# inverse function for big 'x'

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

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.
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