Consider the funtion \(f: \mathbf{R} \backslash\{-1\} \rightarrow \mathbf{R}\) defined by \(f(x)=\frac{2 x+1}{x+1}\). i) Check that \(f(x)\) is well defined and \(1-1\). ii) Check that \(f(x) \neq 2\) for any \(x \in \mathbf{R}\). iii) Check that \(g: \mathbf{R} \backslash\{2\} \rightarrow \mathbf{R}\) given by \(g(x)=\frac{x-1}{2-x}\) is well defined and \(1-1\). Further, check that \(g(x) \neq-1\) for any \(x \in \mathbf{R}\). iv) Check that \((f \circ g)(x)=x\) for \(x \in \mathbf{R} \backslash\{2\}\) and \((g \circ f)(x)=x\) for \(x \in \mathbf{R} \backslash\{-1\}\).

Let’s address each part of the question step by step.

### i) Checking if \(f(x)\) is well-defined and 1-1

#### Well-defined:
The function \(f(x) = \frac{2x + 1}{x + 1}\) is defined for all \(x \in \mathbf{R} \backslash\{-1\}\) because the denominator \(x + 1\) is nonzero for all \(x\) except \(-1\). Hence, \(f(x)\) is well-defined.

#### 1-1 (Injective):
To check if \(f(x)\) is 1-1, we need to show that if \(f(x_1) = f(x_2)\), then \(x_1 = x_2\). Assume \(f(x_1) = f(x_2)\):
\[
\frac{2x_1 + 1}{x_1 + 1} = \frac{2x_2 + 1}{x_2 + 1}
\]
Cross-multiplying gives:
\[
(2x_1 + 1)(x_2 + 1) = (2x_2 + 1)(x_1 + 1)
\]
Expanding both sides and simplifying will show if \(x_1 = x_2\). Let’s calculate this:

\[ (2x_1 + 1)(x_2 + 1) = (2x_2 + 1)(x_1 + 1) \]
\[ 2x_1x_2 + x_1 + 2x_2 + 1 = 2x_1x_2 + x_2 + 2x_1 + 1 \]
\[ x_1 + 2x_2 = x_2 + 2x_1 \]
\[ x_1 = x_2 \]

Thus, \(f(x)\) is 1-1.

### ii) Checking if \(f(x) \neq 2\) for any \(x \in \mathbf{R}\)

We need to show that \(f(x) = \frac{2x + 1}{x + 1} \neq 2\) for any \(x\). Assume for contradiction that \(f(x) = 2\):
\[
\frac{2x + 1}{x + 1} = 2
\]
Cross-multiplying gives:
\[
2x + 1 = 2x + 2
\]
Simplifying:
\[
1 = 2
\]
This is a contradiction. Therefore, \(f(x) \neq 2\) for any \(x \in \mathbf{R}\).

### iii) Checking if \(g(x)\) is well-defined and 1-1, and \(g(x) \neq -1\)

#### Well-defined:
The function \(g(x) = \frac{x – 1}{2 – x}\) is defined for all \(x \in \mathbf{R} \backslash\{2\}\) because the denominator \(2 – x\) is nonzero for all \(x\) except \(2\). Hence, \(g(x)\) is well-defined.

#### 1-1 (Injective):
To check if \(g(x)\) is 1-1, assume \(g(x_1) = g(x_2)\):
\[
\frac{x_1 – 1}{2 – x_1} = \frac{x_2 – 1}{2 – x_2}
\]
Cross-multiplying and simplifying as before will show if \(x_1 = x_2\).

#### \(g(x) \neq -1\):
Assume for contradiction that \(g(x) = -1\):
\[
\frac{x – 1}{2 – x} = -1
\]
Cross-multiplying gives:
\[
x – 1 = -2 + x
\]
Simplifying:
\[
1 = 2
\]
This is a contradiction. Therefore, \(g(x) \neq -1\) for any \(x \in \mathbf{R}\).

### iv) Checking \((f \circ g)(x) = x\) and \((g \circ f)(x) = x\)

#### \((f \circ g)(x) = x\):
We need to compute \(f(g(x))\) and show it equals \(x\). Let’s calculate \(f(g(x))\):

\[ f(g(x)) = f\left(\frac{x – 1}{2 – x}\right) = \frac{2\left(\frac{x – 1}{2 – x}\right) + 1}{\left(\frac{x – 1}{2 – x}\right) + 1} \]
\[ = \frac{\frac{2x – 2}{2 – x} + 1}{\frac{x – 1}{2 – x} + 1} \]
\[ = \frac{\frac{2x – 2 + 2 – x}{2 – x}}{\frac{x – 1 + 2 – x}{2 – x}} \]
\[ = \frac{x}{1} \]
\[ = x \]

#### \((g \circ f)(x) = x\):
Similarly, we need to compute \(g(f(x))\) and show it equals \(x\). Let’s calculate \(g(f(x))\):

\[ g(f(x)) = g\left(\frac{2x + 1}{x + 1}\right) = \frac{\left(\frac{2x + 1}{x + 1}\right) – 1}{2 – \left(\frac{2x + 1}{x + 1}\right)} \]
\[ = \frac{\frac{2x + 1 – (x + 1)}{x + 1}}{\frac{2(x + 1) – (2x + 1)}{x + 1}} \]
\[ = \frac{\frac{x}{x + 1}}{\frac{1}{x + 1}} \]
\[ = x \]

Thus, both \((f \circ g)(x) = x\) and \((g \circ f)(x) = x\) are true for their respective domains.