We are asked to simplify the expression:

$\frac{f(x+h) - f(x)}{h}$

where $f(x) = \frac{1}{2}x$.

Step 1: Substitute the function $f(x) = \frac{1}{2}x$ into the expression

We begin by substituting $f(x)$ and $f(x+h)$:

$f(x+h) = \frac{1}{2}(x+h)$

Now, substitute this and $f(x)$ into the original expression:

$\frac{f(x+h) - f(x)}{h} = \frac{\frac{1}{2}(x+h) - \frac{1}{2}x}{h}$

Step 2: Simplify the numerator

Simplify the terms in the numerator:

$\frac{1}{2}(x+h) - \frac{1}{2}x = \frac{1}{2}x + \frac{1}{2}h - \frac{1}{2}x = \frac{1}{2}h$

Step 3: Simplify the whole expression

Now, the expression becomes:

$\frac{\frac{1}{2}h}{h}$

Since $h \neq 0$, we can cancel $h$ from the numerator and denominator:

$\frac{1}{2}$

Final Answer:

Thus, the simplified expression is:

$\frac{f(x+h) - f(x)}{h} = \frac{1}{2}$

Let me know if you'd like more explanation or if you have further questions!

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Here are some follow-up questions that expand on this topic:

1. What is the significance of this expression in calculus (hint: it’s related to the derivative)?
2. How would the expression change if $f(x) = x^2$?
3. What happens when you take the limit of $\frac{f(x+h) - f(x)}{h}$ as $h$ approaches 0?
4. What is the derivative of $f(x) = \frac{1}{2}x$?
5. How do you simplify this expression for a more complex function, like $f(x) = x^3 + 2x$?

Tip: When working with expressions like this, remember to cancel out like terms early to avoid dealing with unnecessary complexity!