Question

find the difference quotient $\frac{f(x + h)-f(x)}{h}$ for the function and simplify it. $f(x)=3x^{2}+6x$ the difference quotient is $square$. (simplify your answer. do not factor.)

Explanation:

Step1: Find f(x + h)

Substitute \(x+h\) into \(f(x)=3x^{2}+6x\).
\[

$$\begin{align*} f(x + h)&=3(x + h)^{2}+6(x + h)\\ &=3(x^{2}+2xh+h^{2})+6x + 6h\\ &=3x^{2}+6xh+3h^{2}+6x + 6h \end{align*}$$

\]

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

\[

$$\begin{align*} f(x + h)-f(x)&=(3x^{2}+6xh+3h^{2}+6x + 6h)-(3x^{2}+6x)\\ &=3x^{2}+6xh+3h^{2}+6x + 6h - 3x^{2}-6x\\ &=6xh+3h^{2}+6h \end{align*}$$

\]

Step3: Calculate the difference quotient

\[

$$\begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{6xh+3h^{2}+6h}{h}\\ &=\frac{h(6x + 3h+6)}{h}\\ &=6x+3h + 6 \end{align*}$$

\]

Answer:

\(6x+3h + 6\)