If $f$ is "almost" $x^k$, is $f^{-1}$ "almost" $x^{1/k}$?

Question

Write $g(x) = \mathcal{O}(x^k)$ if $g(x) = h(x)x^k$, with $h(x) \to 0$, as $x \to 0$. Then $f(x) = x^k + \mathcal{O}(x^k)$ means that $f$ "is almost" $x^k$ near $x$. Assume that $f$ is strictly increasing (and thus invertible). Does this imply that $f^{-1}(x) = x^{1/k}+ \mathcal{O}(x^{1/k})$? This would seem sensible, but I can't come up with a nice short argument.