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Is there a function f , diﬀerentiable for all real x, such that | f (x) |< 2 and f (x)f ′ (x) ≥ sin(x)?

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).

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

I honestly have no idea if such a function even exists. Does anyone know what to do?

f i do what Dick said, and 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. I am not sure what this tells me. Do I need to find a function with this property? I also need | f (x) |< 2

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).

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

I honestly have no idea if such a function even exists. Does anyone know what to do?

f i do what Dick said, and 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. I am not sure what this tells me. Do I need to find a function with this property? I also need | f (x) |< 2

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