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: 1

Answer

a) The slope of the tangent to the graph of $f\left( x \right)=2x+3$ at $\left( 1,5 \right)$ is $2$. b) The slope intercept equation of the tangent line is $y=2x+3$.

Work Step by Step

(a) Consider the function $f\left( x \right)=2x+3$ and the point $\left( 1,5 \right)$ Here, $\left( a,f\left( a \right) \right)=\left( 1,5 \right)$ Now, compute the slope of the tangent line for the function $f\left( x \right)=2x+3$ as follows: Put $a=4$ , ${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 1+h \right)-f\left( 1 \right)}{h}$ To compute $f\left( 1+h \right)$, substitute $x=1+h$ in the function $f\left( x \right)=2x+3$. ${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ 2\left( 1+h \right)+3 \right]-\left[ 2\left( 1 \right)+3 \right]}{h}$ Now, simplify $2\left( 1+h \right)$ by using the distributive property $A\left( B+C \right)=AB+AC$. ${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ 2+2h+3 \right]-5}{h}$ Combine the like terms in the numerator. ${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{2h}{h}$ Cancel out h and apply the limits. $\begin{align} & {{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,2 \\ & =2 \end{align}$ A window showing the graph of the function and the tangent will appear as shown below. The graph of the function $f\left( x \right)=2x+3$ is a line and there is no tangent drawn in the graph because a tangent line to a straight line is the line itself. Thus, the slope of the tangent to the graph of $f\left( x \right)=2x+3$ at $\left( 1,5 \right)$ is $2$. (b) Consider the function $f\left( x \right)=2x+3$ and the point $\left( 1,5 \right)$ From part (a), the slope of the tangent to the graph of $f\left( x \right)=2x+3$ at $\left( 1,5 \right)$ is $2$. Now compute the slope intercept equation of the tangent line as follows: Here, the tangent line passes through the point $\left( 1,5 \right)$ which implies ${{x}_{1}}=1$ and ${{y}_{1}}=5$ Substitute $m=2,{{x}_{1}}=1,{{y}_{1}}=5$ in the equation $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$. $\begin{align} & y-5=2\left( x-1 \right) \\ & y-5=2x-2 \\ & y=2x-2+5 \\ & y=2x+3 \end{align}$ The tangent line of the function $f\left( x \right)=2x+3$ is same as the function itself which is shown in the graph. Thus, the slope intercept equation of the tangent line is $y=2x+3$.
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