Math Problem Statement

Analyze and solve the function f(x) = x / (x^2 + 10x + 21) for domain, critical points, increasing/decreasing intervals, and relative extrema.

Solution

Let's break down the solution and explain it step by step.

(a) Domain of f(x)f(x)

The function f(x)=xx2+10x+21f(x) = \frac{x}{x^2 + 10x + 21} is undefined where the denominator equals zero: x2+10x+21=0x^2 + 10x + 21 = 0 Factoring: (x+7)(x+3)=0(x + 7)(x + 3) = 0 Thus, x=7x = -7 and x=3x = -3 are the points where the function is undefined. Therefore, the domain of f(x)f(x) in interval notation is: (,7)(7,3)(3,)(-\infty, -7) \cup (-7, -3) \cup (-3, \infty)


(b) Critical Numbers

Critical numbers occur where the derivative of f(x)f(x), f(x)f'(x), is zero or undefined. Calculating f(x)f'(x) using the quotient rule: f(x)=(1)(x2+10x+21)x(2x+10)(x2+10x+21)2f'(x) = \frac{(1)(x^2 + 10x + 21) - x(2x + 10)}{(x^2 + 10x + 21)^2} Simplify the numerator: (x2+10x+21)(2x2+10x)=x2+21(x^2 + 10x + 21) - (2x^2 + 10x) = -x^2 + 21 So, f(x)=x2+21(x2+10x+21)2f'(x) = \frac{-x^2 + 21}{(x^2 + 10x + 21)^2} Set the numerator equal to zero to find critical points: x2+21=0x2=21x=±21-x^2 + 21 = 0 \quad \Rightarrow \quad x^2 = 21 \quad \Rightarrow \quad x = \pm \sqrt{21} Thus, the critical numbers are: x=21,x=21x = -\sqrt{21}, \quad x = \sqrt{21}


(c) Increasing and Decreasing Intervals

Analyze the sign of f(x)f'(x) around the critical points x=±21x = \pm \sqrt{21} and where f(x)f(x) is undefined (x=7,3x = -7, -3).

  1. Test intervals:

    • (,21)(-\infty, -\sqrt{21})
    • (21,21)(-\sqrt{21}, \sqrt{21})
    • (21,)(\sqrt{21}, \infty)
  2. Sign of f(x)f'(x):

    • f(x)>0f'(x) > 0 on (21,21)(-\sqrt{21}, \sqrt{21}) (increasing).
    • f(x)<0f'(x) < 0 on (,21)(21,)(-\infty, -\sqrt{21}) \cup (\sqrt{21}, \infty) (decreasing).

So, f(x)f(x) is:

  • Increasing on (21,21)(-\sqrt{21}, \sqrt{21}).
  • Decreasing on (,21)(21,)(-\infty, -\sqrt{21}) \cup (\sqrt{21}, \infty).

(d) Relative Extrema (First Derivative Test)

  1. At x=21x = -\sqrt{21}:

    • f(x)f'(x) changes from negative to positive.
    • Thus, x=21x = -\sqrt{21} is a relative minimum.
  2. At x=21x = \sqrt{21}:

    • f(x)f'(x) changes from positive to negative.
    • Thus, x=21x = \sqrt{21} is a relative maximum.

Summary:

  • Relative maxima: x=21x = \sqrt{21}
  • Relative minima: x=21x = -\sqrt{21}

Recap:

  1. Domain: (,7)(7,3)(3,)(-\infty, -7) \cup (-7, -3) \cup (-3, \infty)
  2. Critical numbers: x=±21x = \pm \sqrt{21}
  3. Increasing: (21,21)(-\sqrt{21}, \sqrt{21})
  4. Decreasing: (,21)(21,)(-\infty, -\sqrt{21}) \cup (\sqrt{21}, \infty)
  5. Relative maxima: x=21x = \sqrt{21}
  6. Relative minima: x=21x = -\sqrt{21}

Do you want a detailed explanation of any specific part? Here are some questions for further exploration:

  1. How is the derivative computed using the quotient rule?
  2. Why do critical points only occur where f(x)=0f'(x) = 0 or undefined?
  3. How do we test intervals to determine increasing or decreasing behavior?
  4. What happens to f(x)f(x) near undefined points (x=7,3x = -7, -3)?
  5. Can we verify these results graphically?

Tip: Always factorize carefully and check for undefined points in rational functions.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Critical Points
Quotient Rule
Increasing and Decreasing Functions
Relative Extrema

Formulas

Quotient Rule: (u/v)' = (u'v - uv') / v^2
Derivative of f(x) = x / (x^2 + 10x + 21)
f'(x) = (-x^2 + 21) / (x^2 + 10x + 21)^2

Theorems

First Derivative Test
Zero Derivative Rule for Critical Points

Suitable Grade Level

Grades 11-12 or College Calculus I