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The given problem asks to find \( v'(8) \), where \( v(x) = f(x)g(x) \), using the graphs of functions \( f(x) \) and \( g(x) \). To solve this, we need to apply the product rule for differentiation:

\[
v'(x) = f'(x)g(x) + f(x)g'(x)
\]

We are asked specifically to compute \( v'(8) \), so let's break this down:

1. **Find \( f(8) \) and \( g(8) \):**
   From the graph:
   - \( f(8) = -1 \)
   - \( g(8) = 3 \)

2. **Find \( f'(8) \) and \( g'(8) \):**
   The derivatives \( f'(x) \) and \( g'(x) \) represent the slopes of the respective graphs at \( x = 8 \).
   - For \( f(x) \), between \( x = 7 \) and \( x = 9 \), the points are \( (7, -2) \) and \( (9, -4) \). The slope is:
     \[
     f'(8) = \frac{-4 - (-2)}{9 - 7} = \frac{-2}{2} = -1
     \]
   - For \( g(x) \), between \( x = 7 \) and \( x = 9 \), the points are \( (7, 3) \) and \( (9, -1) \). The slope is:
     \[
     g'(8) = \frac{-1 - 3}{9 - 7} = \frac{-4}{2} = -2
     \]

3. **Apply the product rule:**
   Using the values:
   \[
   v'(8) = f'(8)g(8) + f(8)g'(8)
   \]
   Substituting the values:
   \[
   v'(8) = (-1)(3) + (-1)(-2) = -3 + 2 = -1
   \]

Thus, \( v'(8) = -1 \).

Would you like further details or explanations?

Here are 5 related questions to think about:
1. What is the product rule for three functions?
2. How do you estimate the slope of a graph visually?
3. How does the chain rule differ from the product rule?
4. Can you apply the product rule to second derivatives?
5. How would \( v'(x) \) behave if \( f(x) \) or \( g(x) \) were constant?

**Tip:** Always double-check the slopes by reviewing the change in \( y \)-values and \( x \)-values carefully!

Learn how to find the derivative of v(x) = f(x)g(x) at x = 8 using the product rule. This calculus problem involves differentiation and understanding the slopes of graphs. Suitable for students in Grades 11-12.

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