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

Question

Accepted Answer

To determine the set $ S = \{ x : f(x) = f^{-1}(x) \} $ for the function $ f(x) = (x+1)^2 - 1 $ with the domain $ x \ge -1 $, we need to find the inverse function $ f^{-1}(x) $ and solve the equation $ f(x) = f^{-1}(x) $.

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

The function $ f(x) = (x+1)^2 - 1 $ can be rewritten as $ y = (x+1)^2 - 1 $. To find the inverse, solve for $ x $ in terms of $ y $:

1. Add 1 to both sides: $ y + 1 = (x+1)^2 $.
2. Take the square root: $ \sqrt{y + 1} = x + 1 $.
3. Solve for $ x $: $ x = \sqrt{y + 1} - 1 $.

Thus, the inverse function is $ f^{-1}(x) = \sqrt{x + 1} - 1 $.

### Step 2: Solve $ f(x) = f^{-1}(x) $

Set the original function equal to its inverse:

$$
(x+1)^2 - 1 = \sqrt{x + 1} - 1
$$

Simplify and solve:

1. Add 1 to both sides: $(x+1)^2 = \sqrt{x + 1}$.
2. Square both sides to eliminate the square root: $(x+1)^4 = x + 1$.

Now, solve the equation $(x+1)^4 = x + 1$:

- Substitute $ z = x + 1 $, then $ z^4 = z $.
- This simplifies to $ z^4 - z = 0 $.
- Factor: $ z(z^3 - 1) = 0 $.
- Solve: $ z = 0 $ or $ z^3 = 1 $.

For $ z = 0 $, $ x + 1 = 0 $ gives $ x = -1 $.

For $ z^3 = 1 $, the solutions are $ z = 1, \frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2} $.

- $ z = 1 $ gives $ x + 1 = 1 $, so $ x = 0 $.

The complex solutions $ \frac{-1 + i\sqrt{3}}{2} $ and $ \frac{-1 - i\sqrt{3}}{2} $ do not yield real $ x $ values within the domain $ x \ge -1 $.

### Conclusion

The real solutions within the domain $ x \ge -1 $ are $ x = 0 $ and $ x = -1 $. Therefore, the set $ S = \{0, -1\} $.

### Explanation of Options

- **Option 1 (empty)**: Incorrect because we found real solutions $ x = 0 $ and $ x = -1 $.
- **Option 2 (\{0, -1\})**: Correct as derived above.
- **Option 3 (\{0, 1, -1\})**: Incorrect because $ x = 1 $ is not a solution to the equation $ f(x) = f^{-1}(x) $.
- **Option 4 (\{0, -1, \frac{-3+i\sqrt{3}}{2}, \frac{-3-i\sqrt{3}}{2}\})**: Incorrect because the complex numbers are not within the domain $ x \ge -1 $.

Thus, the correct answer is Option 2: $\{0, -1\}$.