Answer
Because \(H(H(x)) = x\) for all \(x\neq 1\) and the range never hits 1, \(H\) acts as its own inverse on \(\mathbb{R}\setminus\{1\}\). Thus the compositions
\[
H \circ H = \mathrm{Id}
\quad\text{and}\quad
H^{-1} = H
\] hold, confirming that \(H\) and \(H^{-1}\) (as given) are indeed inverse functions.
Work Step by Step
We have a function
\[
H : \mathbb{R}\setminus\{1\} \;\to\; \mathbb{R}\setminus\{1\},
\quad
H(x) \;=\; \frac{x+1}{\,x-1\,}.
\]
The statement says \(H\) and \(H^{-1}\) are **both** given by that same formula, meaning \(H\) is its own inverse. To verify this, we must check that
\[
H\bigl(H(x)\bigr) \;=\; x
\quad
\text{for all } x \in \mathbb{R}\setminus\{1\},
\]
and also confirm that the values never map to 1 (so the expression remains in the codomain \(\mathbb{R}\setminus\{1\}\)).
---
## 1. Show \(H \circ H = \mathrm{Id}\)
Let \(y = H(x) = \frac{x+1}{x-1}\). Then
1. Compute \(y + 1\):
\[
y + 1
\;=\;
\frac{x+1}{\,x-1\,} \;+\; 1
\;=\;
\frac{x+1}{\,x-1\,} \;+\; \frac{x-1}{\,x-1\,}
\;=\;
\frac{(x+1) + (x-1)}{\,x-1\,}
\;=\;
\frac{2x}{\,x-1\,}.
\]
2. Compute \(y - 1\):
\[
y - 1
\;=\;
\frac{x+1}{\,x-1\,} \;-\; 1
\;=\;
\frac{x+1}{\,x-1\,} \;-\; \frac{x-1}{\,x-1\,}
\;=\;
\frac{(x+1) - (x-1)}{\,x-1\,}
\;=\;
\frac{2}{\,x-1\,}.
\]
Hence,
\[
H\bigl(H(x)\bigr)
\;=\;
H(y)
\;=\;
\frac{\,y+1\,}{\,y-1\,}
\;=\;
\frac{\tfrac{2x}{x-1}}{\tfrac{2}{x-1}}
\;=\;
\left(\frac{2x}{\,x-1\,}\right)
\Bigl/\!
\left(\frac{2}{\,x-1\,}\right)
\;=\;
\frac{2x}{\,x-1\,}\;\times\;\frac{x-1}{2}
\;=\;
x.
\]
Thus \(\bigl(H \circ H\bigr)(x) = x\). This shows \(H\) is indeed its own inverse on the domain \(\mathbb{R}\setminus\{1\}\).
---
## 2. Confirm the Codomain Condition
We also need to check that \(H(x)\neq 1\) for all \(x\neq 1\). Suppose for contradiction
\[
\frac{x+1}{\,x-1\,} = 1.
\]
Cross‐multiplying,
\[
x+1 = x-1
\quad\Longrightarrow\quad
1 = -1,
\]
a contradiction. So \(H(x)\neq 1\). Hence \(H(x)\) indeed stays in \(\mathbb{R}\setminus\{1\}\).
