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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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    I was wondering: assume that $\int_0^x g(t)\, dt=0$ for every $x \geq 0$. Can we deduce anything about $g$?2012-07-05
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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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    but differentiating two equal integrals doesn't necessarily make them equal right ?2012-07-05
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    If two functions are equal, their derivatives are also equal ;)2012-07-05
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    Well if two functions are equal, they are the SAME function...2012-07-05
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    How is the derivative of a function defined, @The-Ever-Kid? It is defined in turns of the values of the function. If two functions are equal at all points, then of course their derivatives are equal.2012-07-05
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    The converse is not true, of course - if two functions have the same derivative, they are not necessarily equal. That might be what is confusing you?2012-07-05
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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