find a function such that it is symmetric across $y=1-x$ passing through $(0,0)$ and $(1,1)$, (not homework) I am attempting to characterize a specific type of function. The function would be such that it is symmetric across the line $y=1-x$ This function would be a mapping $f:[0,1]\rightarrow [0,1]$ such that $f(0)=0$, $f(1)=1$.
This is being used for a robust measure of copula dependence and for detection of concordance unusual events in the bivariate case with possible future generalization to multivariate case. A function that I came up with was 
$$f(x; a,b) = \frac{b^{ax}-1}{b^a-1}$$
I don not know if there exist a base $b$ such that this function will ever be symmetric across the line $y=1-x$ A plot is given below of $f$ and $f^{-1}$ from 0 to 1 for base being the golden ratio and the parameter $a$ being 10. My goal is to have a be a parameter such that, in some limit the function and its inverse will engulf the entire square.
http://i.imgur.com/98FGdNt.jpg
edit I think it boils down to finding a function such that $f^{-1}(x) = 1-f(x)$
 A: Try $f(x;a)=1-(1-x^{a})^{1/a}$ for some $a\geqslant1$. Then $f$ and $f^{-1}$ indeed engulf the entire unit square $(0,1)^2$ when $a\to+\infty$. The area enclosed between their graphs starts at the line $y=x$ when $a=1$ and increases to being the full square $(0,1)^2$ when $a\to+\infty$. 
The symmetry with respect to $x+y=1$ is obvious since the graph of $f$ is the South-East quarter of the curve $|x|^{a}+|y-1|^{a}=1$. More generally, every function $f$ defined by $\varphi(x)+\varphi(1-f(x))=0$, for some function $\varphi$, would satisfy this symmetry.
A: Define $g(x)=f(1-x)$. Then $g(x)$ is symmetric about $x=y$ if, and only if $f(x)$ is symmetric about $y=1-x$. This condition is equivalent to $(g\circ g)(x)=:g^2(x)=x$. Also, it implies that $g$ is invertible (if there were $x_1,x_2$ such that $g(x_1)=g(x_2)=y$, then $x_1=g^2(x_1)=g(y)=g^2(x_2)=x_2$ which is obviously impossible).
For your particular case we have:
$$g(x) = \frac{b^{a(1-x)}-1}{b^a-1}$$
We compute
$$g^2(x) = \frac{b^{a\left(1-\frac{b^{a(1-x)}-1}{b^a-1}\right)}-1}{b^a-1}=\frac{b^{a\frac{b^a(1-b^{a(-x)})}{b^a-1}}-1}{b^a-1}$$
Now I don't think it possible to find parameters $a$ and $b$ such that this function is the identity...
