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: 29

Answer

a) See below b) The slope intercept equation of the tangent line is $y=2x-5$.

Work Step by Step

(b) Consider the function, $f\left( x \right)={{\left( x-2 \right)}^{2}}$ The $x$ coordinate for the tangent is $3$. Substitute $x=3$ in the function $f\left( x \right)={{\left( x-2 \right)}^{2}}$. $\begin{align} & f\left( 3 \right)={{\left( 3-2 \right)}^{2}} \\ & ={{1}^{2}} \\ & =1 \end{align}$ 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 $f\left( x \right)={{\left( x-2 \right)}^{2}}$ and replace $f\left( 3 \right)=1$. $\begin{align} & {{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ {{\left( \left( 3+h \right)-2 \right)}^{2}} \right]-1}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ {{\left( 1+h \right)}^{2}} \right]-1}{h} \end{align}$ Use the property${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$. ${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ 1+2h+{{h}^{2}} \right]-1}{h}$ Combine the like terms. ${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{h}^{2}}+2h}{h}$ Make a factor of the numerator and simplify. $\begin{align} & {{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{h\left( h+2 \right)}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\left( h+2 \right) \end{align}$ Apply the limits. $\begin{align} & {{m}_{\tan }}=0+2 \\ & =2 \end{align}$ Thus, the slope of the tangent to the graph of $f\left( x \right)={{\left( x-2 \right)}^{2}}$ at $\left( 3,1 \right)$ is $2$. Substitute ${{x}_{1}}=3$, ${{y}_{1}}=1$ and $m=2$ in the equation $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$. $\begin{align} & y-1=2\left( x-3 \right) \\ & y=2x-6+1 \\ & y=2x-5 \end{align}$ Therefore, the slope intercept equation of the tangent line is $y=2x-5$.
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