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.