Find $f(5)$ where $f$ satisfies $f(x)+f(1/(1-x))=x $

Question

Question:

How do you Find $f(5)$ in which the function satisfies $$f(x)+f\left(\frac{1}{1-x}\right)=x $$ where $x\in\Bbb{R}$ and $x\neq 0,1$?

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My steps:

Step 1)

Substitute $5$ into the equation to get:

$$f(5)+f\left(\frac{1}{-4}\right)=5$$

But then I had gotten stuck there and I could not find $f(5)$

Please write detailed steps.

Answer 1

by using $x=5,4/5,-1/4$ we have: $$f\left(-\frac 1 4\right)+f\left(\frac 4 5\right)=-\frac 1 4$$ $$ f(5)+f\left(\frac 4 5\right)=\frac 4 5$$ $$f(5)+f\left(-\frac 1 4\right)=5$$

Then sum the last two expressions and subtract the first to get: $$ 2f(5)+f\left(\frac 4 5\right)+f\left(-\frac 1 4\right)-f\left(-\frac 1 4\right)-f\left(\frac 4 5\right)=5+\frac 45+\frac 1 4$$ hence $2f(5)=6.05$ and then $f(5)=3.025$.

Answer 2

$$f\left(5\right)+f\left(-\frac14\right)=5$$

If we know $f\left(-\frac14\right)$, we can solve the problem.

$$f\left(-\frac14\right)+f\left(\frac45\right)=-\frac14$$

If we know $f\left(\frac45 \right)$, we can solve the problem.

$$f\left(\frac45\right)+f\left(5\right)=\frac45$$

Why don't we just solve the linear system? Are you able to solve it?

Answer 3

You can use the fact that$$\left( \frac{1}{1-x} \right)^{-1}=1-\frac{1}{x},$$where the exponent $-1$ stands for the reverse. If you substitute $1/(1-x)$ and $1-1/x$ in the functional equation and solve three simultaneous equations, you can find general form of $f(x)$.