Question

Let \[F(x)\] be the real-valued function defined for all real \[x\] except for \[x = 0\] and \[x = 1\] and satisfying the functional equation \[F(x) + F\left(\frac{x-1}x\right) = 1+x.\]Find the \[F(x)\] satisfying these conditions

Answer 1

part A)
\(let ~~t=\dfrac{x-1}{x}=1-\dfrac{1}{x}\), then \(\dfrac{1}{x}=1-t\rightarrow x=\dfrac{1}{1-t}\)
\(F(x) + F(\dfrac{x-1}{x})= 1+x ~~becomes~~F(\dfrac{1}{1-t})+F(t) =1+\dfrac{1}{1-t}=\dfrac{2-t}{1-t}\)
Let's call it is A

Answer 2

if A is that, just replace variable t by x, then we have
\(F\left(\dfrac{1}{1-x}\right)+F(x)= \dfrac{2-x}{1-x}\)
Again, We have A part.

Answer 3

Now, part B)
\(let~ t = \dfrac{1}{1-x}\rightarrow \dfrac{1}{t}=1-x\), then \(x= 1-\dfrac{1}{t}=\dfrac{t-1}{t}\)

\(F(x) +F\left(\dfrac{x-1}{x}\right)=1+x\)

\(becomes~~F\left(\dfrac{t-1}{t}\right)+F\left(\dfrac{1}{t-1}\right)=1+\dfrac{t-1}{t}=\dfrac{2t-1}{t}\)

Answer 4

Now, just replace variable t by x, we have

\(F\left(\dfrac{x-1}{x}\right)+F\left(\dfrac{1}{x-1}\right)=\dfrac{2x-1}{x}\)
Call it B

Answer 5

Take original + A + B and solve for \(F\left(\dfrac{1}{x-1}\right)\)
then plug back A to get \(F(x)\)
good luck

Answer 6

can you elaborate a bit more? It is kind of confusing.

Answer 7

I am not sure what he is doing either, maybe one of the 5 likes that understood it enough to click best answer can help you.

Answer 8

\[original ~one~~F(x)+F\left(\dfrac{x-1}{x}\right)=1+x\\+A, A ~~is ~~~~~F\left(\dfrac{1}{1-x}\right)+F(x)=\dfrac{2-x}{1-x}\\
+B,B~~is~~~~~~~F\left(\dfrac{x-1}{x}\right)+F\left(\dfrac{1}{x-1}\right)=\dfrac{2x-1}{x}\\--------------------------------------------------------------------\\2\left(F(x)+F\left(\dfrac{x-1}{x}\right)+F\left(\dfrac{1}{x-1}\right)\right) =RHS\]
you just add the right hand sides together

Answer 9

But as original given, \(F(x) +F\left(\dfrac{x-1}{x}\right)=1+x\)
Hence the \(LHS=2\left((1+x)+F\left(\dfrac{1}{x-1}\right)\right)= RHS\)

Answer 10

Now, you can solve for \(F\left(\dfrac{1}{x-1}\right)\)
When you get it, replace back A to get \(F(x)\)