To find $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$, we need to go through the following steps:

Step 1: Find $f(x+h)$

We are given the function $f(x) = -3x + 10$. Now, substitute $x+h$ into the function:

$f(x+h) = -3(x+h) + 10 = -3x - 3h + 10$

Step 2: Compute $f(x+h) - f(x)$

Now, subtract $f(x)$ from $f(x+h)$:

$f(x+h) - f(x) = (-3x - 3h + 10) - (-3x + 10)$

Simplify the expression:

$f(x+h) - f(x) = -3x - 3h + 10 + 3x - 10 = -3h$

Step 3: Set up the difference quotient

Now, divide $f(x+h) - f(x)$ by $h$:

$\frac{f(x+h) - f(x)}{h} = \frac{-3h}{h}$

Simplify:

$\frac{f(x+h) - f(x)}{h} = -3$

Step 4: Take the limit as $h \to 0$

The next step is to take the limit of the difference quotient as $h \to 0$:

$f'(x) = \lim_{h \to 0} -3 = -3$

Final Result:

The derivative of the function is constant:

$f'(x) = -3$

Would you like further details or clarifications?

Here are 5 related questions for further practice:

1. What is the geometric interpretation of the derivative $f'(x)$?
2. How would the derivative change if the function were $f(x) = -3x + c$ with a different constant $c$?
3. What is the second derivative $f''(x)$ of $f(x) = -3x + 10$?
4. How does the derivative help in understanding the behavior of linear functions?
5. What happens to the slope when you modify the coefficient of $x$ in a linear function?

Tip: For linear functions of the form $f(x) = mx + b$, the derivative $f'(x)$ is always equal to the slope $m$.