h(x) = cos (3x/2), 0 < x < 2pi

Question

Identify the open interval on which the function is increasing or decreasing. (Using interval notation.)

h(x) = cos (3x/2), 0 < x < 2pi

 

So far for increasing I have (2pi/3, 4pi/3)

Answer 1

h(x) = cos(3x/2) 

 

Find critical points:

 

   h'(x) = -(3/2)sin(3x/2) = 0 when 3x/2 = kπ, k = 0, ±1, ±2, ...

 

         If k = 0, then x = 0

 

         If k = 1, then 3x/2 = π. So, x = 2π/3.

 

         If k = 2, then 3x/2 = 2π. So, x = 4π/3.

 

         If k = 3, then 3x/2 = 3π. So, x = 2π.

 

When 0 < x < 2π/3, h'(x) < 0. So, h(x) is decreasing.

 

When 2π/3 < x < 4π/3, h'(x) > 0, So, h(x) is increasing.

 

When 4π/3 < x< 2π, h'(x) < 0. So, h(x) is decreasing. 

 

h(x) is increasing on (2π/3, 4π/3).

 

h(x) is decreasing on (0, 2π/3) ∪ (4π/3, 2π)