Precalculus (6th Edition) Blitzer

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

Chapter 1 - Section 1.3 - More on Functions and Their Graphs - Exercise Set - Page 200: 124

Answer

The difference quotient of a function is obtained by substituting the values of $f\left( x \right)\text{ and }f\left( x+h \right)$ in the difference quotient formula and solving the equation.

Work Step by Step

The difference quotient of a function is obtained by substituting the values of $f\left( x \right)\text{ and }f\left( x+h \right)$ in the difference quotient formula and solving the equation. Consider the given function $f\left( x \right)$ and the difference quotient formula $\frac{f\left( x+h \right)-f\left( x \right)}{h}$. To obtain the difference quotient formula of a function, substitute $x=x+h$ and then put the function in the formula. $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ Example: Consider the following function as an example: $f\left( x \right)={{x}^{2}}+x+1$ Substitute $x+h$ in the place of $x$ as follows: $\begin{align} & f\left( x+h \right)={{\left( x+h+1 \right)}^{2}}+\left( x+h \right)+1 \\ & f\left( x \right)={{x}^{2}}+x+1 \end{align}$ Now, substitute the values of $f\left( x \right)\text{ and }f\left( x+h \right)$ in the difference quotient formula as follows: $\begin{align} & \frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{\left( {{\left( x+h+1 \right)}^{2}}+\left( x+h \right)+1 \right)-\left( {{x}^{2}}+x+1 \right)}{h} \\ & =\frac{{{h}^{2}}+2hx+3h+2x+1}{h} \end{align}$ Thus, to get the difference quotient formula, the value of x is substituted.
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