Finding an inverse function (sum of non-integer powers) I have a function:
$$f(x)=x^{2.2} + (1-x)^{2.2}$$
It is defined on the interval $[0,1]$. Minimum: $x=0.5, y=2*0.5^{2.2} = 2^{-1.2}$.
I want to find an inverse for it. Since the function has two "wings", inverse will be a family of two functions.
After some tinkering, I crafted something that looks like a very good approximation of an inverse function:
$$ g(x)=\frac{1}{2} \left( 1 \pm \left(\frac{x-2^{-1.2}}{1-2^{-1.2}}\right)^{0.504288} \right) $$
The number $0.504288 \approx 1 / 1.9829939 $ was found experimentally by substituting $g(x)$ into $f(x)$ and tweaking it to make it look as straight as possible:
$$ p(x) = f(g(x)) \approx x $$
Illustration: https://www.geogebra.org/graphing/zgzafsk4 (Might be a bit slow. Image substitute just in case.)
And now it bothers me if I'm just one step away from the exact solution.
So the question is: is it possible to express the exact power in $g(x)$ to get the equality $p(x) = x$ and what that value will be?
Update:
OK, people seem to focus on using usual numeric tools to get an arbitrarily close approximation. But this is not what the question was about. I have an approximation that is good enough for my purposes.
The question is about this particular special case. There is a power function added to reversed and shifted copy of itself.
Inverse function for a power function $y = x^{2.2}$ will be just the power reversed $x = y^{1/2.2}$. Since we adding an increasing and a decreasing function, the resulting curvature has changed. And it raises the suspicion that there might even be an exact power value, smaller than the original 2.2...
After writing this, I realized that the problem can be expressed in a different way. What I actually did is that I made an inverse function for an approximation of $f(x)$:
$$f_{approx}(x) = 2^{-1.2}+ (1-2^{-1.2}) (2x-1)^{1.983}$$
Now I made a different illustration: https://www.geogebra.org/graphing/msfzaqah (image).
There is also $h(x) = \frac{f_{approx}(x)}{f(x)}$ on the illustration. It clearly has some extremes, and changing the power just pushes them around. So the answer to the original question must be: this approximation doesn't fit the function exactly, so there is no exact number to put in there.
Now the question is: can the original function be expressed as something invertible? Same shape functions with an integer power are invertible. What stands in the way for a function with non-integer (fractional) power to be invertible too?
Note:
For powers 2 and 3, similar functions can be expressed in a clearly invertible form:
$$x^2+(1-x)^2 = \frac{1}{2} + \frac{1}{2}(2x - 1)^2$$
$$x^3+(1-x)^3 = \frac{1}{4} + \frac{3}{4}(2x - 1)^2$$
For the power of 4 and above WolframAlpha doesn't provide a form like this (single power), but still able to construct inverse functions, albeit more and more complicated.
Interesting that for powers of 2 and 3 the resulting function has the power of 2. And this fact seems to persist for higher powers - a sum of (2n+1) power functions will be a (2n) power function. But that's a digression.
Update 2:
I really appreciate the answers about Tailor series expansion. But I'm still concerned: is it the best we can do?
 A: Too long for a comment.
Concerning what you wrote in the note about the invertible forms
$$x^2+(1-x)^2 = \frac{1}{2} + \frac{1}{2}(2x - 1)^2$$
$$x^3+(1-x)^3 = \frac{1}{4} + \frac{3}{4}(2x - 1)^2$$
consider
$$y=x^k+(1-x)^k$$ and let $x=\frac {1+u}2$ that is to say $u=(2x-1)$. This makes
$$y=2^{-k} \left((1+u)^k+(1-u)^k\right)\implies 2^k y=(1+u)^k+(1-u)^k$$
Now, expand as Taylor series around $u=0$ to get
$$2^k y-2=(k-1) k u^2+\frac{(k-3) (k-2) (k-1) k}{12}  u^4+O\left(u^6\right)$$Notice that, since $u \leq \frac 12$, the ratio of the second term to the first term is at most $\frac{(k-3) (k-2)}{48} $ which, for $2\leq k \leq 3$, is, in absolute value,  less that $\frac 1{192}$. This justifies the trunction to $O\left(u^4\right)$ and the formula you propose before adjusting the power for a better fit (for the compensation of the neglected terms).
A: There may not be an exact expression in terms of elementary functions for the inverse $f^{-1}$ of a function $f$.  However, there are several ways you can systematically approximate the inverse.  One might be this.  Choose several data points $(x,f(x))$ for your function $f$ (where the inverse is defined).  Then, the inverse satisfies $(x,f^{-1}(x)) = (f(x),x)$.  Once you find these data points (basically by flipping $x$ and $f(x)$), you can fit an accurate regression curve which closely approximates your data $(x,f^{-1}(x))$.  If your function is a polynomial, you might choose a polynomial basis (e.g. {1,x,x^2,x^3,...x^n}).  Then, your approximate inverse will have the form
$$f^{-1}(x) = a_0+a_1x+a_2x^2 + \cdots a_nx^n.$$
A: In addition to the pertinent answer from Claude Leibovici, this is an alternative way for series expansion, with a more general power ($a$ instead of $2.2$ ).
$$f(x)=x^a+(1-x)^a \tag 1$$
Obviously the curve which represent $f(x)$ is symmetrical with respect to $x=\frac12$. The minimum of $f$ is $2^{1-a}$. This suggests the change of variables : 
$$\begin{cases}
x=\frac12+X \\
f=2^{1-a}(1+Y^2)^a
\end{cases} \quad\implies\quad \left(\frac12+X\right)^a+\left(\frac12-X\right)^a =2^{1-a}(1+Y^2)^a \tag 2$$
For the inverse function we look for a series expansion of the form :
$$X=c_0+c_1Y+c_2Y^2+...+c_kY^k+...$$
In Eq.$(2)$ we replace $X$ by the series of powers of $Y$ and we identify the coefficients. The result is :
$$c_0=c_2=c_4=c_6=0$$
$$c_1=\frac{1}{\sqrt{2(a-1)}}$$
$$c_3=\frac{\sqrt{2}}{24}\frac{(a+1)(2a-3)}{(a-1)^{5/2}}$$
$$c_5=\frac{\sqrt{2}}{2880}\frac{(a+1)(12a^3-40a^2+47a-45}{(a-1)^{5/2}}$$
The calculus was done only up to $c_5$ which is sufficient for a very good accuracy.
The inverse function is :
$$\boxed{x\simeq \frac12+c_1Y+c_3Y^3+c_5Y^5 \quad\text{with}\quad Y=\pm\sqrt{(2^{a-1}f)^{1/a}-1}}$$ 
RESULT in case of $a=2.2$ :
$$a_1=0.645497224368$$
$$a_3=0.200821358692$$
$$a_5=-0.007395326225$$
In the tables below $f_k$ are the given values. $x_k$ are the computed values.
On can compare the given $f_k$ with the computed $f_k\simeq (x_k)^{2.2}+(1-x_k)^{2.2}$.
Branch $+$ of the square root :

Branch $-$ of the square root :

A: This is not a useful answer, but I wanted to try it anyway.
Let's try to perform an inversion operation in a domain where it is trivial - in polar coordinates.
$$y = x^{2.2}+(1-x)^{2.2}\tag{1}$$
$$r\sin{\theta} = (r\cos{\theta})^{2.2}+(1-r\cos{\theta})^{2.2}\tag{2}$$
Inversion is a rotation around a line $\theta = \pi/4$. We get the rotated equation by replacing $\theta$ with $(\frac{\pi}{2} - \theta)$. We can also notice that
$$\sin{(\frac{\pi}{2} - \theta)} = \cos{\theta}$$
$$\cos{(\frac{\pi}{2} - \theta)} = \sin{\theta}$$
And the rotated equation in polar coordinates looks like this:
$$r\cos{\theta} = (r\sin{\theta})^{2.2}+(1-r\sin{\theta})^{2.2}\tag{3}$$
Very neat, isn't it?
Now let's try to get back to Cartesian coordinates. (Note: we only concerned about the first quadrant of polar plane.)
$$r = \sqrt{x^2 + y^2}$$
$$\theta = \tan^{-1}{(y/x)}$$
$$\cos{(\tan^{-1}{(y/x)})} = \frac{1}{\sqrt{(y/x)^2+1}}$$
$$\sin{(\tan^{-1}{(y/x)})} = \frac{y/x}{\sqrt{(y/x)^2+1}}$$
$$\frac{\sqrt{x^2 + y^2}}{\sqrt{(y/x)^2 + 1}} = \left(\frac{y}{x}\frac{\sqrt{x^2 + y^2}}{\sqrt{(y/x)^2 + 1}}\right)^{2.2} + \left(1 - \frac{y}{x}\frac{\sqrt{x^2 + y^2}}{\sqrt{(y/x)^2 + 1}}\right)^{2.2}\tag{4}$$
We can then simplify this equation to the following:
$$x = y^{2.2}+(1-y)^{2.2}\tag{5}$$
Which is both silly and obvious. (Like Captain Obvious level of obvious...)
And we are back to where we started.
