Math Problem Statement

Let u(x) = f(x)g(x) and v(x) = f(x)/g(x). (a) Find u'(1). (b) Find v'(5).

Solution

Let's analyze the graph and solve the given problems step by step. The image shows the graphs of two functions, f(x)f(x) and g(x)g(x), and the problems involve calculating the derivatives of the functions u(x)=f(x)g(x)u(x) = f(x) \cdot g(x) and v(x)=f(x)g(x)v(x) = \frac{f(x)}{g(x)}.

Given:

  • u(x)=f(x)g(x)u(x) = f(x) \cdot g(x)
  • v(x)=f(x)g(x)v(x) = \frac{f(x)}{g(x)}

We need to find:

  1. u(1)u'(1)
  2. v(5)v'(5)

Step 1: Derivative of u(x)=f(x)g(x)u(x) = f(x)g(x)

Using the product rule: u(x)=f(x)g(x)+f(x)g(x)u'(x) = f'(x)g(x) + f(x)g'(x) Now we need the values of f(1)f(1), g(1)g(1), f(1)f'(1), and g(1)g'(1) from the graph.

From the graph:

  • f(1)=2f(1) = 2, g(1)=1g(1) = 1
  • f(1)f'(1): The slope of the function f(x)f(x) at x=1x = 1 can be calculated. Between x=0x = 0 and x=2x = 2, the line rises by 2 and runs by 2, so: f(1)=2020=1f'(1) = \frac{2 - 0}{2 - 0} = 1
  • g(1)g'(1): The slope of g(x)g(x) between x=0x = 0 and x=2x = 2 rises by 2 and runs by 2, so: g(1)=2020=1g'(1) = \frac{2 - 0}{2 - 0} = 1

Now substitute into the product rule: u(1)=f(1)g(1)+f(1)g(1)=(1)(1)+(2)(1)=1+2=3u'(1) = f'(1)g(1) + f(1)g'(1) = (1)(1) + (2)(1) = 1 + 2 = 3

Step 2: Derivative of v(x)=f(x)g(x)v(x) = \frac{f(x)}{g(x)}

Using the quotient rule: 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} Now we need the values of f(5)f(5), g(5)g(5), f(5)f'(5), and g(5)g'(5) from the graph.

From the graph:

  • f(5)=1f(5) = 1, g(5)=2g(5) = 2
  • f(5)f'(5): The slope of f(x)f(x) between x=4x = 4 and x=6x = 6 decreases by 2, so: f(5)=0264=22=1f'(5) = \frac{0 - 2}{6 - 4} = \frac{-2}{2} = -1
  • g(5)g'(5): The slope of g(x)g(x) between x=4x = 4 and x=6x = 6 increases by 2, so: g(5)=4264=22=1g'(5) = \frac{4 - 2}{6 - 4} = \frac{2}{2} = 1

Now substitute into the quotient rule: v(5)=f(5)g(5)f(5)g(5)(g(5))2=(1)(2)(1)(1)(2)2=214=34v'(5) = \frac{f'(5)g(5) - f(5)g'(5)}{(g(5))^2} = \frac{(-1)(2) - (1)(1)}{(2)^2} = \frac{-2 - 1}{4} = \frac{-3}{4}

Final Answers:

(a) u(1)=3u'(1) = 3

(b) v(5)=34v'(5) = -\frac{3}{4}

Would you like more details or have any further questions?

Related Questions:

  1. What is the product rule for derivatives, and how does it apply to other functions?
  2. Can you explain how the quotient rule for derivatives is derived?
  3. What is the significance of the slope of a function in relation to its derivative?
  4. How do you determine the derivative of a piecewise linear function?
  5. What are some common mistakes to avoid when using the product and quotient rules?

Tip:

When analyzing graphs of functions, always double-check slopes by calculating the change in yy-value over the change in xx-value to ensure accuracy in the derivative.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Product Rule
Quotient Rule
Graph Interpretation

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

High School or College (Calculus Level)