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The problem asks us to apply the difference quotient to the function \( f(x) = 2x - 5 \).

### Step 1: Simplify \( f(x + h) \)
Given that \( f(x) = 2x - 5 \), let's compute \( f(x + h) \):

\[
f(x + h) = 2(x + h) - 5 = 2x + 2h - 5
\]

So, \( f(x + h) = 2x + 2h - 5 \).

### Step 2: Simplify the Difference Quotient
The difference quotient is:

\[
\frac{f(x+h) - f(x)}{h}
\]

Now substitute \( f(x + h) = 2x + 2h - 5 \) and \( f(x) = 2x - 5 \):

\[
\frac{(2x + 2h - 5) - (2x - 5)}{h}
\]

Simplify the numerator:

\[
\frac{2x + 2h - 5 - 2x + 5}{h} = \frac{2h}{h}
\]

Finally, simplify the expression:

\[
\frac{2h}{h} = 2
\]

Thus, the simplified difference quotient is 2.

Would you like more details on this or have any further questions?

### 5 Related Questions:
1. What does the difference quotient represent in calculus?
2. How does the difference quotient relate to the derivative of a function?
3. What is the geometric interpretation of the difference quotient?
4. How would the difference quotient change if the function \( f(x) \) were non-linear?
5. Can you explain why the difference quotient for a linear function always simplifies to a constant?

### Tip:
The difference quotient is foundational to understanding derivatives—it's essentially the formula for the slope of the secant line between two points on a curve.

Simplify the expression f(x + h) for f(x) = 2x - 5, and apply the difference quotient formula.

This problem involves simplifying f(x + h) for the function f(x) = 2x - 5, followed by applying the difference quotient formula. It covers algebraic manipulation and key calculus concepts, such as the difference quotient, suitable for high school students (Grades 9-12).

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