What is the value of $x$, given that $f\left(x\right)+f\left(\frac{1}{1-x}\right)=x$ and $f^{-1}\left(x\right)=2$?

Question

I tried to approach the problem by using the equation $f\left(2\right)=x$ but I always get stuck in the middle of the process.

Answer 1

Try sub in $x=x$,$x=\frac{1}{1-x}$,$x=\frac{x-1}{x}$.

you should get $3$ equations with $3$ variables, namely $f(x),f(\frac{1}{1-x}), f(\frac{x-1}{x})$.

From which you can find $f(x)$ (left as an exercise for you), then sub $x=2$.

Answer 2

I think a function $f$ doesn't have an inverse in $\mathbb{R}$, This is my attempt :

We have $$f(x)+f\left(\frac{1}{1-x}\right)=x $$

If we change variable $x \to\ 1-\frac{1}{x}$, we have $$f\left(1-\frac{1}{x}\right)+f(x)=1-\frac{1}{x}$$

Also, $x\to\frac{1}{1-x}$ from original one, we have $$f\left(\frac{1}{1-x}\right)+f\left(1-\frac{1}{x}\right)=\frac{1}{1-x}$$

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From these, we can get \begin{eqnarray} f\left(\frac{1}{1-x}\right)-f(x)&=&\frac{1}{1-x}+\frac{1}{x}-1 \\ f\left(\frac{1}{1-x}\right)+f(x)&=&x \, \text{(This is just the original one.)} \end{eqnarray}

And you can get original function $f$ from subtracting these.

Answer 3

$$f(2)+f(-1)=2,$$ $$f(-1)+f\left(\frac{1}{2}\right)=-1$$ and $$f\left(\frac{1}{2}\right)+f(2)=\frac{1}{2},$$ which gives $$f(2)+f(-1)+f\left(\frac{1}{2}\right)=\frac{3}{4}$$ and $$\frac{3}{4}-f(2)=-1$$ or $$f(2)=\frac{7}{4}.$$

Answer 4

We have

$$x=2\Rightarrow f(2)+f(-1)=2$$

$$x=1/2\Rightarrow f(1/2)+f(2)=1/2$$

$$x=-1\Rightarrow f(-1)+f(1/2)=-1$$

This is a system of three equations with three unknowns. If it helps, we can rewrite it as

$$s+y=2$$

$$s+z=1/2$$

$$y+z=-1$$

where $f(2)=s$, $f(-1)=y$, and $f(1/2)=z$. We can easily solve this by noting that

$$2+1/2=(s+y)+(s+z)=2s+(y+z)=2s-1$$

$$7/2=2s$$

$$s=7/4$$

We conclude $f(2)=7/4$ and therefore $x=7/4$.