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: 20

Answer

a) The derivative of $f\left( x \right)={{x}^{2}}-4x+7$ at x is $f'\left( x \right)=2x-4$. b) The slope of the tangent line to the graph of $f\left( x \right)={{x}^{2}}-4x+7$ at $x=\frac{3}{2}$ is $f'\left( \frac{3}{2} \right)=-1$ and at $x=2$ is $f'\left( 2 \right)=0$.

Work Step by Step

(a) Consider the function, $f\left( x \right)={{x}^{2}}-4x+7$ Now, compute the derivate of $f\left( x \right)={{x}^{2}}-4x+7$ using the formula $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$ To compute $f\left( x+h \right)$, substitute $x=x+h$ in the function $f\left( x \right)={{x}^{2}}-4x+7$. $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ {{\left( x+h \right)}^{2}}-4\left( x+h \right)+7 \right]-\left( {{x}^{2}}-4x+7 \right)}{h}$ Now, simplify ${{\left( x+h \right)}^{2}}$ by using the property ${{\left( A+B \right)}^{2}}={{A}^{2}}+2AB+{{B}^{2}}$ and $-3\left( x+h \right)$ by using the distributive property $A\left( B+C \right)=AB+AC$. $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}+2xh+{{h}^{2}}-4x-4h+7-{{x}^{2}}+4x-7}{h}$ Combine the like terms in the numerator; this gives, $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{2xh+{{h}^{2}}-4h}{h}$ Divide numerator and denominator by h. $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 2x+h-4 \right)$ Apply the limits, $\begin{align} & f'\left( x \right)=2x+0-4 \\ & =2x-4 \end{align}$ Thus, the derivative of $f\left( x \right)={{x}^{2}}-4x+7$ at x is $f'\left( x \right)=2x-4$. (b) Consider the function, $f\left( x \right)={{x}^{2}}-4x+7$ From part (a), the derivative of $f\left( x \right)={{x}^{2}}-4x+7$ at x is $f'\left( x \right)=2x-4$. To compute the slope of the tangent line at $x=\frac{3}{2}$, substitute $x=\frac{3}{2}$ in $f'\left( x \right)$. $\begin{align} & f'\left( \frac{3}{2} \right)=2\left( \frac{3}{2} \right)-4 \\ & =3-4 \\ & =-1 \end{align}$ To compute the slope of the tangent line at $x=2$, substitute $x=2$ in $f'\left( x \right)$. $\begin{align} & f'\left( 2 \right)=2\left( 2 \right)-4 \\ & =4-4 \\ & =0 \end{align}$
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