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: 34

Answer

b) The slope intercept equation of the tangent line is $y=3x-4$.
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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 $\left( 1,-1 \right)$. Now, compute the slope of the tangent line for the function $f\left( x \right)={{x}^{3}}-2$ as follows: 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 $f\left( 1 \right)=-1$. $\begin{align} & {{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ {{\left( 1+h \right)}^{3}}-2 \right]-\left[ -1 \right]}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{{{\left( 1+h \right)}^{3}}-2+1}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{{{\left( 1+h \right)}^{3}}-1}{h} \end{align}$ 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 $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$. $\begin{align} & y-\left( -1 \right)=3\left( x-1 \right) \\ & y+1=3x-3 \\ & y=3x-3-1 \\ & y=3x-4 \end{align}$ Therefore, the slope intercept equation of the tangent line is $y=3x-4$.
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