Math Problem Statement

Find the intervals on which f is increasing and decreasing for f(x) = -3x^4 + 2x^2 + 6.

Solution

To find the intervals on which the function f(x)=3x4+2x2+6f(x) = -3x^4 + 2x^2 + 6 is increasing or decreasing, we need to follow these steps:

  1. Find the derivative f(x)f'(x): This derivative will help us determine where the function is increasing or decreasing.

    f(x)=12x3+4xf'(x) = -12x^3 + 4x

  2. Set the derivative equal to zero to find critical points:

    12x3+4x=0-12x^3 + 4x = 0

    Factor out 4x4x:

    4x(3x2+1)=04x(-3x^2 + 1) = 0

    This gives two cases:

    • 4x=0x=04x = 0 \Rightarrow x = 0
    • 3x2+1=0x2=13x=±33-3x^2 + 1 = 0 \Rightarrow x^2 = \frac{1}{3} \Rightarrow x = \pm \frac{\sqrt{3}}{3}

    So, the critical points are x=0x = 0, x=33x = \frac{\sqrt{3}}{3}, and x=33x = -\frac{\sqrt{3}}{3}.

  3. Determine the sign of f(x)f'(x) on the intervals defined by these critical points:

    Test intervals around x=33x = -\frac{\sqrt{3}}{3}, x=0x = 0, and x=33x = \frac{\sqrt{3}}{3} to see if f(x)f'(x) is positive or negative in each interval.

    • For x<33x < -\frac{\sqrt{3}}{3}, choose a test point like x=1x = -1: f(1)=12(1)3+4(1)=124=8f'(-1) = -12(-1)^3 + 4(-1) = 12 - 4 = 8 (positive).
    • For 33<x<0-\frac{\sqrt{3}}{3} < x < 0, choose x=0.2x = -0.2: f(0.2)=12(0.2)3+4(0.2)f'(-0.2) = -12(-0.2)^3 + 4(-0.2) which gives a negative result.
    • For 0<x<330 < x < \frac{\sqrt{3}}{3}, choose x=0.2x = 0.2: f(0.2)=12(0.2)3+4(0.2)f'(0.2) = -12(0.2)^3 + 4(0.2) which gives a positive result.
    • For x>33x > \frac{\sqrt{3}}{3}, choose x=1x = 1: f(1)=12(1)3+4(1)=12+4=8f'(1) = -12(1)^3 + 4(1) = -12 + 4 = -8 (negative).
  4. Determine intervals of increase and decrease:

    • Increasing on (,33)(-\infty, -\frac{\sqrt{3}}{3}) and (0,33)(0, \frac{\sqrt{3}}{3}).
    • Decreasing on (33,0)(-\frac{\sqrt{3}}{3}, 0) and (33,)(\frac{\sqrt{3}}{3}, \infty).

Thus, the correct answer would be:

  • Increasing on (,33)(0,33)(-\infty, -\frac{\sqrt{3}}{3}) \cup (0, \frac{\sqrt{3}}{3})
  • Decreasing on (33,0)(33,)(-\frac{\sqrt{3}}{3}, 0) \cup (\frac{\sqrt{3}}{3}, \infty)

This matches answer choice A.

Would you like further details or explanations on any part of this solution?


Here are five related questions that might deepen your understanding:

  1. How do you interpret the behavior of a polynomial function based on its degree and leading coefficient?
  2. What does the sign of the first derivative indicate about the function’s behavior on an interval?
  3. How would the analysis change if the leading coefficient of f(x)f(x) were positive?
  4. Can a function be both increasing and decreasing at a single point?
  5. How do the critical points relate to the function’s local maxima and minima?

Tip: When finding intervals of increase or decrease, always test points within each interval created by the critical points to confirm the sign of the derivative.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivative
Increasing and Decreasing Functions
Critical Points

Formulas

First Derivative Test

Theorems

First Derivative Test
Critical Points of a Function

Suitable Grade Level

Grades 11-12