- #1

christos.miko

- 2

- 0

I noticed that [f(x)*f(x)]' = 2f(x)*f'(x) = [f(x)]^2

So I tried multiplying that inequality by 2.

2f (x)f ′ (x) ≥ 2sin(x)

Then I tried integrating both sides.

[f(x)]^2 ≥ -2cos(x).

If I integrate both sides from 0 to Pi

We will get [f(pi)]^2-[f(0)]^2 ≥ -2[cos(pi) - cos(0)]

This gives us [f(pi)]^2-[f(0)]^2 ≥ 4.

This gives us [f(pi)]^2-[f(0)]^2 ≥ 4. I am not sure what this tells me. Do I need to find a function with this property? I also need | f (x) |< 2