Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Section 11.3 - Limits and Continuity - Exercise Set - Page 1163: 67


The value of $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ is $3{{x}^{2}}+3xh+{{h}^{2}}$.

Work Step by Step

Consider the function $ f\left( x \right)={{x}^{3}}$, Find the value of $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ using the definition of the function. $\begin{align} & \frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{{{\left( x+h \right)}^{3}}-{{x}^{3}}}{h} \\ & =\frac{{{x}^{3}}+{{h}^{3}}+3xh\left( x+h \right)-{{x}^{3}}}{h} \\ & =\frac{{{h}^{3}}+3xh\left( x+h \right)}{h} \end{align}$ Factoring out h from the numerator, $\begin{align} & \frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{h\left( {{h}^{2}}+3x\left( x+h \right) \right)}{h} \\ & ={{h}^{2}}+3x\left( x+h \right) \\ & ={{h}^{2}}+3{{x}^{2}}+3xh \end{align}$ Thus, the value of $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ is $3{{x}^{2}}+3xh+{{h}^{2}}$.
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