I have an equation: $f(x) = \frac{x^n}{x^n + (1-x)^n} $ where $n$ has a range between [2-infinity].
For example, considering $n=2$, the first derivative (slope of the above equation) is $\frac{2x(1-x)}{(2x^2-2x+1)^2}$.
My aim: determine the value of $x$ given a known RHS value which is a tangential function.
Example: $\frac{2x(1-x)}{(2x^2-2x+1)^2}$ = 2.
Questions:
How can I solve for $x$ in this case the equation being a 4th order polynomial? I know the answer is 0.5 for this case but dont know any further if the RHS is other than 2.
Based on my explanation $n$ has an order between 2 to infinity. If $n=3$, the equation to be solved would be of 6th order and so on..
I heard from a friend few solvers like "Maple/ Mathematica" can solve these but are not open source. Can anyone help me in this regard?
Thanks!!
Elaborating on your example :
$$\frac{2x(1-x)}{(2x^2-2x+1)^2} = 2 \Leftrightarrow 2x(1-x)=2(2x^2-2x+1)^2 $$
$$\Leftrightarrow$$
$$2x(1-x) = 2[2x(x-1) +1]^2 \Leftrightarrow2x(1-x) = 2[4x^2(x-1)^2+4x(x-1) + 1]$$
$$\Leftrightarrow$$
$$2x(1-x)=8x^2(x-1)^2 + 8x(x-1) + 2 \Leftrightarrow8x^2(x-1)^2 -6x(x-1)+2=0$$
Let $y=x(x-1)$. Then, the equation becomes :
$$8y^2-6y+2 =0$$
You can now solve this equation easily to find $y$ and then substitute the values of $y$ you found back into $y=x(x-1)$ to find the values of $x$.
Generally : You should always look for such tricks (this was an easy one to see that both the LHS and RHS had a common term after factorizing. If such a manipulation is not possible, then the only way to solve such polynomials is referring to Horner's Rule. For higher order polynomials that have really weird roots it's almost impossible to solve them exactly by hand and an online calculator is the only thing that could help for exact roots.