Determine the value of the difference quotient (the average rate of change) near the point
(
3
,
f
(
3
)
)
when:
f
(
x
)
=
x
2
−
4
x
and when the length of
h
is
0.5
.

−
17.5

1

1.25

2

2.5

To determine the value of the difference quotient (which is the average rate of change of the function \( f(x) \) near the point \( (3, f(3)) \)), we'll use the formula for the difference quotient:

\[
\text{Difference Quotient} = \frac{f(3 + h) - f(3)}{h}
\]

Given:
- \( f(x) = x^2 - 4x \)
- \( h = 0.5 \)

### Step 1: Calculate \( f(3) \)
\[
f(3) = 3^2 - 4(3) = 9 - 12 = -3
\]

### Step 2: Calculate \( f(3 + h) \)
\[
f(3 + 0.5) = f(3.5)
\]
\[
f(3.5) = (3.5)^2 - 4(3.5) = 12.25 - 14 = -1.75
\]

### Step 3: Compute the difference quotient
\[
\frac{f(3.5) - f(3)}{0.5} = \frac{-1.75 - (-3)}{0.5} = \frac{-1.75 + 3}{0.5} = \frac{1.25}{0.5} = 2.5
\]

So, the value of the difference quotient (average rate of change) near the point \( (3, f(3)) \) is **2.5**.

If you have any questions or would like further details, feel free to ask!

Here are some related questions:
1. What is the difference quotient for \( f(x) = x^2 - 4x \) at \( x = 4 \) and \( h = 0.5 \)?
2. How does the value of \( h \) affect the difference quotient?
3. What happens to the difference quotient as \( h \) approaches zero?
4. Can the difference quotient be used to approximate the derivative?
5. How do you interpret the difference quotient in the context of a graph?

**Tip:** The difference quotient is a fundamental concept in calculus, as it leads to the definition of the derivative, representing the instantaneous rate of change at a point.

Learn how to calculate the difference quotient (average rate of change) for the function f(x) = x^2 - 4x near the point (3, f(3)). Detailed step-by-step solution provided. Suitable for Grades 11-12.

Math Problem Statement

Determine the value of the difference quotient (the average rate of change) near the point ( 3 , f ( 3 ) ) when: f ( x )

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