Let `f (x) = (x+1)^2 - 1,(x ge - 1)` then the set S={x:f (x) = `f^(-1)` (x)} is

Options

A. `{0,-1,(-3 +- 2 sqrt(3))/(2)}`
B. `(0,1,-1)`
C. ` {0,-1}`
D. empty

Answer

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

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

1. Start with the equation \( y = f(x) = (x + 1)^2 - 1 \).
2. Rearrange this to express \( x \) in terms of \( y \):
\[ y + 1 = (x + 1)^2 \]
\[ \sqrt{y + 1} = x + 1 \quad \text{(taking the positive root since } x \geq -1\text{)} \]
\[ x = \sqrt{y + 1} - 1 \]
3. Thus, the inverse function is:
\[ f^{-1}(x) = \sqrt{x + 1} - 1 \]

### Step 2: Set up the equation \( f(x) = f^{-1}(x) \)

1. We need to solve:
\[ (x + 1)^2 - 1 = \sqrt{x + 1} - 1 \]
2. Simplifying gives:
\[ (x + 1)^2 = \sqrt{x + 1} \]

### Step 3: Square both sides to eliminate the square root

1. Squaring both sides:
\[ ((x + 1)^2)^2 = x + 1 \]
\[ (x + 1)^4 = x + 1 \]

### Step 4: Rearranging the equation

1. Rearranging gives:
\[ (x + 1)^4 - (x + 1) = 0 \]
2. Factor out \( (x + 1) \):
\[ (x + 1)((x + 1)^3 - 1) = 0 \]

### Step 5: Solve the factors

1. From \( (x + 1) = 0 \):
\[ x = -1 \]
2. From \( (x + 1)^3 - 1 = 0 \):
\[ (x + 1)^3 = 1 \]
\[ x + 1 = 1 \implies x = 0 \]

### Step 6: Check for additional solutions

1. The cubic equation \( (x + 1)^3 - 1 = 0 \) can also be solved by:
\[ (x + 1) = 1 \implies x = 0 \]
The cubic can also be factored or solved using the Rational Root Theorem, but we already found the relevant roots.

### Step 7: Compile the set \( S \)

1. The solutions we found are \( x = -1 \) and \( x = 0 \).
2. Thus, the set \( S \) is:
\[ S = \{-1, 0\} \]

### Final Answer

The set \( S = \{-1, 0\} \).

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