Math Problem Statement

Identify the open intervals on which the function is increasing or decreasing. (Enter your answers using interval notation.) h(x) = cos(3x/2)

, 0 < x < 2𝜋

Solution

To determine the intervals where h(x)=cos(3x2)h(x) = \cos\left(\frac{3x}{2}\right) is increasing or decreasing, we need to analyze the behavior of its derivative within the given interval 0<x<2π0 < x < 2\pi.

Step 1: Differentiate h(x)h(x)

The derivative of h(x)h(x) with respect to xx is: h(x)=sin(3x2)32=32sin(3x2).h'(x) = -\sin\left(\frac{3x}{2}\right) \cdot \frac{3}{2} = -\frac{3}{2} \sin\left(\frac{3x}{2}\right).

Step 2: Set the Derivative Equal to Zero

To find the critical points where the function could change from increasing to decreasing or vice versa, we set h(x)=0h'(x) = 0: 32sin(3x2)=0.-\frac{3}{2} \sin\left(\frac{3x}{2}\right) = 0. This simplifies to: sin(3x2)=0.\sin\left(\frac{3x}{2}\right) = 0.

Within the interval 0<x<2π0 < x < 2\pi, we find the values of xx for which sin(3x2)=0\sin\left(\frac{3x}{2}\right) = 0: 3x2=nπ,nZ.\frac{3x}{2} = n\pi, \quad n \in \mathbb{Z}. Solving for xx, we get: x=2nπ3.x = \frac{2n\pi}{3}. Now, find the values of xx within 0<x<2π0 < x < 2\pi by substituting integer values for nn:

  1. n=1n = 1: x=2π3x = \frac{2\pi}{3}
  2. n=2n = 2: x=4π3x = \frac{4\pi}{3}
  3. n=3n = 3: x=2πx = 2\pi

Thus, the critical points within the interval 0<x<2π0 < x < 2\pi are x=2π3x = \frac{2\pi}{3} and x=4π3x = \frac{4\pi}{3}.

Step 3: Determine the Sign of h(x)h'(x) in Each Interval

To identify the intervals where h(x)h(x) is increasing or decreasing, we examine the sign of h(x)=32sin(3x2)h'(x) = -\frac{3}{2} \sin\left(\frac{3x}{2}\right) in each subinterval:

  1. 0<x<2π30 < x < \frac{2\pi}{3}
  2. 2π3<x<4π3\frac{2\pi}{3} < x < \frac{4\pi}{3}
  3. 4π3<x<2π\frac{4\pi}{3} < x < 2\pi

Since h(x)h'(x) is a product of 32-\frac{3}{2} and sin(3x2)\sin\left(\frac{3x}{2}\right), the sign of h(x)h'(x) depends on the sign of sin(3x2)\sin\left(\frac{3x}{2}\right):

  • In (0,2π3)\left(0, \frac{2\pi}{3}\right), sin(3x2)>0\sin\left(\frac{3x}{2}\right) > 0 (positive), so h(x)<0h'(x) < 0 (negative), meaning h(x)h(x) is decreasing.
  • In (2π3,4π3)\left(\frac{2\pi}{3}, \frac{4\pi}{3}\right), sin(3x2)<0\sin\left(\frac{3x}{2}\right) < 0 (negative), so h(x)>0h'(x) > 0 (positive), meaning h(x)h(x) is increasing.
  • In (4π3,2π)\left(\frac{4\pi}{3}, 2\pi\right), sin(3x2)>0\sin\left(\frac{3x}{2}\right) > 0 (positive), so h(x)<0h'(x) < 0 (negative), meaning h(x)h(x) is decreasing.

Final Answer

The function h(x)=cos(3x2)h(x) = \cos\left(\frac{3x}{2}\right) is:

  • Decreasing on (0,2π3)\left(0, \frac{2\pi}{3}\right) and (4π3,2π)\left(\frac{4\pi}{3}, 2\pi\right).
  • Increasing on (2π3,4π3)\left(\frac{2\pi}{3}, \frac{4\pi}{3}\right).

Would you like further details, or do you have any questions?


  1. How does the derivative of a trigonometric function determine intervals of increase or decrease?
  2. What is the impact of different constants in the argument of trigonometric functions on their period?
  3. How can we verify intervals of increase or decrease graphically?
  4. What is the relationship between critical points and intervals of monotonicity?
  5. How do changes in the domain affect the analysis of periodic functions?

Tip: Use the sign of the derivative to determine if a function is increasing or decreasing within specific intervals.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Calculus
Derivative
Increasing and Decreasing Functions

Formulas

Derivative of cos(kx) is -k*sin(kx)
Set derivative equal to zero to find critical points

Theorems

First Derivative Test

Suitable Grade Level

Grades 10-12