Let $f:[0, 1] \rightarrow [0, \infty]$ be a function of $x$, with a parameter $\theta > 0$, such that
- $f$ is continuous
- $f$ is strictly decreasing
- $f(0) = \infty$
- $f(1) = 0$
For example, $f(x) = (- \log(x))^{\theta - 1}$.
For a given $\theta$, is there a way to approximate $f^{-1}(2 f(0.5))$ for any such $f$?
That value could be almost anything. Take for $\theta > 0$: $$ f_\theta(x) = x^{- \theta} - 1 $$ Then $f_\theta(1/2) = 2^\theta - 1$, and $$f^{-1}_\theta(2 f_\theta(1/2)) = \left( 2(2^\theta - 1) + 1 \right)^{1/\theta} = (2^{\theta + 1} - 1)^{1/\theta} \approx 2^{(\theta + 1)/\theta} \left(1 - \frac{2^{- \theta - 1}}{\theta} + \ldots\right)$$ by using the first two terms of the binomial expansion.
As $f_\theta(0) \to \infty$ and $f_\theta(1) = 0$, one could also scale this function at will, and mangle it in a myriard other ways.