Question

On the closed interval, [0, 2*pi], find the absolute maximum of the function f(x)= sin^2(x)

Answer 1

LOL, |sin x| is always </= 1
it is at + pi/2 and 3 pi/2
where cos x , the derivative of sin x, is 0

Answer 2

To show how I got the answer, is the work sufficient?

f(x)= sin^2(x)
f'(x)=2sinx(cosx)
= cosx(2sinx)

cosx=0 sinx=0


f(0)=0
f(pi)=0
f(pi/2)=1
f(3pi/2)=1

Absolute Maximum is at pi/2 and 3pi/2

Answer 3

You mean cos x sin x = 0 when cos x = 0 or when sin x = 0
when cos x = 0, then sin x = 1 and you have a max
when sin x = 0, then cos x = 1 and you have a min

Answer 4

Yes, so what work do I need to show how I got a Absolute Maximum at pi/2 and 3pi/2

Answer 5

You already did it.
f(0) = 0
f(pi) = 1 etc, at the four points where the derivative is zero

Answer 6

So don't write cos x = 0 and sin x = 0 ?

Answer 7

The max is where |sin x| = 1, cos x = 0
the min is where sin x = 0, |cos x| = 1

but really all you really have to say is
|sin x | </= 1
so
|sin^2 x| </= 1
so max is 1 and that is when |sin x|=1