Math Problem Statement

[ f(x)=\frac{x}{x^{2}+15 x+50} ]a) Give the domain of ( f ) (in interval notation) ( \square )b) Find the critical numbers of ( f ). ( \square ) (Separate multiple answers by commas.)c) Determine the intervals on which ( f ) is increasing and decreasing.( f ) is increasing on: ( \square ) and( f ) is decreasing on: ( \square ) . ( \square ) , and ( \square )d) Use the First Derivative Test to determine whether each critical point is a local maximum, minimum, or neither.Local maxima occur at ( x= ) ( \square ) (Separate multiple answers by commas.)Local minima occur at ( x= ) ( \square ) (Separate multiple answers by commas.)

Solution

Let's solve this step by step:

Function Definition:

f(x)=xx2+15x+50f(x) = \frac{x}{x^2 + 15x + 50}


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

The function f(x)f(x) is undefined where the denominator equals 0. Solve: x2+15x+50=0x^2 + 15x + 50 = 0 Factorize: (x+10)(x+5)=0(x + 10)(x + 5) = 0 Thus, the denominator is 0 at x=10x = -10 and x=5x = -5. Therefore, the domain is: (,10)(10,5)(5,)(-\infty, -10) \cup (-10, -5) \cup (-5, \infty)


(b) Critical Numbers of f(x)f(x)

To find the critical numbers, compute the derivative f(x)f'(x) using the quotient rule: f(x)=(x2+15x+50)(1)x(2x+15)(x2+15x+50)2f'(x) = \frac{(x^2 + 15x + 50)(1) - x(2x + 15)}{(x^2 + 15x + 50)^2} Simplify the numerator: (x2+15x+50)(2x2+15x)=x2+50(x^2 + 15x + 50) - (2x^2 + 15x) = -x^2 + 50 Thus: f(x)=x2+50(x2+15x+50)2f'(x) = \frac{-x^2 + 50}{(x^2 + 15x + 50)^2}

Set f(x)=0f'(x) = 0 to find critical points: x2+50=0-x^2 + 50 = 0 x2=50    x=±50=±52x^2 = 50 \quad \implies \quad x = \pm \sqrt{50} = \pm 5\sqrt{2}

Critical numbers are: x=52,x=52x = 5\sqrt{2}, \, x = -5\sqrt{2}


(c) Intervals of Increase and Decrease

The critical points and points where the function is undefined divide the number line into intervals: (,10),(10,52),(52,5),(5,52),(52,)(-\infty, -10), (-10, -5\sqrt{2}), (-5\sqrt{2}, -5), (-5, 5\sqrt{2}), (5\sqrt{2}, \infty)

Test the sign of f(x)f'(x) in each interval by choosing a test point:

  1. In (,10)(-\infty, -10): Choose x=11x = -11 f(x)=(11)2+50((11)2+15(11)+50)2=negativef'(x) = \frac{-(-11)^2 + 50}{((-11)^2 + 15(-11) + 50)^2} = \text{negative}

  2. In (10,52)(-10, -5\sqrt{2}): Choose x=8x = -8 f(x)=positivef'(x) = \text{positive}

  3. In (52,5)(-5\sqrt{2}, -5): Choose x=6x = -6 f(x)=negativef'(x) = \text{negative}

  4. In (5,52)(-5, 5\sqrt{2}): Choose x=0x = 0 f(x)=positivef'(x) = \text{positive}

  5. In (52,)(5\sqrt{2}, \infty): Choose x=10x = 10 f(x)=negativef'(x) = \text{negative}

From this, f(x)f(x) is:

  • Increasing on (10,52)(5,52)(-10, -5\sqrt{2}) \cup (-5, 5\sqrt{2})
  • Decreasing on (,10)(52,5)(52,)(-\infty, -10) \cup (-5\sqrt{2}, -5) \cup (5\sqrt{2}, \infty)

(d) Local Maxima and Minima (First Derivative Test)

  • At x=52x = -5\sqrt{2}: f(x)f'(x) changes from positive to negative → local maximum.
  • At x=52x = 5\sqrt{2}: f(x)f'(x) changes from positive to negative → local maximum.
  • At x=5x = -5: f(x)f'(x) changes from negative to positive → local minimum.

Thus:

  • Local maxima: x=52,52x = -5\sqrt{2}, 5\sqrt{2}
  • Local minima: x=5x = -5

Final Answers

(a) Domain: (,10)(10,5)(5,)(-\infty, -10) \cup (-10, -5) \cup (-5, \infty)

(b) Critical numbers: x=52,x=52x = -5\sqrt{2}, \, x = 5\sqrt{2}

(c) Intervals: Increasing: (10,52)(5,52)\text{Increasing: } (-10, -5\sqrt{2}) \cup (-5, 5\sqrt{2}) Decreasing: (,10)(52,5)(52,)\text{Decreasing: } (-\infty, -10) \cup (-5\sqrt{2}, -5) \cup (5\sqrt{2}, \infty)

(d) Local maxima: x=52,52x = -5\sqrt{2}, 5\sqrt{2} Local minima: x=5x = -5


Would you like further clarification or a graph of f(x)f(x) to visualize these results?

5 Related Questions:

  1. How do we apply the second derivative test for concavity analysis?
  2. Can f(x)f(x) have absolute extrema on its domain? Why or why not?
  3. How do vertical asymptotes affect the behavior of f(x)f(x)?
  4. What happens to f(x)f(x) as xx \to \infty or xx \to -\infty?
  5. How do the critical points influence the shape of f(x)f(x)?

Tip: When analyzing functions, always start with the domain to avoid undefined operations!

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Math Problem Analysis

Mathematical Concepts

Functions
Critical Points
Intervals of Increase and Decrease
First Derivative Test

Formulas

Quotient Rule for Derivatives: f'(x) = (g(x)h'(x) - g'(x)h(x)) / (h(x))^2
Factoring Quadratic Equations

Theorems

First Derivative Test

Suitable Grade Level

Grades 11-12