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