Math Problem Statement

Find the intervals on which f is increasing and the intervals on which it is decreasing. ​f(x)equals=negative 3 cosine squared x−3cos2x on ​[minus−piπ​,piπ​

Solution

We are tasked with determining the intervals on which the function f(x)=3cos2(x)3cos(2x)f(x) = -3\cos^2(x) - 3\cos(2x) is increasing and decreasing on the interval [π,π][- \pi, \pi].

Step 1: Differentiate the function

To determine where the function is increasing or decreasing, we need to first find its derivative f(x)f'(x).

Using the chain rule and the derivative of trigonometric functions:

  • Derivative of 3cos2(x)-3\cos^2(x): ddx(3cos2(x))=32cos(x)(sin(x))=6cos(x)sin(x)=3sin(2x)\frac{d}{dx}\left(-3\cos^2(x)\right) = -3 \cdot 2\cos(x) \cdot (-\sin(x)) = 6\cos(x)\sin(x) = 3\sin(2x) (using the double-angle identity 2sin(x)cos(x)=sin(2x)2\sin(x)\cos(x) = \sin(2x)).

  • Derivative of 3cos(2x)-3\cos(2x): ddx(3cos(2x))=3(2sin(2x))=6sin(2x)\frac{d}{dx} \left(-3\cos(2x)\right) = -3 \cdot (-2\sin(2x)) = 6\sin(2x)

Now, summing up both derivatives: f(x)=3sin(2x)+6sin(2x)=9sin(2x)f'(x) = 3\sin(2x) + 6\sin(2x) = 9\sin(2x)

Step 2: Set the derivative equal to zero

To find critical points, we solve: 9sin(2x)=09\sin(2x) = 0 This simplifies to: sin(2x)=0\sin(2x) = 0

The general solution for sin(2x)=0\sin(2x) = 0 is 2x=nπ2x = n\pi, where nn is an integer. Dividing by 2, we get: x=nπ2x = \frac{n\pi}{2}

Since we are working on the interval [π,π][-\pi, \pi], the solutions for xx are: x=π,π2,0,π2,πx = -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi

Step 3: Analyze the sign of f(x)f'(x)

Now, we analyze the sign of f(x)=9sin(2x)f'(x) = 9\sin(2x) in the intervals determined by the critical points x=π,π2,0,π2,πx = -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi.

  • On (π,π2)(-\pi, -\frac{\pi}{2}), 2x(2π,π)2x \in (-2\pi, -\pi), and sin(2x)<0\sin(2x) < 0, so f(x)<0f'(x) < 0 (decreasing).
  • On (π2,0)(-\frac{\pi}{2}, 0), 2x(π,0)2x \in (-\pi, 0), and sin(2x)>0\sin(2x) > 0, so f(x)>0f'(x) > 0 (increasing).
  • On (0,π2)(0, \frac{\pi}{2}), 2x(0,π)2x \in (0, \pi), and sin(2x)>0\sin(2x) > 0, so f(x)>0f'(x) > 0 (increasing).
  • On (π2,π)(\frac{\pi}{2}, \pi), 2x(π,2π)2x \in (\pi, 2\pi), and sin(2x)<0\sin(2x) < 0, so f(x)<0f'(x) < 0 (decreasing).

Step 4: Conclusion

  • The function is increasing on (π2,0)(0,π2)(-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}).
  • The function is decreasing on (π,π2)(π2,π)(-\pi, -\frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi).

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

Here are some related questions you might consider:

  1. How do you determine the concavity of the function f(x)f(x)?
  2. What happens at the critical points in terms of local maxima or minima?
  3. How do you find inflection points for this function?
  4. Can you apply this process to other trigonometric functions?
  5. How does the double-angle identity help simplify the derivative?

Tip: Always check the endpoints of the interval when analyzing increasing/decreasing behavior over a restricted domain.

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Math Problem Analysis

Mathematical Concepts

Calculus
Trigonometry
Derivative
Increasing and Decreasing Intervals

Formulas

f'(x) = 9sin(2x)
sin(2x) = 2sin(x)cos(x)

Theorems

First Derivative Test
Double-Angle Identity

Suitable Grade Level

Grades 11-12