ariannamcqueen6
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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

Respuesta :

Versatilequeen
The answer is x= π/2 to x= π
(just did this question & got it right)

Answer: Option (A) is correct.

Step-by-step explanation: The given function is

[tex]f(x)=4\cos (2x-\pi).[/tex]

We are to select the interval in which the above function is decreasing.

For the interval [tex]x=\dfrac{\pi}{2}~\textup{to}~x=\pi,[/tex]

[tex]f(\dfrac{\pi}{2})=4\cos(2\times \dfrac{\pi}{2}-\pi)}=4\cos(\pi-\pi)=4\cos 0=4(1)=4,\\\\f(\pi)=4\cos(2\pi-\pi)=4\cos \pi=4(-1)=-4.[/tex]

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

For the interval [tex]x=0~\textup{to}~x=\dfrac{\pi}{2}[/tex],

[tex]f(0)=4\cos(2\times 0-\pi)}=4\cos(0-\pi)=4\cos pi=4(-1)=-4,\\\\f(\dfrac{\pi}{2})=4\cos(2\times \dfrac{\pi}{2}-\pi)=4\cos (\pi-\pi)=4(1)=4.[/tex]

So, f(x) isnot decreasing in this interval.

For the interval [tex]x=\dfrac{\pi}{2}~\textup{to}~x=\dfrac{3\pi}{2},[/tex]

[tex]f(\dfrac{\pi}{2})=4\cos(2\times \dfrac{\pi}{2}-\pi)}=4\cos(\pi-\pi)=4\cos 0=4(1)=4,\\\\f(\dfrac{3\pi}{2})=4\cos(2\times \dfrac{3\pi}{2}-\pi)=4\cos (3\pi-\pi)=4\cos 2\pi=4(1)=4.[/tex]

So, f(x) is not decreasing in this interval.

For the interval [tex]x=\pi~\textup{to}~x=\dfrac{3\pi}{2},[/tex]

[tex]f(\dfrac{\pi}{2})=4\cos(2\times \dfrac{\pi}{2}-\pi)=4\cos (\pi-\pi)=4(1)=4,\\\\f(\dfrac{3\pi}{2})=4\cos(2\times \dfrac{3\pi}{2}-\pi)=4\cos (3\pi-\pi)=4\cos 2\pi=4(1)=4.[/tex]

So, f(x) is not decreasing.

Thus, (A) is the correct option.

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