Showing $F(x)=\frac{x+1}{x-1}$ is its own functional inverse

Question

If $F(x) = \frac{x+1}{x-1}$, how can I show that $F(F(x))= x$ ? I tried doing in inverse and all but how I can solve problems as these in $F(F(x))$?

Answer 1

Assuming you mean that $f(x)=\frac{x+1}{x-1}$ and you ask what $f(f(x))$ is, note that $f(x)$ can be rewritten as $1+\frac{2}{x-1}$. This wasn't strictly necessary to notice, but it helps simplify the algebra later on.

To see this, remember that you can always add zero or multiply by one and it won't change the value of anything. The trick is just how zero or one looks like in your particular context. Here, I added and subtracted $2$ in the numerator to get $f(x)=\frac{x+1}{x-1} = \frac{x+1+0}{x-1}=\frac{x+1+2-2}{x-1}=\frac{x-1+2}{x-1}=\frac{x-1}{x-1}+\frac{2}{x-1}=1+\frac{2}{x-1}$

From here, it is a simple matter of algebraic manipulation. Remember what it means to have a function and giving it an input. In your case, you have $f(\color{red}{x})=1+\frac{2}{\color{red}{x}-1}$, and if you were to plug in a specific value in place of $x$, that value just takes its place in the expression on the right, so for instance $f(\color{red}{5}) = 1+\frac{2}{\color{red}{5}-1}$ or $f(\color{red}{10000})=1+\frac{2}{\color{red}{10000}-1}$.

This remains true even in the event that the value you plug in is more complicated and involves variables, such as $f(\color{red}{y})=1+\frac{2}{\color{red}{y}-1}$ or $f(\color{red}{x^2})=1+\frac{2}{\color{red}{x^2}-1}$ or even $f(\color{red}{f(x)})=1+\frac{2}{\color{red}{f(x)}-1}$

You would have then $f(f(x)) = 1+\frac{2}{f(x)-1} = 1+\frac{2}{1+\frac{2}{x-1} - 1}$

Continuing the simplification, we have:

$1+\frac{2}{1+\frac{2}{x-1} - 1} = 1+\frac{2}{~\frac{2}{x-1}~} = 1+x-1 = x$

Answer 2

\begin{equation} F(F(x)) = F(\frac{x+1}{x-1}) = \frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1} = \frac{2x}{2} = x \end{equation} You can also use $F(x) = y = \frac{x+1}{x-1} \implies x = \frac{y+1}{y-1}$ Therefore $x = F^{-1}(y) = \frac{y+1}{y-1} = F(y) $.

Answer 3

$$F(F(x)) = \frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1} = \frac{x+1+x-1}{x+1-(x-1)} = \frac{2x}{2} =x$$