On which of the following intervals is the function f(x) = 4 cos(2x - π) decreasing?
x = pi over 2 to x = π
x = 0 to x = pi over 2
x = pi over 2 to x = 3 pi over 2
x = π to x = 3 pi over 2

Solution:

Given, the function is f(x) = 4 cos(2x - π)

We have to find the interval on which the function is decreasing.

1) For the interval x = π/2 to x = π

\(\\f(x)=4cos(2x-\pi )\\=4cos(2(\frac{\pi }{2})-\pi )\\=4cos(\pi- \pi )\\=4cos(0)\\=4(1)\\=4\)

\(\\f(x)=4cos(2x-\pi )\\=4cos(2\pi -\pi )\\=4cos(\pi)\\=4(-1)\\=-4\)

Therefore, f(x) is decreasing in this interval.

2) For the interval x = 0 to x = π/2

\(\\f(x)=4cos(2x-\pi )\\=4cos(2(0) -\pi )\\=4cos(-\pi)\\=4(-1)\\=-4\)

\(\\f(x)=4cos(2x-\pi )\\=4cos(2(\frac{\pi }{2})-\pi )\\=4cos(\pi- \pi )\\=4cos(0)\\=4(1)\\=4\)

f(x) is not decreasing in this interval.

3) For the interval x = π/2 to x = 3π/2

\(\\f(x)=4cos(2x-\pi )\\=4cos(2(\frac{\pi }{2})-\pi )\\=4cos(\pi- \pi )\\=4cos(0)\\=4(1)\\=4\)

\(\\f(x)=4cos(2x-\pi )\\=4cos(2(\frac{3\pi }{2})-\pi )\\=4cos(3\pi- \pi )\\=4cos(2\pi )\\=4(1)\\=4\)

f(x) is not decreasing in this interval.

4) For the interval x = π to x = 3π/2

\(\\f(x)=4cos(2x-\pi )\\=4cos(2\pi -\pi )\\=4cos(\pi)\\=4(-1)\\=-4\)

\(\\f(x)=4cos(2x-\pi )\\=4cos(2(\frac{3\pi }{2})-\pi )\\=4cos(3\pi- \pi )\\=4cos(2\pi )\\=4(1)\\=4\)

f(x) is not decreasing in this interval.

Therefore, f(x) is decreasing in the interval x = π/2 to x = π.

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On which of the following intervals is the function f(x) = 4 cos(2x - π) decreasing?

Summary:

The function f(x) = 4 cos(2x - π) is decreasing in the interval x = π/2 to x = π.