confusion with minimum and maximum of trig function

[cotfw]

sin(x)^2 * cos(x)^2

I was supposed to give the x values at the minimum and maximums of this for 0 < x < pi.

I found the first and second derivative correctly, but I don't understand the book's answer for the minimum and maximum. It only listed a maximum when x = pi/4, and no minimum. But that isn't what I got, and when I entered the equation in a graphing calculator, it also showed a high at 3*pi/4, and a minimum at pi/2. I am confused

 

[Ishuda]

sin(x)^2 * cos(x)^2

I was supposed to give the x values at the minimum and maximums of this for 0 < x < pi.

I found the first and second derivative correctly, but I don't understand the book's answer for the minimum and maximum. It only listed a maximum when x = pi/4, and no minimum. But that isn't what I got, and when I entered the equation in a graphing calculator, it also showed a high at 3*pi/4, and a minimum at pi/2. I am confused

Well
f(x) = sin²(x) cos²(x) = [(1/2) * 2 sin(x) cos(x)]² = 0.25 sin²(2x)
and
f'(x) = sin(2x) cos(2x) = 0.5 sin(4x)
f''(x) = 2 cos(2x)

So f'(x) = 0 at 4x = \(\displaystyle n \pi\) or
\(\displaystyle x_n = \frac{n \pi}{4}, n = 0, \pm 1, \pm 2, ...\)

You might check that I made no misteaks

 

[cotfw]

Well
f(x) = sin²(x) cos²(x) = [(1/2) * 2 sin(x) cos(x)]² = 0.25 sin²(2x)
and
f'(x) = sin(2x) cos(2x) = 0.5 sin(4x)
f''(x) = 2 cos(2x)

So f'(x) = 0 at 4x = \(\displaystyle n \pi\) or
\(\displaystyle x_n = \frac{n \pi}{4}, n = 0, \pm 1, \pm 2, ...\)

You might check that I made no misteaks

Yep your solution is correct. I think it's just the books answers that are wrong. I looked it up on the internet and apparently it's got quite a few errors in it

Thanks for your help