# Identify the open interval on which the function cos (4x/3) is increasing or decreasing (0 < x < 2π).

Increasing functions are those that continuously increase in a particular domain. For example, y = x is an increasing function. Decreasing functions are those that continuously decreases in a particular domain. For example, y = - x is a decreasing function.

## Answer: The function cos (4x/3) is increasing in the domain 3π/4 < x < 3π/2 and decreasing in the domain (0 < x < 3π/4)** **∪ (3π/2 < x < 2π).

Let's understand the solution in detail.

**Explanation:**

We are given the function as f(x) = cos (4x/3).

To find its monotonicity, we first calculate the first derivative of the function.

Hence, f'(x) = -4/3 sin (4x/3) (using the chain rule of differentiation)

Now we equate this derivative to zero.

Hence, -4/3 sin (4x/3) = 0

For the above expression to be zero, sin (4x/3) must be zero.

And if sin (4x/3) = 0, then 4x/3 = nπ (since the sine function is always zero in the integral multiples of π).

Hence, x = 3nπ/4

Putting n = 0, we get x = 0, hence f'(x) = -4/3 sin (4(0) / 3) = -4/3 sin (0) = 0

Putting n = 1, we get x = 3π/4, hence f'(x) = -4/3 sin (4 (3π/4) / 3) = -4/3 sin (π) = 0

Similarly, when we put n = 2, we get x = 3π/2, hence f'(3π/2) = 0

If we put n = 3, then x = 3π which is not in the given domain. So we stop here.

Now, we check the sign of f'(x) in the following intervals:

**0 < x < 3π/4**: For this interval, let's take a random value of x, say π/4. Now, we calculate f'(π/4) =-4/3 sin (4 (π/4) / 3) = -4/3 × √3/2 = -2/√3, which is less than zero. Hence, f(x) is decreasing in this interval.**3π/4 < x < 3π/2**: For this interval, let's take a random value of x, say π. Now, we calculate f'(π) =-4/3 sin (4 (π) / 3) = -4/3 × (-√3/2) = 2/√3, which is more than zero. Hence, f(x) is increasing in this interval.**3π/2 < x < 2π**: For this interval, let's take a random value of x, say 7π/4. Now, we calculate f'(π/4) =-4/3 sin (4 (7π/4) / 3) = -4/3 × √3/2 = -2/√3, which is less than zero. Hence, f(x) is decreasing in this interval.