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: 31


b) The slope intercept equation of the tangent line is $y=\frac{1}{2}x-1$.

Work Step by Step

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