Math Problem Statement

Given the function f, prime, of, x, equals, x, cubed, minus, 15, x, commaf ′ (x)=x 3 −15x, determine all intervals on which ff is decreasing.

Solution

To determine the intervals on which f(x)f(x) is decreasing, we need to analyze the behavior of the derivative f(x)=x315xf'(x) = x^3 - 15x. Here's the step-by-step approach:

  1. Find Critical Points: Set f(x)=0f'(x) = 0 to find critical points, as these points can indicate changes in the increasing or decreasing behavior. x315x=0x^3 - 15x = 0 Factor out xx: x(x215)=0x(x^2 - 15) = 0 This gives: x=0orx=±15x = 0 \quad \text{or} \quad x = \pm \sqrt{15}

  2. Analyze the Sign of f(x)f'(x): To determine the intervals where f(x)f(x) is decreasing, check the sign of f(x)f'(x) on each interval defined by the critical points x=15,0,15x = -\sqrt{15}, 0, \sqrt{15}.

    Let's analyze f(x)=x315xf'(x) = x^3 - 15x over each interval:

    • For x<15x < -\sqrt{15}: Choose a test point (e.g., x=4x = -4).

      • f(4)=(4)315(4)=64+60=4f'(-4) = (-4)^3 - 15(-4) = -64 + 60 = -4 (negative).
    • For 15<x<0-\sqrt{15} < x < 0: Choose a test point (e.g., x=1x = -1).

      • f(1)=(1)315(1)=1+15=14f'(-1) = (-1)^3 - 15(-1) = -1 + 15 = 14 (positive).
    • For 0<x<150 < x < \sqrt{15}: Choose a test point (e.g., x=1x = 1).

      • f(1)=(1)315(1)=115=14f'(1) = (1)^3 - 15(1) = 1 - 15 = -14 (negative).
    • For x>15x > \sqrt{15}: Choose a test point (e.g., x=4x = 4).

      • f(4)=(4)315(4)=6460=4f'(4) = (4)^3 - 15(4) = 64 - 60 = 4 (positive).
  3. Determine Intervals of Decrease: From the sign analysis:

    • f(x)<0f'(x) < 0 for x(,15)x \in (-\infty, -\sqrt{15}) and x(0,15)x \in (0, \sqrt{15}).

    Therefore, f(x)f(x) is decreasing on the intervals: (,15)and(0,15)(-\infty, -\sqrt{15}) \quad \text{and} \quad (0, \sqrt{15})


Would you like a breakdown of any step or a graph of f(x)f(x) to illustrate? Here are some related questions to deepen understanding:

  1. How would we find intervals where f(x)f(x) is increasing?
  2. What if f(x)f(x) were given as x415xx^4 - 15x instead—how would the intervals change?
  3. What role does the second derivative f(x)f''(x) play in determining the concavity of f(x)f(x)?
  4. How can we confirm the behavior of f(x)f(x) at the critical points themselves?
  5. What effect would changing the constant in f(x)f(x) have on these intervals?

Tip: Always test points within intervals to confirm behavior, especially with higher-order polynomials, where behavior can change multiple times.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Intervals of Increase and Decrease
Critical Points

Formulas

Derivative: f'(x) = x^3 - 15x

Theorems

First Derivative Test

Suitable Grade Level

Grades 11-12