Are the following functions invertible in their respective domains ? If so, find the inverse in each case :
(i) f(x) = x + 1
(ii) `f(x)=(x-1)/(x+1),xne-1`.

Answer

To determine if the functions are invertible and to find their inverses, we will analyze each function step by step.

### Function (i): \( f(x) = x + 1 \)

1. **Check if the function is one-to-one (injective)**:
- A function is one-to-one if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).
- Assume \( f(x_1) = f(x_2) \):
\[ x_1 + 1 = x_2 + 1 \]
Subtracting 1 from both sides gives:
\[ x_1 = x_2 \]
- Thus, \( f(x) = x + 1 \) is one-to-one.

2. **Check if the function is onto (surjective)**:
- The function maps all real numbers \( x \) to \( y \) in \( \mathbb{R} \) since for any \( y \in \mathbb{R} \), we can find \( x = y - 1 \) such that \( f(x) = y \).
- Therefore, \( f(x) \) is onto.

3. **Conclusion**:
- Since \( f(x) \) is both one-to-one and onto, it is invertible.

4. **Find the inverse**:
- To find the inverse, we replace \( f(x) \) with \( y \):
\[ y = x + 1 \]
- Solve for \( x \):
\[ x = y - 1 \]
- Thus, the inverse function is:
\[ f^{-1}(x) = x - 1 \]

### Function (ii): \( f(x) = \frac{x - 1}{x + 1}, x \neq -1 \)

1. **Check the domain**:
- The function is defined for all \( x \in \mathbb{R} \) except \( x = -1 \).

2. **Check if the function is one-to-one (injective)**:
- Assume \( f(x_1) = f(x_2) \):
\[ \frac{x_1 - 1}{x_1 + 1} = \frac{x_2 - 1}{x_2 + 1} \]
- Cross-multiplying gives:
\[ (x_1 - 1)(x_2 + 1) = (x_2 - 1)(x_1 + 1) \]
- Expanding both sides:
\[ x_1 x_2 + x_1 - x_2 - 1 = x_1 x_2 + x_2 - x_1 - 1 \]
- Simplifying leads to:
\[ 2x_1 = 2x_2 \implies x_1 = x_2 \]
- Thus, \( f(x) \) is one-to-one.

3. **Check if the function is onto (surjective)**:
- To check if \( f(x) \) is onto, we need to find if for every \( y \in \mathbb{R} \), there exists an \( x \) such that:
\[ y = \frac{x - 1}{x + 1} \]
- Rearranging gives:
\[ y(x + 1) = x - 1 \implies yx + y = x - 1 \implies x - yx = y - 1 \implies x(1 - y) = y - 1 \]
- Thus:
\[ x = \frac{y - 1}{1 - y} \quad \text{(as long as \( y \neq 1 \))} \]
- Therefore, \( f(x) \) is onto.

4. **Conclusion**:
- Since \( f(x) \) is both one-to-one and onto, it is invertible.

5. **Find the inverse**:
- We have:
\[ y = \frac{x - 1}{x + 1} \]
- Rearranging gives:
\[ y(x + 1) = x - 1 \implies yx + y = x - 1 \implies x - yx = y - 1 \implies x(1 - y) = y - 1 \]
- Thus:
\[ x = \frac{y - 1}{1 - y} \]
- Therefore, the inverse function is:
\[ f^{-1}(x) = \frac{x - 1}{1 - x}, \quad x \neq 1 \]

### Summary of Results:
- For \( f(x) = x + 1 \):
- Invertible: Yes
- Inverse: \( f^{-1}(x) = x - 1 \)

- For \( f(x) = \frac{x - 1}{x + 1}, x \neq -1 \):
- Invertible: Yes
- Inverse: \( f^{-1}(x) = \frac{x - 1}{1 - x}, x \neq 1 \)