Math Problem Statement

The graphs of the function f (given in blue) and g (given in red) are plotted above. Suppose that u(x)=f(x)g(x) and v(x)=f(x)/g(x) . Find each of the following:

u′(1)

v′(1)

The graphs of the function f (blue) and g (red) are plotted above. Suppose u(x)=f(x)g(x) and v(x)=f(x)/g(x). Find u′(1) and v′(1).

Solution

The given problem asks for the derivatives u(1)u'(1) and v(1)v'(1), where u(x)=f(x)g(x)u(x) = f(x)g(x) and v(x)=f(x)g(x)v(x) = \frac{f(x)}{g(x)}. To find these derivatives, we will need to use the product rule and the quotient rule, respectively. First, let's break down the information we can extract from the graph:

Step 1: Reading f(1)f(1), g(1)g(1), f(1)f'(1), and g(1)g'(1)

We will extract the following values from the graph at x=1x = 1:

  1. For f(x)f(x) (blue graph):

    • f(1)f(1) is the value of the blue function at x=1x = 1.
    • f(1)f'(1) is the slope of the blue function at x=1x = 1.
  2. For g(x)g(x) (red graph):

    • g(1)g(1) is the value of the red function at x=1x = 1.
    • g(1)g'(1) is the slope of the red function at x=1x = 1.

Let’s extract these values from the graph visually:

  • At x=1x = 1, f(1)=2f(1) = 2 (from the blue graph).
  • At x=1x = 1, g(1)=4g(1) = 4 (from the red graph).

Next, we'll estimate the slopes:

  • For f(1)f'(1), the slope between x=0x = 0 and x=1x = 1 appears to be 20=22 - 0 = 2, so f(1)=2f'(1) = 2.
  • For g(1)g'(1), the slope between x=0x = 0 and x=1x = 1 is 45=14 - 5 = -1, so g(1)=1g'(1) = -1.

Step 2: Calculating u(1)u'(1) and v(1)v'(1)

Product Rule for u(x)=f(x)g(x)u(x) = f(x)g(x):

The derivative of the product u(x)=f(x)g(x)u(x) = f(x)g(x) is given by: u(x)=f(x)g(x)+f(x)g(x)u'(x) = f'(x)g(x) + f(x)g'(x) At x=1x = 1: u(1)=f(1)g(1)+f(1)g(1)u'(1) = f'(1)g(1) + f(1)g'(1) Substitute the values: u(1)=(2)(4)+(2)(1)=82=6u'(1) = (2)(4) + (2)(-1) = 8 - 2 = 6

Quotient Rule for v(x)=f(x)g(x)v(x) = \frac{f(x)}{g(x)}:

The derivative of the quotient v(x)=f(x)g(x)v(x) = \frac{f(x)}{g(x)} is given by: v(x)=f(x)g(x)f(x)g(x)[g(x)]2v'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} At x=1x = 1: v(1)=(2)(4)(2)(1)[4]2=8+216=1016=58v'(1) = \frac{(2)(4) - (2)(-1)}{[4]^2} = \frac{8 + 2}{16} = \frac{10}{16} = \frac{5}{8}

Final Answers:

u(1)=6u'(1) = 6 v(1)=58v'(1) = \frac{5}{8}

Would you like more details or clarification on any steps?

Here are some related questions you may explore:

  1. How does the product rule work for the derivative of two functions?
  2. What are some examples of applying the quotient rule in different contexts?
  3. How can you estimate the slope of a curve at a specific point visually?
  4. What are other applications of product and quotient rules in calculus?
  5. How do these rules extend to higher-order derivatives?

Tip: When working with graphs, make sure to accurately estimate function values and slopes by carefully observing the grid and scale.

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Math Problem Analysis

Mathematical Concepts

Derivatives
Product Rule
Quotient Rule
Graphical Estimation

Formulas

Product Rule: u'(x) = f'(x)g(x) + f(x)g'(x)
Quotient Rule: v'(x) = [f'(x)g(x) - f(x)g'(x)] / g(x)^2

Theorems

Product Rule
Quotient Rule

Suitable Grade Level

Grades 11-12 or Early College