Let's analyze and solve the problem step by step:

Given:

• Two functions $f(x)$ and $g(x)$ are represented graphically.
• We need to find:
  1. $u'(3)$, where $u(x) = f(x)g(x)$.
  2. $v'(-2)$, where $v(x) = f(f(x))$.

Part (a) — Finding $u'(3)$

We are given $u(x) = f(x)g(x)$. To find $u'(x)$, we apply the product rule for differentiation: $u'(x) = f'(x)g(x) + f(x)g'(x)$

Now, we need to compute $u'(3)$, so we'll evaluate both $f'(3)$, $g'(3)$, $f(3)$, and $g(3)$.

From the graph:

• $f(3)$: The value of $f(x)$ at $x = 3$ is 2.
• $g(3)$: The value of $g(x)$ at $x = 3$ is 1.
• $f'(3)$: $f(x)$ appears linear between $x = 2$ and $x = 4$, with a slope of $\frac{4 - 0}{4 - 2} = 2$.
• $g'(3)$: $g(x)$ also appears linear, with a slope of $\frac{0 - 4}{4 - 2} = -2$.

Using these, we can compute $u'(3)$: $u'(3) = f'(3)g(3) + f(3)g'(3)$ $u'(3) = (2)(1) + (2)(-2) = 2 - 4 = -2$

Part (b) — Finding $v'(-2)$

We are given $v(x) = f(f(x))$. To find $v'(x)$, we apply the chain rule: $v'(x) = f'(f(x)) \cdot f'(x)$

Now, we need to compute $v'(-2)$, so we'll evaluate $f'(-2)$, $f(-2)$, and $f'(f(-2))$.

From the graph:

• $f(-2)$: The value of $f(x)$ at $x = -2$ is $-3$.
• $f'(-2)$: The slope of $f(x)$ for $x < 0$ is $\frac{-2 - (-4)}{0 - (-2)} = 1$.
• **f'(f(-2)) = f'(-3):** \( f(x) at $x = -3$ also has a slope of 1 (same slope as in that region).

Now, using the chain rule: $v'(-2) = f'(f(-2)) \cdot f'(-2) = f'(-3) \cdot f'(-2) = 1 \cdot 1 = 1$

Final Answers:

• (a) $u'(3) = -2$
• (b) $v'(-2) = 1$

Would you like any further clarification or a more detailed explanation on any part of the solution?

Here are 5 related questions to explore further:

1. How would the answer change if the graph of $f(x)$ or $g(x)$ had a different slope?
2. Can you apply the product rule in other real-world scenarios?
3. What happens when combining chain and product rules in the same problem?
4. How do inflection points of functions affect their derivatives?
5. What is the significance of second derivatives in understanding graph curvature?

Tip: When using the product or chain rule, always ensure you compute the values of the functions and their derivatives at the specific point required.