Math Problem Statement

find the intervals of concavity y=sin(2x) [-pi,pi]

Solution

To determine the intervals of concavity for the function y=sin(2x)y = \sin(2x) over the interval [π,π][-\pi, \pi], we need to find the second derivative of the function and analyze its sign.

Step 1: Find the first and second derivatives

  • First derivative yy': y=ddxsin(2x)=2cos(2x)y' = \frac{d}{dx} \sin(2x) = 2 \cos(2x)

  • Second derivative yy'': y=ddx(2cos(2x))=4sin(2x)y'' = \frac{d}{dx} \left( 2 \cos(2x) \right) = -4 \sin(2x)

Step 2: Determine where the second derivative changes sign

The concavity of the function is determined by the sign of the second derivative:

  • If y>0y'' > 0, the function is concave up.
  • If y<0y'' < 0, the function is concave down.

Since y=4sin(2x)y'' = -4 \sin(2x), we analyze where sin(2x)\sin(2x) is positive or negative.

  • y>0y'' > 0 when sin(2x)<0\sin(2x) < 0 (concave up).
  • y<0y'' < 0 when sin(2x)>0\sin(2x) > 0 (concave down).

Step 3: Solve for critical points where y=0y'' = 0

4sin(2x)=0    sin(2x)=0-4 \sin(2x) = 0 \implies \sin(2x) = 0 The solutions to sin(2x)=0\sin(2x) = 0 are: 2x=nπfornZ2x = n\pi \quad \text{for} \quad n \in \mathbb{Z} x=nπ2x = \frac{n\pi}{2} On the interval [π,π][-\pi, \pi], the critical points are: x=π2,0,π2x = -\frac{\pi}{2}, 0, \frac{\pi}{2}

Step 4: Analyze the sign of yy'' on intervals

Now, we check the sign of y=4sin(2x)y'' = -4 \sin(2x) on the intervals determined by the critical points x=π2,0,π2x = -\frac{\pi}{2}, 0, \frac{\pi}{2}:

  • Interval (π,π2)(-\pi, -\frac{\pi}{2}): For x(π,π2)x \in (-\pi, -\frac{\pi}{2}), 2x(2π,π)2x \in (-2\pi, -\pi), and sin(2x)\sin(2x) is negative. Therefore, y>0y'' > 0, so the function is concave up on this interval.

  • Interval (π2,0)(-\frac{\pi}{2}, 0): For x(π2,0)x \in (-\frac{\pi}{2}, 0), 2x(π,0)2x \in (-\pi, 0), and sin(2x)\sin(2x) is positive. Therefore, y<0y'' < 0, so the function is concave down on this interval.

  • Interval (0,π2)(0, \frac{\pi}{2}): For x(0,π2)x \in (0, \frac{\pi}{2}), 2x(0,π)2x \in (0, \pi), and sin(2x)\sin(2x) is positive. Therefore, y<0y'' < 0, so the function is concave down on this interval.

  • Interval (π2,π)(\frac{\pi}{2}, \pi): For x(π2,π)x \in (\frac{\pi}{2}, \pi), 2x(π,2π)2x \in (\pi, 2\pi), and sin(2x)\sin(2x) is negative. Therefore, y>0y'' > 0, so the function is concave up on this interval.

Step 5: Conclusion

The concavity of y=sin(2x)y = \sin(2x) on [π,π][-\pi, \pi] is:

  • Concave up on (π,π2)(-\pi, -\frac{\pi}{2}) and (π2,π)(\frac{\pi}{2}, \pi).
  • Concave down on (π2,0)(-\frac{\pi}{2}, 0) and (0,π2)(0, \frac{\pi}{2}).

Summary of concavity intervals:

  • Concave up: (π,π2)(π2,π)(-\pi, -\frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi)
  • Concave down: (π2,0)(0,π2)(-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2})

Would you like more details on the steps or a different aspect of this problem? Feel free to ask!


Here are some related questions to explore:

  1. How do the concavity intervals change for a function with a higher-degree sine or cosine term?
  2. What is the relationship between the second derivative and inflection points?
  3. How do you find the concavity of trigonometric functions involving other angles, such as cos(3x)\cos(3x)?
  4. How does the graph of y=sin(2x)y = \sin(2x) look in terms of concavity?
  5. Can the second derivative help in identifying local maxima and minima in the function?

Tip: When analyzing concavity, always check the sign of the second derivative at the critical points, and test within the intervals defined by these points to determine the concavity behavior!

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Math Problem Analysis

Mathematical Concepts

Calculus
Concavity
Derivatives
Trigonometric Functions

Formulas

First derivative: y' = d/dx(sin(2x)) = 2cos(2x)
Second derivative: y'' = d/dx(2cos(2x)) = -4sin(2x)

Theorems

Second Derivative Test
Concavity and the Second Derivative

Suitable Grade Level

Grades 11-12