Question

$\frac{1}{n}\sum _{i=1}^n{f}^{-1}\left(x_i\right)=f\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)$

Answer 1

The condition,
$\sum _{i=1}^n{f}^{-1}\left(x_i\right)=\sum _{i=1}^n{x}_i$ can be written as;
$\frac{1}{n}\sum _{i=1}^n{f}^{-1}\left(x_i\right)=\frac{1}{n}\sum _{i=1}^n{f}^{-1}{x}_i$
ie, AM of y-coordinates of $f^{-1}=AM$ of x coordinates of f.
The given two conditions hold if and only if $x^2+3x-3=x$ (point where f and $f^{-1}$ meet).
$\Rightarrow x^2+2x-3=0$
So, $x=-3,+1$
But $x\ge 0\Rightarrow x=1$ (neglecting $x=-3$)
Hence, we can write $\frac{1}{n}\sum _{i=1}^n{x}_i=1$ which is the required result.