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let be a function so

$ f(x)= x^{2}+h(x) $ , here $h(x) $ is a function so $ h(x) = O(logx)$

it is clear that $ f(x) \sim x^{2} $ for big 'x' so the inverse

$ f^{-1}(x)\sim x^{1/2}$ for big 'x' is this correct, does this mean that the function $ f(x)$ has an inverse and that this inverse will be asymptotic to $ x^{1/2} $ ?

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You have to specify the the $x$-domain you have in mind. –  Christian Blatter Jul 11 '12 at 11:04
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No, for instance, if $$ h(x)=\begin{cases}x & \text{if $\lvert x\rvert<2$}\\ 0&\text{else}\end{cases} $$ then $f(0)=f(-1)$, hence $f$ is not invertible. This counterexample could be turned into a continuous or smooth one easily.

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so your function is $ y=|x| $ the inverse is then $ x=|y| $ and i can draw it numerically –  Jose Garcia Jul 11 '12 at 11:44
    
No, my function is $f(x)=x^2+x$ if $x$ has modulus smaller than two and it is $f(x)=x^2$ if $x$ has modulus greater or equal to two. –  Rasmus Jul 11 '12 at 13:25
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