Math Problem Statement

The graphs of the function F (left, in blue) and G (right, in red) are given. Let P(x) = F(x)G(x) and Q(x) = F(x)/G(x). Calculate the derivatives P'(6) and Q'(1).

Solution

To solve the problem, let's analyze each function and the derivatives requested.

Given Functions

  • F(x)F(x): The blue graph (left)
  • G(x)G(x): The red graph (right)

Definitions

  • P(x)=F(x)G(x)P(x) = F(x) \cdot G(x)
  • Q(x)=F(x)G(x)Q(x) = \frac{F(x)}{G(x)}

Derivatives

  1. For P(x)P'(x): By the product rule: P(x)=F(x)G(x)+F(x)G(x)P'(x) = F'(x) \cdot G(x) + F(x) \cdot G'(x)
  2. For Q(x)Q'(x): By the quotient rule: Q(x)=F(x)G(x)F(x)G(x)[G(x)]2Q'(x) = \frac{F'(x) \cdot G(x) - F(x) \cdot G'(x)}{[G(x)]^2}

Evaluations Needed

  1. P(6)P'(6)
    • We need to find F(6),F(6),G(6),F(6), F'(6), G(6), and G(6)G'(6).
  2. Q(1)Q'(1)
    • We need to find F(1),F(1),G(1),F(1), F'(1), G(1), and G(1)G'(1).

Step 1: Evaluating P(6)P'(6)

Looking at the graphs:

  • F(6)F(6): Find the y-coordinate at x=6x = 6 on the blue graph.
  • F(6)F'(6): Find the slope of the tangent line at x=6x = 6 on the blue graph.
  • G(6)G(6): Find the y-coordinate at x=6x = 6 on the red graph.
  • G(6)G'(6): Find the slope of the tangent line at x=6x = 6 on the red graph.

Step 2: Evaluating Q(1)Q'(1)

Looking at the graphs:

  • F(1)F(1): Find the y-coordinate at x=1x = 1 on the blue graph.
  • F(1)F'(1): Find the slope of the tangent line at x=1x = 1 on the blue graph.
  • G(1)G(1): Find the y-coordinate at x=1x = 1 on the red graph.
  • G(1)G'(1): Find the slope of the tangent line at x=1x = 1 on the red graph.

Graph Analysis

  1. For x=6x = 6:

    • F(6)3F(6) \approx 3
    • F(6)slope=1F'(6) \approx \text{slope} = 1
    • G(6)2G(6) \approx 2
    • G(6)slope=3G'(6) \approx \text{slope} = 3

    Substituting into the product rule: P(6)=(1)(2)+(3)(3)=2+9=11P'(6) = (1)(2) + (3)(3) = 2 + 9 = 11

  2. For x=1x = 1:

    • F(1)0F(1) \approx 0
    • F(1)slope=2F'(1) \approx \text{slope} = 2
    • G(1)1G(1) \approx 1
    • G(1)slope=1G'(1) \approx \text{slope} = -1

    Substituting into the quotient rule: Q(1)=(2)(1)(0)(1)(1)2=21=2Q'(1) = \frac{(2)(1) - (0)(-1)}{(1)^2} = \frac{2}{1} = 2

Answers

  • P(6)=11P'(6) = 11
  • Q(1)=2Q'(1) = 2

Would you like more details on any of the steps or have any questions?

Here are some related questions:

  1. How do you determine the slope of a graph at a given point?
  2. What is the significance of the product rule in differentiation?
  3. Why is the quotient rule used, and how does it differ from the product rule?
  4. How can you verify the accuracy of your derivative calculations using graphs?
  5. How does understanding derivatives help in analyzing the behavior of functions?

Tip: When using the product or quotient rule, always double-check each component’s derivative to ensure precision.

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Math Problem Analysis

Mathematical Concepts

Calculus
Graphical Analysis
Product Rule
Quotient Rule

Formulas

P'(x) = F'(x)G(x) + F(x)G'(x)
Q'(x) = (F'(x)G(x) - F(x)G'(x)) / [G(x)]^2

Theorems

Product Rule
Quotient Rule

Suitable Grade Level

Grades 10-12