Question

a) find the simplified form of the difference quotient for the function ( f(x) = 2x^2 - 3x ). b) complete the following table.

( x )	( h )	( \frac{f(x + h) - f(x)}{h} )
6	1
6	0.1
6	0.01

a) the difference quotient is ( square ).
(simplify your answer. do not factor)

Explanation:

Part (a)

Step1: Recall the difference quotient formula

The difference quotient of a function \( f(x) \) is given by \( \frac{f(x + h)-f(x)}{h} \), where \( h
eq0 \).

Step2: Compute \( f(x + h) \)

Given \( f(x)=2x^{2}-3x \), we substitute \( x + h \) into the function:
\[

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

\]

Step3: Compute \( f(x + h)-f(x) \)

Subtract \( f(x) \) from \( f(x + h) \):
\[

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

\]

Step4: Divide by \( h \) to get the difference quotient

Divide \( f(x + h)-f(x) \) by \( h \) ( \( h
eq0 \)):
\[
\frac{f(x + h)-f(x)}{h}=\frac{4xh + 2h^{2}-3h}{h}=\frac{h(4x + 2h-3)}{h}=4x + 2h-3
\]

Answer:

\( 4x + 2h-3 \)

Part (b)

We use the simplified difference quotient \( \frac{f(x + h)-f(x)}{h}=4x + 2h-3 \) with \( x = 6 \).

For each value of \( h \):

• When \( h = 2 \):

\( 4(6)+2(2)-3=24 + 4-3=25 \)

• When \( h = 1 \):

\( 4(6)+2(1)-3=24 + 2-3=23 \)

• When \( h = 0.1 \):

\( 4(6)+2(0.1)-3=24+0.2 - 3=21.2 \)

• When \( h = 0.01 \):

\( 4(6)+2(0.01)-3=24 + 0.02-3=21.02 \)

The completed table is:

\( x \)	\( h \)	\( \frac{f(x + h)-f(x)}{h} \)
6	1	23
6	0.1	21.2
6	0.01	21.02