Let's break down the problem shown in the image.

We are given two functions $f(x)$ and $g(x)$, and we are tasked with finding:

(a) $u'(1)$, where $u(x) = f(x) g(x)$.

(b) $v'(5)$, where $v(x) = \frac{f(x)}{g(x)}$.

Part (a): Finding $u'(1)$

For $u(x) = f(x)g(x)$, we need to apply the product rule: $u'(x) = f'(x)g(x) + f(x)g'(x)$ At $x = 1$, we need to find $f(1)$, $f'(1)$, $g(1)$, and $g'(1)$.

From the graph:

• $f(1) = 0$ (since the function $f(x)$ crosses the x-axis at $x = 1$),
• $g(1) = 0$ (the graph of $g(x)$ also crosses the x-axis at $x = 1$),
• $f'(1)$, which is the slope of the line at $x = 1$, is negative (approx. $-2$),
• $g'(1)$ is positive (approx. $2$).

Using the product rule at $x = 1$: $u'(1) = f'(1)g(1) + f(1)g'(1) = (-2)(0) + (0)(2) = 0$ So, $u'(1) = 0$, which is correctly marked in your solution.

Part (b): Finding $v'(5)$

For $v(x) = \frac{f(x)}{g(x)}$, we need to apply the quotient rule: $v'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$ At $x = 5$, we need to find $f(5)$, $f'(5)$, $g(5)$, and $g'(5)$.

From the graph:

• $f(5) = 1$,
• $g(5) = -2$,
• $f'(5)$ (the slope of $f(x)$ at $x = 5$) is approximately $0.5$,
• $g'(5)$ (the slope of $g(x)$ at $x = 5$) is $1$.

Now, applying the quotient rule: $v'(5) = \frac{(0.5)(-2) - (1)(1)}{(-2)^2} = \frac{-1 - 1}{4} = \frac{-2}{4} = -\frac{1}{2}$

Thus, $v'(5) = -\frac{1}{2}$.

Let me know if you need more details, or have further questions! Here are some related questions for you:

1. What is the difference between the product rule and quotient rule?
2. How do you determine the slope of a linear function from a graph?
3. Can the derivative of a product ever be zero? Why or why not?
4. How does the sign of the slope affect the behavior of the derivative?
5. What does a zero derivative tell you about the function at that point?

Tip: Always verify the slopes carefully from the graph when calculating derivatives using product or quotient rules.