Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Section 11.4 - Introduction to Derrivatives - Exercise Set - Page 1174: 36

Answer

b) The slope intercept equation of the tangent line is $y=x-4$.

Work Step by Step

b) Consider the function, $f\left( x \right)=-\frac{1}{x-2}$ The $x$ coordinate for the tangent is $3$. Substitute $x=3$ in the function $f\left( x \right)=-\frac{1}{x-2}$. $\begin{align} & f\left( 3 \right)=-\frac{1}{3-2} \\ & =-\frac{1}{1} \\ & =-1 \end{align}$ Thus, the tangent line passes through $\left( 3,-1 \right)$. The value of $a$ from the point $\left( a,f\left( a \right) \right)$ or $\left( 3,-1 \right)$ is $3$. Substitute $a=3$ in ${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}$. ${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 3+h \right)-f\left( 3 \right)}{h}$ Substitute $x=3+h$ in the function $f\left( x \right)=-\frac{1}{x-2}$ and replace $f\left( 3 \right)=-1$. $\begin{align} & {{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ -\frac{1}{\left( 3+h \right)-2} \right]-\left( -1 \right)}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ -\frac{1}{h+1} \right]+1}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{-1+h+1}{h\left( h+1 \right)} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{\left( h+1 \right)} \end{align}$ Apply the limits and simplify. $\begin{align} & {{m}_{\tan }}=\frac{1}{\left( 0+1 \right)} \\ & =1 \end{align}$ Thus, the slope of the tangent to the graph of $f\left( x \right)=-\frac{1}{x-2}$ at $\left( 3,-1 \right)$ is $1$. Substitute ${{x}_{1}}=3$, ${{y}_{1}}=-1$ and $m=1$ in the equation $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$. $\begin{align} & y-\left( -1 \right)=1\left( x-3 \right) \\ & y+1=x-3 \\ & y=x-3-1 \\ & y=x-4 \end{align}$ Therefore, the slope intercept equation of the tangent line is $y=x-4$.
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