Question

find the difference quotient $\frac{f(x + h)-f(x)}{h}$, where $h
eq0$, for the function below.

$f(x)=2x^{2}+6$

simplify your answer as much as possible.

$\frac{f(x + h)-f(x)}{h}=square$

Explanation:

Step1: Find $f(x + h)$

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

$$\begin{align*} f(x + h)&=2(x + h)^2+6\\ &=2(x^{2}+2xh+h^{2})+6\\ &=2x^{2}+4xh + 2h^{2}+6 \end{align*}$$

\]

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

\[

$$\begin{align*} f(x + h)-f(x)&=(2x^{2}+4xh + 2h^{2}+6)-(2x^{2}+6)\\ &=2x^{2}+4xh + 2h^{2}+6 - 2x^{2}-6\\ &=4xh+2h^{2} \end{align*}$$

\]

Step3: Find the difference quotient

\[

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

\]

Answer:

$4x + 2h$