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I have a question about finding the value of a certain function that I cannot wrap my head around.

The question is: Given a function $f(x)$ satisfying $$f(x) + 2f\left(\frac{1}{1-x}\right) = x,$$ Then find $f(2).$

So far, I have tried plugging 2 into the original equation to yield $f(2) + 2f(-1) = 2.$

Next, I plugged $\displaystyle \frac{1}{2}$ into the original equation to yield $\displaystyle f\left(\frac{1}{2}\right) + 2f(2) = \frac{1}{2}$.

However, I do not know how to solve this system of equations for $f(2).$

Please let me know of any hints you may have. Many thanks.

juantheron
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Linear Pedant
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  • Some hints for formatting in MathJax are here. It would be much easier to read the question if you do that. Your approach of finding good values to plug in for $x$ is a good one. I wrote an answer, but T. Boners supplied more info. – Ross Millikan Jul 16 '16 at 03:36
  • Thank you, Ross. I am new to the website and did not know how MathJax worked, so I appreciate your direction to the tips page. – Linear Pedant Jul 16 '16 at 03:42

2 Answers2

3

You have two equations, but three variables: Namely, $f(2), f(-1)$ and $f(1/2)$. Try setting $x = -1$ to also get the equation

$$f(-1) + 2 f\left( \frac 1 2\right) = \frac 1 2$$

This is a third equation, from which you can solve the system of three.

2

Given $$f(x)+2f\left(\frac{1}{1-x}\right) = x...................(1)$$

Replace $\displaystyle x\rightarrow \frac{1}{1-x}\;,$ We get

$$f\left(\frac{1}{1-x}\right)+2f\left(\frac{x-1}{x}\right) = \frac{1}{1-x}.......(2)$$

Again replace $\displaystyle x\rightarrow \frac{1}{1-x}\;,$ We get

$$f\left(\frac{1}{1-x}\right)+2f(x) = \frac{x-1}{x}..............(3)$$

Now from $(1)$ and $(2)$

$$f(x)-4f\left(\frac{x-1}{x}\right)=x-\frac{2}{1-x}..........(4)$$

Multiply $(3)$ by $4\;,$ We get

$$4f\left(\frac{1}{1-x}\right)+8f(x) = \frac{4(x-1)}{x}..........(5)$$

Add $(4)$ and $(5)$

$$9f(x) = x-\frac{2}{1-x}+\frac{4(x-1)}{x}$$

So $$9f(2)=2+2+2=6\Rightarrow f(2) = \frac{2}{3}$$

juantheron
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