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$\int^x_0\frac1{f'(t)}dt = \int^x_02f(t)dt \tag{1}$ where $0 \leq x \leq 1$ and $f(0) = 0$

I need to prove that $f(\frac1{\sqrt{2}})> \frac1{\sqrt{2}}$

$f(\tan (x))> \tan(x) > x , x \in (0,\frac{\pi}{4}) $ $f(e^{-x^2})\geq e^{-x^2}$

The problem is that i dont know how to derive the function it self.

All I could do was say that $\frac{d\left(\int^x_0\frac1{f~'(t)}dt)\right )}{dx} = \frac1{f~'(x)}$

and $\frac{d\left(\int^x_02f(t)dt\right )}{dx} =2f(x)$

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    There is no *functional inequality* here.2012-07-05

1 Answers 1

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You get

$\frac{1}{f'(x)}=2f(x) $

Thus

$2f'(x)f(x)=1 \,.$ or

$\left( f(x)^2 \right)' =1 \,.$

Integrate, find the constant and you are done.

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    Thanks this is what confused me coz i remembered my teacher saying somthing like diffrentiating inequalities doesnt always satisfy it though integration does and thought that it applies to equalities to.....silly me :p2012-07-05