Let's break down the given problem step by step using the provided graph of the functions $f(x)$ and $g(x)$.

Part (a) $u(x) = f(x)g(x)$, find $u'(3)$:

To differentiate $u(x)$, we apply the product rule: $u'(x) = f'(x)g(x) + f(x)g'(x)$ Now, we need to find $u'(3)$, which means we evaluate the above expression at $x = 3$.

• From the graph:
  • $f(3) = 2$ (the red line at $x = 3$),
  • $g(3) = 0$ (the blue line at $x = 3$),
  • $f'(x)$ (the slope of $f(x)$) is constant and equal to 1 (since it is a straight line with a slope of 1),
  • $g'(x)$ (the slope of $g(x)$) at $x = 3$ can be found by observing the change in $g(x)$. It appears that $g(x)$ has a slope of $-1$.

Now apply these values: $u'(3) = f'(3)g(3) + f(3)g'(3) = (1)(0) + (2)(-1) = -2$

Part (b) $v(x) = f(f(x))$, find $v'(-2)$:

To differentiate $v(x)$, we apply the chain rule: $v'(x) = f'(f(x)) \cdot f'(x)$ Now, we need to find $v'(-2)$.

• From the graph:
  • $f(-2) = -2$ (the red line at $x = -2$),
  • $f'(x) = 1$ (again, since $f(x)$ has a constant slope of 1 for all $x$).

Thus, $v'(-2) = f'(f(-2)) \cdot f'(-2) = f'(-2) \cdot 1 = 1$

Final Answers:

• (a) $u'(3) = -2$
• (b) $v'(-2) = 1$

Would you like more details on any step or have any other questions?

Here are 5 related questions for further practice:

1. How would you find $u'(x)$ for any general $x$?
2. What would be $u'(x)$ if both functions were exponential?
3. How do you compute $v'(x)$ if $f(x)$ were a quadratic function?
4. How can we interpret the slopes $f'(x)$ and $g'(x)$ geometrically from the graph?
5. What is the significance of the product rule in real-world applications?

Tip: When working with derivatives from a graph, always focus on the slope and function values at the point of interest.