Animated Solution for Mathematics - Functions: Let $f:\mathbb{R}-\{0,1\}\to \mathbb{R}$ be a function such that $f(x)+f\left(\frac{1}{1-x}\right)=1+x$. Then $f(2)$ is equal to :

Welcome to the arena of functional equations. We are given the equation:

At first glance, the term $f\left(\frac{1}{1-x}\right)$ suggests that the function is linked to its own transformation. The secret to this problem is to exploit the cyclic nature of the transformation $g(x)=\frac{1}{1-x}$.

Let us examine the behavior of the transformation $g(x)=\frac{1}{1-x}$ starting at $x=2$:

1. Applying $g(x)$ once: $g(2)=\frac{1}{1-2}=-1$.

2. Applying $g(x)$ again: $g(-1)=\frac{1}{1-(-1)}=\frac{1}{2}$.

3. Applying $g(x)$ a third time: $g\left(\frac{1}{2}\right)=\frac{1}{1-1/2}=2$.

We have returned to our starting point. This confirms that the transformation has a periodicity of 3, creating a cycle: $2\to -1\to \frac{1}{2}\to 2$.

We translate this journey into algebra by substituting these three values into our original equation $f(x)+f\left(\frac{1}{1-x}\right)=1+x$:

For $x=2$:

For $x=-1$:

For $x=\frac{1}{2}$:

We now have a system of three linear equations with three unknowns: $f(2)$, $f(-1)$, and $f\left(\frac{1}{2}\right)$.

To isolate $f(2)$, we first add equations (A) and (C):

Notice that the sum $f(-1)+f\left(\frac{1}{2}\right)$ is exactly the left side of equation (B). Subtracting equation (B) from our sum yields:

The terms $f(-1)$ and $f\left(\frac{1}{2}\right)$ cancel out, leaving us with:

Dividing by 2, we arrive at the final result:

Through the power of cyclic substitution and linear algebra, we have conquered the problem. Remember this technique for any functional equation involving a cyclic transformation.