I have the function

$f(x)=\frac{(1+x)^k}{1+ax}$, where $x>0, 0<a<k<1$.

The function has only one maximum at $x_0=\frac{a-k}{a(k-1)}$, increases on the left of $x_0$ and decreases on the right of $x_0$.

I want to find an approximation for $f(x)$ on the right of $x_0$. There are two constraints of the approximation:

  1. It only needs to work for $f(x)>1$.

  2. The approximation should have fairly simple (preferably unique) inverse function.

Any hint is appreciated.

  • 1
    $\begingroup$ Could you explain the purpose of the approximation and/or provide criteria for determining how good it is? Otherwise there are far too many solutions to describe and they're all arbitrary. $\endgroup$ – whuber May 17 '18 at 20:25
  • $\begingroup$ Actually I want to find a simple expression with respect to $x$ for $x_r=f^{-1}_r(f(x))$, where $0<x<x_0$ and $f_r^{-1}$ is the inverse function of $f(x)$ on the right of $x_0$. I intend to find an approximation for $f(x)$ on the left and an approximation for $f(x)$ on the right of $x_0$. I think the left can be approximated by a parabolic curve when $a$ is close to $k$. This question is for the right part. $\endgroup$ – Tri Nguyen May 17 '18 at 20:59

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