Math Problem Statement

find the intervals on which f is increasing and decreasing. f(x)=-4sin^2x on [negative pi, pi]

Solution

To find the intervals where the function f(x)=4sin2xf(x) = -4 \sin^2 x is increasing and decreasing on the interval [π,π][-\pi, \pi], we need to follow these steps:

  1. Find the derivative: f(x)=42sinxcosx=8sinxcosx=4sin(2x)f'(x) = -4 \cdot 2 \sin x \cos x = -8 \sin x \cos x = -4 \sin(2x)

  2. Set the derivative to zero to find critical points: 4sin(2x)=0    sin(2x)=0-4 \sin(2x) = 0 \implies \sin(2x) = 0 This occurs when: 2x=nπfor nZ    x=nπ22x = n\pi \quad \text{for } n \in \mathbb{Z} \implies x = \frac{n\pi}{2}

    Within the interval [π,π][-\pi, \pi], the critical points are:

    • x=πx = -\pi
    • x=π2x = -\frac{\pi}{2}
    • x=0x = 0
    • x=π2x = \frac{\pi}{2}
    • x=πx = \pi
  3. Test the intervals determined by the critical points: The intervals to test are:

    • (π,π2)(-\pi, -\frac{\pi}{2})
    • (π2,0)(-\frac{\pi}{2}, 0)
    • (0,π2)(0, \frac{\pi}{2})
    • (π2,π)(\frac{\pi}{2}, \pi)

    Choose test points from each interval:

    • For x=3π4x = -\frac{3\pi}{4} (in (π,π2)(-\pi, -\frac{\pi}{2})): f(3π4)=4sin(3π2)=4(1)=4(increasing)f'(-\frac{3\pi}{4}) = -4 \sin(-\frac{3\pi}{2}) = -4 \cdot (-1) = 4 \quad (\text{increasing})

    • For x=π4x = -\frac{\pi}{4} (in (π2,0)(-\frac{\pi}{2}, 0)): f(π4)=4sin(π2)=4(1)=4(increasing)f'(-\frac{\pi}{4}) = -4 \sin(-\frac{\pi}{2}) = -4 \cdot (-1) = 4 \quad (\text{increasing})

    • For x=π4x = \frac{\pi}{4} (in (0,π2)(0, \frac{\pi}{2})): f(π4)=4sin(π2)=41=4(decreasing)f'(\frac{\pi}{4}) = -4 \sin(\frac{\pi}{2}) = -4 \cdot 1 = -4 \quad (\text{decreasing})

    • For x=3π4x = \frac{3\pi}{4} (in (π2,π)(\frac{\pi}{2}, \pi)): f(3π4)=4sin(3π2)=4(1)=4(increasing)f'(\frac{3\pi}{4}) = -4 \sin(\frac{3\pi}{2}) = -4 \cdot (-1) = 4 \quad (\text{increasing})

  4. Summary of intervals:

    • Increasing on: (π,π2)(-\pi, -\frac{\pi}{2}) and (π2,0)(-\frac{\pi}{2}, 0) and (π2,π)(\frac{\pi}{2}, \pi)
    • Decreasing on: (0,π2)(0, \frac{\pi}{2})

Conclusion

  • f(x)f(x) is increasing on the intervals (π,π2)(-\pi, -\frac{\pi}{2}), (π2,0)(-\frac{\pi}{2}, 0), and (π2,π)(\frac{\pi}{2}, \pi).
  • f(x)f(x) is decreasing on the interval (0,π2)(0, \frac{\pi}{2}).

Would you like further details or clarifications? Here are some questions you might consider:

  1. How does the second derivative test confirm the nature of the critical points?
  2. What is the maximum and minimum value of the function on the interval?
  3. How does the periodic nature of the sine function affect this analysis?
  4. What is the behavior of the function outside the interval [π,π][-π, π]?
  5. Can you graph the function to visualize the increasing and decreasing intervals?

Tip: Always check the endpoints of your interval to determine if they are included in increasing or decreasing intervals.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Trigonometric Functions

Formulas

f'(x) = -4sin(2x)
Critical points where f'(x) = 0

Theorems

First Derivative Test

Suitable Grade Level

Grades 11-12