Let 
P(x) = F(x)G(x)
 and 
Q(x) = 
F(x)
G(x)
,
 where F and G are the functions whose graphs are shown.

The x y coordinate plane is given. There are 2 functions on the graph.
The function labeled F enters the window near point (0.3, 6), goes down and right becoming less steep, passess through the point (1, 4), changes direction at the point (2, 3), goes up and right becoming more steep, sharply changes direction at the point (3, 4), goes linearly up and right, passes through the point (7, 5), and ends near the point (8, 5.3).
The function labeled G consists of two line segments. Function G starts at the point (0, 1), goes linearly up and right, passes through the point (2, 2), sharply changes direction at the point (4, 3), goes linearly down and right, passes through the point (7, 1), and ends near the point (8, 0.3).
(a)
Find 
P'(2).
1
 
Incorrect: Your answer is incorrect.
(b)
Find 
Q'(7).


Let's address this problem step by step. We are given two functions, \( F(x) \) and \( G(x) \), and we need to find derivatives of the composite products \( P(x) = F(x)G(x) \) and \( Q(x) = \frac{F(x)}{G(x)} \). 

To solve the problem:

### (a) Find \( P'(2) \)
Given that \( P(x) = F(x)G(x) \), we use the **product rule** for differentiation:

\[
P'(x) = F'(x)G(x) + F(x)G'(x)
\]

We are tasked with finding \( P'(2) \), so we need to evaluate \( F(2), G(2), F'(2), \) and \( G'(2) \).

From the graph descriptions:
- \( F(2) = 3 \) (since the function passes through \( (2, 3) \)).
- \( G(2) = 2 \) (since the function passes through \( (2, 2) \)).

Now, to estimate the derivatives \( F'(2) \) and \( G'(2) \), we need to use the graph's behavior around \( x = 2 \):
- \( F'(2) \) can be approximated by the slope of the tangent line at \( x = 2 \). The graph indicates that the function changes direction at \( (2, 3) \), suggesting the slope is zero at this point. Hence, \( F'(2) = 0 \).
- \( G'(2) \) can be calculated by the slope of the line segment between \( (0, 1) \) and \( (2, 2) \), which gives a slope of:

\[
G'(2) = \frac{2 - 1}{2 - 0} = \frac{1}{2}
\]

Using these values in the product rule:

\[
P'(2) = F'(2)G(2) + F(2)G'(2) = (0)(2) + (3)\left( \frac{1}{2} \right) = 0 + \frac{3}{2} = \frac{3}{2}
\]

Thus, the correct value is:

\[
P'(2) = \frac{3}{2}
\]

### (b) Find \( Q'(7) \)
Given that \( Q(x) = \frac{F(x)}{G(x)} \), we use the **quotient rule** for differentiation:

\[
Q'(x) = \frac{F'(x)G(x) - F(x)G'(x)}{(G(x))^2}
\]

We are tasked with finding \( Q'(7) \), so we need to evaluate \( F(7), G(7), F'(7), \) and \( G'(7) \).

From the graph descriptions:
- \( F(7) = 5 \) (since the function passes through \( (7, 5) \)).
- \( G(7) = 1 \) (since the function passes through \( (7, 1) \)).

Now, we estimate the derivatives \( F'(7) \) and \( G'(7) \):
- \( F'(7) \) can be approximated by the slope of the line segment around \( x = 7 \). Based on the graph description, the function is increasing linearly between \( x = 3 \) and \( x = 7 \), so we can calculate the slope:

\[
F'(7) = \frac{5 - 4}{7 - 3} = \frac{1}{4}
\]

- \( G'(7) \) can be calculated by the slope of the line segment between \( (4, 3) \) and \( (7, 1) \), which gives a slope of:

\[
G'(7) = \frac{1 - 3}{7 - 4} = \frac{-2}{3}
\]

Using these values in the quotient rule:

\[
Q'(7) = \frac{F'(7)G(7) - F(7)G'(7)}{(G(7))^2} = \frac{\left( \frac{1}{4} \right)(1) - (5)\left( \frac{-2}{3} \right)}{(1)^2}
\]

Simplifying:

\[
Q'(7) = \frac{\frac{1}{4} + \frac{10}{3}}{1} = \frac{1}{4} + \frac{10}{3}
\]

To add these fractions, we find a common denominator:

\[
Q'(7) = \frac{3}{12} + \frac{40}{12} = \frac{43}{12}
\]

Thus, the correct value is:

\[
Q'(7) = \frac{43}{12}
\]

### Recap:
- (a) \( P'(2) = \frac{3}{2} \)
- (b) \( Q'(7) = \frac{43}{12} \)

Would you like further details or clarification on any part? 

Here are 5 related questions to expand on these concepts:
1. How does the product rule apply when multiplying more than two functions?
2. Can you explain how to estimate derivatives from a graph more precisely?
3. How would \( P(x) \) and \( Q(x) \) change if one of the functions were a constant?
4. How does the quotient rule differ from the product rule in practical applications?
5. What are common scenarios where using the quotient rule becomes challenging?

**Tip:** When using the product and quotient rules, carefully check the signs of your derivatives, as small errors here can significantly affect the final result.

Let P(x) = F(x)G(x) and Q(x) = F(x)/G(x), where F and G are the functions whose graphs are shown. The x y coordinate plane is given. There are 2 functions on the graph. The function labeled F enters the window near point (0.3, 6), goes down and right becoming less steep, passes through the point (1, 4), changes direction at the point (2, 3), goes up and right becoming more steep, sharply changes direction at the point (3, 4), goes linearly up and right, passes through the point (7, 5), and ends near the point (8, 5.3). The function labeled G consists of two line segments. Function G starts at the point (0, 1), goes linearly up and right, passes through the point (2, 2), sharply changes direction at the point (4, 3), goes linearly down and right, passes through the point (7, 1), and ends near the point (8, 0.3). (a) Find P'(2). (b) Find Q'(7).

Learn how to find the derivatives P'(2) and Q'(7) for composite functions using the product and quotient rules. This problem involves analyzing the behavior of the functions F and G from their graphs and applying key calculus techniques, suitable for high school calculus students.

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