What are the intervals for which the function f(x) = 6cospix is decreasing?

Question

A. [n, n+1], where n is any integer.

B. [2n, 2n+1], where n is any integer.

C. [n, 2n+1], where n is any integer.

D. [n+1, n], where n is any integer.

I don't understand how my teacher solved this question...

Answer 1

f(x) = 6cos(πx)

f is decreasing when f'(x) < 0

f'(x) = -6πsin(πx) = 0 when πx = kπ, where k = 0, 1, -1, 2, -2, ...

So, x = k

Critical points are 0, 1, -1, 2, -2, ...

Between x = 0 and x = 1, 0 < πx < π, so sin(πx) > 0. Therefore, f'(x) < 0. so f is decreasing.

Between x = 1 and x = 2, π < πx < 2π, so sin(πx) < 0. Therefore, f'(x) > 0, so f is increasing.

Between x = 2 and x = 3, 2π < πx < 3π, so sin(πx) > 0. Therefore, f'(x) < 0. so f is decreasing, etc

f is decreasing on [0, 1], [2,3], [4, 5], ... and [[-2, -1], [-4,-3], [-6, -5], ...

So, f is decreasing on [2n, 2n+1] where n = 0, ±1, ±2, ±3, ...

Answer 2