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: 33


b) The slope intercept equation of the tangent line is $y=3x+4$.

Work Step by Step

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