Let `y = f (x)` such that `xy = x+y +1, x in R-{1} and g (x) =x f (x)`
The minimum value of `g (x)` is:

Options

A. A) `3- sqrt2`
B. B) ` 3+sqrt2`
C. C) ` 3-2 sqrt2`
D. D) `3+2 sqrt2`

Answer

To find the minimum value of \( g(x) = x f(x) \) given the equation \( xy = x + y + 1 \), we will follow these steps:

### Step 1: Express \( y \) in terms of \( x \)
We start with the equation:
\[ xy = x + y + 1 \]
Rearranging gives:
\[ xy - y = x + 1 \]
Factoring out \( y \) from the left side:
\[ y(x - 1) = x + 1 \]
Thus, we can express \( y \) as:
\[ y = \frac{x + 1}{x - 1} \]
Since \( y = f(x) \), we have:
\[ f(x) = \frac{x + 1}{x - 1} \]

### Step 2: Substitute \( f(x) \) into \( g(x) \)
Now, we substitute \( f(x) \) into \( g(x) \):
\[ g(x) = x f(x) = x \cdot \frac{x + 1}{x - 1} = \frac{x(x + 1)}{x - 1} = \frac{x^2 + x}{x - 1} \]

### Step 3: Differentiate \( g(x) \)
To find the minimum value, we need to differentiate \( g(x) \):
Using the quotient rule \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \), where \( u = x^2 + x \) and \( v = x - 1 \):
- \( \frac{du}{dx} = 2x + 1 \)
- \( \frac{dv}{dx} = 1 \)

Thus, we have:
\[ g'(x) = \frac{(x - 1)(2x + 1) - (x^2 + x)(1)}{(x - 1)^2} \]

### Step 4: Simplify the derivative
Now, simplifying the numerator:
\[ g'(x) = \frac{(x - 1)(2x + 1) - (x^2 + x)}{(x - 1)^2} \]
Expanding the first term:
\[ = \frac{(2x^2 + x - 2x - 1) - (x^2 + x)}{(x - 1)^2} \]
\[ = \frac{2x^2 - x - 1 - x^2 - x}{(x - 1)^2} \]
\[ = \frac{x^2 - 2x - 1}{(x - 1)^2} \]

### Step 5: Set the derivative to zero
To find critical points, set the numerator equal to zero:
\[ x^2 - 2x - 1 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2} \]

### Step 6: Evaluate \( g(x) \) at critical points
We have two critical points: \( x = 1 - \sqrt{2} \) and \( x = 1 + \sqrt{2} \). We will evaluate \( g(x) \) at \( x = 1 + \sqrt{2} \) (as \( x = 1 - \sqrt{2} \) is less than 1 and not in the domain).

Calculating \( g(1 + \sqrt{2}) \):
\[ g(1 + \sqrt{2}) = \frac{(1 + \sqrt{2})^2 + (1 + \sqrt{2})}{(1 + \sqrt{2}) - 1} \]
Calculating \( (1 + \sqrt{2})^2 \):
\[ = 1 + 2 + 2\sqrt{2} = 3 + 2\sqrt{2} \]
Thus,
\[ g(1 + \sqrt{2}) = \frac{(3 + 2\sqrt{2}) + (1 + \sqrt{2})}{\sqrt{2}} = \frac{4 + 3\sqrt{2}}{\sqrt{2}} = 2\sqrt{2} + 3 \]

### Final Answer
The minimum value of \( g(x) \) is:
\[ \boxed{3 + 2\sqrt{2}} \]