Q.If $f(x)=\dfrac{x+1}{x-1}$, $x\neq 1$, show that $f$ is its own inverse, i.e. $f(f(x))=x$.

Compute $f(f(x))$ by substituting the whole rule into itself and simplifying the resulting compound fraction.

We compute $f(f(x))$, replacing the input of the outer $f$ by $f(x)=\dfrac{x+1}{x-1}$:

Simplify the numerator:

Simplify the denominator:

Divide (the common $(x-1)$ cancels):

$f\big(f(x)\big)=\frac{\dfrac{2x}{x-1}}{\dfrac{2}{x-1}}=\frac{2x}{x-1}\times\frac{x-1}{2}=\frac{2x}{2}=x.$ …

The main difficulty is the compound fraction: combine the numerator and denominator over the common denominator $x-1$ before dividing, then invert-and-multiply. Cancelling too early, …