Question

given the following function, determine the difference quotient, $\frac{f(x + h)-f(x)}{h}$. $f(x)=3x^{2}-2x + 4$

Explanation:

Step1: Find \(f(x + h)\)

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

$$\begin{align*} f(x + h)&=3(x + h)^2-2(x + h)+4\\ &=3(x^{2}+2xh+h^{2})-2x-2h + 4\\ &=3x^{2}+6xh+3h^{2}-2x-2h + 4 \end{align*}$$

\]

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

\[

$$\begin{align*} f(x + h)-f(x)&=(3x^{2}+6xh+3h^{2}-2x-2h + 4)-(3x^{2}-2x + 4)\\ &=3x^{2}+6xh+3h^{2}-2x-2h + 4-3x^{2}+2x - 4\\ &=6xh+3h^{2}-2h \end{align*}$$

\]

Step3: Find the difference quotient \(\frac{f(x + h)-f(x)}{h}\)

\[

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

\]

Answer:

\(6x+3h - 2\)