Question

find the difference quotient of f; that is, find $\frac{f(x + h)-f(x)}{h}$, $h
eq0$, for the following function. $f(x)=2x^{2}-x - 1$ $\frac{f(x + h)-f(x)}{h}=square$ (simplify your answer.)

Explanation:

Step1: Find f(x + h)

Substitute x + h into f(x):
\[

$$\begin{align*} f(x + h)&=2(x + h)^2-(x + h)-1\\ &=2(x^{2}+2xh+h^{2})-x - h-1\\ &=2x^{2}+4xh+2h^{2}-x - h-1 \end{align*}$$

\]

Step2: Calculate f(x + h)-f(x)

\[

$$\begin{align*} f(x + h)-f(x)&=(2x^{2}+4xh+2h^{2}-x - h-1)-(2x^{2}-x - 1)\\ &=2x^{2}+4xh+2h^{2}-x - h-1 - 2x^{2}+x + 1\\ &=4xh+2h^{2}-h \end{align*}$$

\]

Step3: Find the difference quotient

\[

$$\begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{4xh+2h^{2}-h}{h}\\ &=\frac{h(4x + 2h-1)}{h}\\ &=4x+2h - 1 \end{align*}$$

\]

Answer:

$4x + 2h-1$