Math Problem Statement

Given \( f(x) = \frac{x}{x^2 + 7x + 10} \), find the domain of \( f \) in interval notation. Then, determine the intervals on which \( f \) is increasing and decreasing.

Solution

To solve this problem, let's analyze each part:

Part (a): Domain of f(x)f(x)

Given: f(x)=xx2+7x+10f(x) = \frac{x}{x^2 + 7x + 10}

To find the domain of f(x)f(x), we need to determine when the denominator x2+7x+10x^2 + 7x + 10 is zero, as division by zero is undefined.

  1. Factor the denominator: x2+7x+10=(x+5)(x+2)x^2 + 7x + 10 = (x + 5)(x + 2)

  2. Set each factor to zero: x+5=0x=5x + 5 = 0 \Rightarrow x = -5 x+2=0x=2x + 2 = 0 \Rightarrow x = -2

  3. Therefore, f(x)f(x) is undefined at x=5x = -5 and x=2x = -2.

Thus, the domain of f(x)f(x) in interval notation is: (,5)(5,2)(2,)(-\infty, -5) \cup (-5, -2) \cup (-2, \infty)

Part (b): Intervals of Increase and Decrease

To determine where f(x)f(x) is increasing or decreasing, we need to analyze the derivative f(x)f'(x).

  1. Compute f(x)f'(x) using the quotient rule: f(x)=xx2+7x+10f(x) = \frac{x}{x^2 + 7x + 10} Let u=xu = x and v=x2+7x+10v = x^2 + 7x + 10. Then u=1u' = 1 and v=2x+7v' = 2x + 7. f(x)=vuuvv2=(x2+7x+10)1x(2x+7)(x2+7x+10)2f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2} = \frac{(x^2 + 7x + 10) \cdot 1 - x \cdot (2x + 7)}{(x^2 + 7x + 10)^2}

  2. Simplify f(x)f'(x): Expanding the numerator: f(x)=x2+7x+10(2x2+7x)(x2+7x+10)2=x2+10(x2+7x+10)2f'(x) = \frac{x^2 + 7x + 10 - (2x^2 + 7x)}{(x^2 + 7x + 10)^2} = \frac{-x^2 + 10}{(x^2 + 7x + 10)^2}

  3. Set f(x)=0f'(x) = 0 to find critical points: x2+10=0x2=10x=±10-x^2 + 10 = 0 \Rightarrow x^2 = 10 \Rightarrow x = \pm \sqrt{10}

  4. Test intervals around x=10x = -\sqrt{10} and x=10x = \sqrt{10} to determine the sign of f(x)f'(x).

After analysis, we find that:

  • f(x)f(x) is increasing on (,10)(10,)(-\infty, -\sqrt{10}) \cup (\sqrt{10}, \infty)
  • f(x)f(x) is decreasing on (10,10)(-\sqrt{10}, \sqrt{10})

Final Answers:

  • Domain: (,5)(5,2)(2,)(- \infty, -5) \cup (-5, -2) \cup (-2, \infty)
  • Increasing on: (,10)(10,)(- \infty, -\sqrt{10}) \cup (\sqrt{10}, \infty)
  • Decreasing on: (10,10)(- \sqrt{10}, \sqrt{10})

Would you like more details on any step, or do you have any questions?


Here are five related questions that expand on this information:

  1. How would the domain change if the function were f(x)=xx2+7x10f(x) = \frac{x}{x^2 + 7x - 10}?
  2. What happens to the intervals of increase and decrease if we add a constant to f(x)f(x)?
  3. How does the behavior of the function near its vertical asymptotes affect its graph?
  4. What are some strategies for determining critical points without explicitly finding the derivative?
  5. Can you determine the concavity of f(x)f(x) by using the second derivative?

Tip: When analyzing rational functions, always check for undefined points in the denominator to find the domain first.

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Math Problem Analysis

Mathematical Concepts

Domain of a function
Increasing and decreasing intervals
Rational functions
Derivative

Formulas

Quotient rule for derivatives
Factoring quadratic expressions

Theorems

Critical points
Sign analysis for increasing/decreasing intervals

Suitable Grade Level

Grades 11-12