Question

I need help with this Extrema question involving a trig function.

f(x)=cos^2(2x), which = (cos(2x))^2
f'(x)= 2(cos(2x))(-sin(2x))(2)
=-4cos(2x)sin(2x)
Now, when I have to find the critical number is when I start to have a problem. I know I set f'(x) equal to 0.
cos(2x)=0 sin(2x)=0
cos(u)=0, where u=2x sin(u)=0, where u=2x

u= (π/2),(3π/2) u=π, 2π
2x=(π/2) 2x=(3π/2) 2x=π 2x=2π

x=(π/4),(3π/4) x=(π/2),(π)
So, now I assume that these are the critical numbers, but my text book tells me otherwise.

Answer 1

I really don't see how they got the other critical numbers.

Answer 2

The functions sin and cos have critical numbers the stretch off into infinity. These critical numbers that the solution page has looks like all the ones between the two given intervals.

Answer 3

a plain sin(x) function has two critical points between the interval 0 < x < 2pi. But a sin(2x) functions period is shorter, (it is double to be exact).

Answer 4

So, since cos2x=0 yields x=π/4,3π/4
I grab all the other radians that contain π/4?

Answer 5

until you get to 0 or 2pi