Let's analyze the problem from the image.

The function is given as:

$f(x) = 4 - x^2$

Part (A):

You are asked to evaluate:

$\frac{f(9) - f(5)}{9 - 5}$

First, compute $f(9)$ and $f(5)$.

$f(9) = 4 - 9^2 = 4 - 81 = -77$ $f(5) = 4 - 5^2 = 4 - 25 = -21$

Now, substitute these values into the expression:

$\frac{f(9) - f(5)}{9 - 5} = \frac{-77 - (-21)}{4} = \frac{-77 + 21}{4} = \frac{-56}{4} = -14$

So, the answer to Part (A) is:

$\boxed{-14}$

Part (B):

You are asked to evaluate the following limit-like expression:

$\frac{f(5 + h) - f(5)}{h}$

We already know that $f(5) = -21$. Now, let's compute $f(5 + h)$:

$f(5 + h) = 4 - (5 + h)^2 = 4 - (25 + 10h + h^2) = 4 - 25 - 10h - h^2 = -21 - 10h - h^2$

Now, substitute this into the expression:

$\frac{f(5 + h) - f(5)}{h} = \frac{-21 - 10h - h^2 - (-21)}{h} = \frac{-10h - h^2}{h} = -10 - h$

So, the answer to Part (B) is:

$\boxed{-10 - h}$

Would you like further clarifications or details?

Here are 5 related questions you can explore:

1. What is the derivative of the function $f(x) = 4 - x^2$?
2. How do you interpret the result of Part (B) in terms of calculus (hint: it's a difference quotient)?
3. What happens to the expression in Part (B) as $h \to 0$?
4. Can you generalize the difference quotient for $f(x)$ at any point $a$ instead of just 5?
5. What is the significance of negative values in the function $f(x) = 4 - x^2$?

Tip: The expression in Part (B) is closely related to the concept of the derivative in calculus. Understanding how these difference quotients behave as $h \to 0$ is key to grasping differentiation.