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- Thread starter kkafal
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- Feb 13, 2012

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In term of ODE we can write [if I undestood correctly...] ...f:R->R is differentiable and f'(x)=[x^2/1+(x^2))] and f(0)=0.

check whether 0<=f(x)<=x for all x belonging R.

Thanks.

$\displaystyle y^{\ '}= \frac{x^{2}}{1 + x^{2}}\ ,\ y(0)=0$ (1)

The (1) is an ODE with separable variables and has as solution...

$\displaystyle y= x - \tan^{-1} x$ (2)

... so that Yor relation is satisfied... but only for $x \ge 0$...

Kind regards

$\chi$ $\sigma$