Suppose `f(x)=(x+1)^2forxgeq-1.` If `g(x)` is the function whose graph is the reflection of the graph of `f(x)` with respect to the line `y=x ,` then `g(x)` equal. `a-sqrt(x)-1,xgeq0` (b) `1/((x+1)^2),x >-1` `sqrt(x+1,)xgeq-1` (d) `sqrt(x)-1,xgeq0`

Options

A. `1-sqrt(x)-1, x ge 0`
B. `(1)/((x+1)^(2)),x gt -1 `
C. `sqrt(x+1), x ge -1`
D. `sqrt(x)-1, x ge 0`

Explanation

To solve the problem, we need to find the function \( g(x) \) which is the reflection of the function \( f(x) = (x + 1)^2 \) with respect to the line \( y = x \). This means that \( g(x) \) is the inverse of \( f(x) \).

### Step-by-Step Solution:

1. **Identify the function**: We have \( f(x) = (x + 1)^2 \) for \( x \geq -1 \).

2. **Find the range of \( f(x) \)**:
- Since \( f(x) = (x + 1)^2 \) is a square function, it is always non-negative.
- The minimum value occurs at \( x = -1 \):
\[ f(-1) = (-1 + 1)^2 = 0. \]
- Therefore, the range of \( f(x) \) is \( [0, \infty) \).

3. **Set \( y = f(x) \)**:
\[ y = (x + 1)^2. \]

4. **Solve for \( x \) in terms of \( y \)**:
- Rearranging gives:
\[ x + 1 = \sqrt{y} \quad \text{(since \( y \geq 0 \))}. \]
- Therefore:
\[ x = \sqrt{y} - 1. \]

5. **Express \( g(x) \)**:
- Since \( g(x) \) is the inverse of \( f(x) \), we have:
\[ g(x) = \sqrt{x} - 1 \quad \text{for } x \geq 0. \]

6. **Conclusion**:
- The function \( g(x) \) is:
\[ g(x) = \sqrt{x} - 1 \quad \text{for } x \geq 0. \]
- Thus, the correct option is (d) \( \sqrt{x} - 1, x \geq 0 \).