Let's analyze and explain the content step by step:

Understanding the Problem

The given function is:
$f(x) = (x-4) \sqrt[3]{x+3}.$

You are working with a limit involving $h \to 0$, aiming to compute: $\lim_{h \to 0} \frac{f(5-h) - 2}{h}.$

From the problem, we already know:

• $f(5) = 2$, which was evaluated using the function $f(x)$.

Step 1: Plugging into the Function

To find $f(5)$, substitute $x = 5$ into $f(x)$: [ f(5) = (5-4) \sqrt[3]{5+3} = 1 \cdot \sqrt[3]{8}. ] Since $\sqrt[3]{8} = 2$: $f(5) = 2.$

Thus, $f(5) = 2$ is confirmed.

Step 2: Setting up the Limit

The limit is: $\lim_{h \to 0} \frac{f(5-h) - 2}{h}.$ This represents the derivative of $f(x)$ at $x = 5$, as the definition of the derivative is: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.$

Step 3: Substitute $f(5-h)$

Using the function $f(x)$, compute $f(5-h)$: $f(5-h) = ((5-h) - 4) \cdot \sqrt[3]{(5-h) + 3}.$ Simplify: $f(5-h) = (1-h) \cdot \sqrt[3]{8-h}.$

Now, substitute $f(5-h)$ into the limit: $\lim_{h \to 0} \frac{(1-h) \cdot \sqrt[3]{8-h} - 2}{h}.$

Step 4: Why $f(5) = 2$ Helps

Since $f(5) = 2$, we simplify the numerator by factoring or simplifying $(1-h) \cdot \sqrt[3]{8-h}$ around $h = 0$. The value $f(5) = 2$ confirms continuity and helps us evaluate the derivative.

Final Remarks

This is the derivative of $f(x)$ at $x = 5$. The steps beyond would involve careful expansion and simplifications (e.g., Taylor series or limits).

Would you like me to go further into evaluating the derivative explicitly or clarify a specific part?

Here are five related questions:

1. How do you evaluate cube roots for functions like $f(x)$?
2. How does the definition of the derivative apply here?
3. Why do we subtract $2$ in the numerator?
4. Can we expand $\sqrt[3]{8-h}$ using approximations?
5. How can you find derivatives of products like $(x-4) \cdot \sqrt[3]{x+3}$ directly?

Tip: Always confirm $f(x)$ values at given points to ensure continuity before computing derivatives.