Let `f (x) = (x +1)^(2) -1 ( x ge -1).` Then the number of elements in the set `S = {x :f (x) = f ^(-1)(x)}` is

Answer

To solve the problem, we need to find the number of elements in the set \( S = \{ x : f(x) = f^{-1}(x) \} \) for the function \( f(x) = (x + 1)^2 - 1 \) where \( x \ge -1 \).

### Step 1: Find the function \( f(x) \)

The function is given by:

\[ f(x) = (x + 1)^2 - 1 \]

Expanding this, we have:

\[ f(x) = (x + 1)(x + 1) - 1 = x^2 + 2x + 1 - 1 = x^2 + 2x \]

### Step 2: Find the inverse function \( f^{-1}(x) \)

To find the inverse, we set \( y = f(x) \):

\[ y = x^2 + 2x \]

Rearranging gives us:

\[ x^2 + 2x - y = 0 \]

This is a quadratic equation in \( x \). We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 1, b = 2, c = -y \):

\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-y)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 4y}}{2} = \frac{-2 \pm 2\sqrt{1 + y}}{2} = -1 \pm \sqrt{1 + y} \]

Since \( x \ge -1 \), we take the positive root:

\[ f^{-1}(x) = -1 + \sqrt{1 + x} \]

### Step 3: Set up the equation \( f(x) = f^{-1}(x) \)

Now we set \( f(x) \) equal to \( f^{-1}(x) \):

\[ x^2 + 2x = -1 + \sqrt{1 + x} \]

### Step 4: Rearrange the equation

Rearranging gives us:

\[ x^2 + 2x + 1 = \sqrt{1 + x} \]

This simplifies to:

\[ (x + 1)^2 = \sqrt{1 + x} \]

### Step 5: Square both sides

Squaring both sides results in:

\[ (x + 1)^4 = 1 + x \]

### Step 6: Expand and rearrange

Expanding \( (x + 1)^4 \):

\[ (x + 1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1 \]

Setting this equal to \( 1 + x \):

\[ x^4 + 4x^3 + 6x^2 + 4x + 1 = 1 + x \]

Subtracting \( 1 + x \) from both sides:

\[ x^4 + 4x^3 + 6x^2 + 3x = 0 \]

### Step 7: Factor the equation

Factoring out \( x \):

\[ x(x^3 + 4x^2 + 6x + 3) = 0 \]

This gives us one solution \( x = 0 \). Now we need to solve the cubic equation:

\[ x^3 + 4x^2 + 6x + 3 = 0 \]

### Step 8: Find the roots of the cubic equation

Using the Rational Root Theorem or synthetic division, we can find the roots. Testing \( x = -1 \):

\[ (-1)^3 + 4(-1)^2 + 6(-1) + 3 = -1 + 4 - 6 + 3 = 0 \]

Thus, \( x + 1 \) is a factor. Dividing \( x^3 + 4x^2 + 6x + 3 \) by \( x + 1 \) gives:

\[ x^3 + 4x^2 + 6x + 3 = (x + 1)(x^2 + 3x + 3) \]

### Step 9: Solve the quadratic equation

Now we solve \( x^2 + 3x + 3 = 0 \) using the quadratic formula:

\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 - 12}}{2} = \frac{-3 \pm \sqrt{-3}}{2} \]

This gives us complex roots. Thus, the only real solutions are from \( x = 0 \) and \( x = -1 \).

### Step 10: Count the solutions

The real solutions are \( x = 0 \) and \( x = -1 \). Therefore, the number of elements in the set \( S \) is:

\[ \text{Number of elements in } S = 2 \]

### Final Answer

The number of elements in the set \( S \) is \( \boxed{2} \).