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Let's work through this problem to fill in the missing steps. The function given is:

\[
f : (1, \infty) \to (1, \infty), \quad f(x) = \frac{x}{x - 1}.
\]

The task involves proving that this function is **one-to-one** (injective) and **onto** (surjective).

---

### Part (a) Proving Injectivity (One-to-One)

To show that \( f \) is one-to-one, assume \( f(a) = f(b) \) for some \( a, b \in (1, \infty) \) and show that this implies \( a = b \).

1. We start with:
   \[
   f(a) = f(b) \Rightarrow \frac{a}{a - 1} = \frac{b}{b - 1}.
   \]

2. Cross-multiplying to clear the denominators, we get:
   \[
   a(b - 1) = b(a - 1).
   \]

3. Expanding both sides:
   \[
   ab - a = ba - b.
   \]

4. Subtracting \( ab \) from both sides:
   \[
   -a = -b.
   \]

5. Multiplying both sides by \(-1\):
   \[
   a = b.
   \]

Therefore, \( f \) is one-to-one.

---

### Part (b) Proving Surjectivity (Onto)

To show that \( f \) is onto, let \( b \) be an element in the codomain \( (1, \infty) \). We need to show that there exists an \( a \) in the domain \( (1, \infty) \) such that \( f(a) = b \).

1. Suppose \( f(a) = b \), then:
   \[
   \frac{a}{a - 1} = b.
   \]

2. Multiply both sides by \( a - 1 \):
   \[
   a = b(a - 1).
   \]

3. Rearrange to isolate terms involving \( a \) on one side:
   \[
   a - ba = -b.
   \]

4. Factor out \( a \):
   \[
   a(1 - b) = -b.
   \]

5. Divide by \( 1 - b \) (note that \( b > 1 \), so \( 1 - b < 0 \)):
   \[
   a = \frac{b}{b - 1}.
   \]

Thus, for each \( b \in (1, \infty) \), there exists an \( a \in (1, \infty) \) such that \( f(a) = b \), proving that \( f \) is onto.

Since \( f \) is both one-to-one and onto, it is invertible.

---

### Finding the Inverse \( f^{-1}(x) \)

To find \( f^{-1}(x) \), set \( y = f(x) \) and solve for \( x \) in terms of \( y \):

1. Let \( y = \frac{x}{x - 1} \).
2. Multiply by \( x - 1 \):
   \[
   y(x - 1) = x.
   \]
3. Expand and rearrange to isolate \( x \):
   \[
   yx - y = x.
   \]
4. Collect \( x \)-terms on one side:
   \[
   x(y - 1) = y.
   \]
5. Solve for \( x \):
   \[
   x = \frac{y}{y - 1}.
   \]

Therefore, the inverse function is:
\[
f^{-1}(x) = \frac{x}{x - 1}.
\]

---

### Summary

1. **Injective**: \( f(a) = f(b) \Rightarrow a = b \).
2. **Surjective**: For every \( b \in (1, \infty) \), there exists \( a \in (1, \infty) \) such that \( f(a) = b \).
3. **Inverse function**: \( f^{-1}(x) = \frac{x}{x - 1} \).

---

Would you like further explanation on any of these steps, or do you have any specific questions?

---

### Additional Questions for Practice

1. What is the general method to prove a function is injective?
2. How can you verify that the inverse function \( f^{-1}(x) = \frac{x}{x - 1} \) is correct?
3. Why is the function \( f(x) = \frac{x}{x - 1} \) restricted to the domain \( (1, \infty) \)?
4. How would the proof of surjectivity change if the codomain were different?
5. Can you apply this method of finding the inverse to other rational functions?

### Tip

When proving a function is injective, always start by assuming \( f(a) = f(b) \) and try to derive \( a = b \). This approach is foundational in function analysis.

Prove that the function f(x) = x / (x - 1) is one-to-one and onto, where f: (1, ∞) → (1, ∞). Also, find the inverse of f.

Learn how to prove that the function f(x) = x / (x - 1) is injective and surjective over the domain (1, ∞) and codomain (1, ∞). Detailed steps are provided to show that f is one-to-one and onto, and how to find its inverse function.

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