Let f be defined by:
`f(x) = sqrt(x-ln(1+x))`. The domain of f is

Options

A. `(-1, infty)`
B. `[0,infty)`
C. `[1,infty)`
D. `(-infty,infty)`

Answer

To find the domain of the function \( f(x) = \sqrt{x - \ln(1 + x)} \), we need to ensure that the expression inside the square root is non-negative, as the square root function is only defined for non-negative values. Therefore, we need to solve the inequality:

\[ x - \ln(1 + x) \geq 0 \]

### Step 1: Set up the inequality
We start with the inequality:

\[ x - \ln(1 + x) \geq 0 \]

### Step 2: Rearrange the inequality
Rearranging gives us:

\[ x \geq \ln(1 + x) \]

### Step 3: Analyze the function
Let's define a new function \( g(x) = x - \ln(1 + x) \). We need to find where \( g(x) \geq 0 \).

### Step 4: Find the critical points
To analyze \( g(x) \), we can find its derivative:

\[ g'(x) = 1 - \frac{1}{1 + x} \]

Setting the derivative to zero to find critical points:

\[ 1 - \frac{1}{1 + x} = 0 \implies 1 + x = 1 \implies x = 0 \]

### Step 5: Test intervals around the critical point
We will test intervals around \( x = 0 \):

- For \( x < 0 \) (e.g., \( x = -1 \)):
\[ g(-1) = -1 - \ln(0) \quad \text{(undefined)} \]

- For \( x = 0 \):
\[ g(0) = 0 - \ln(1) = 0 \]

- For \( x > 0 \) (e.g., \( x = 1 \)):
\[ g(1) = 1 - \ln(2) \quad \text{(which is positive since } \ln(2) < 1\text{)} \]

### Step 6: Determine the behavior of \( g(x) \)
Since \( g'(x) > 0 \) for \( x > 0 \), \( g(x) \) is increasing for \( x > 0 \). Thus, \( g(x) \) will be non-negative for all \( x \geq 0 \).

### Step 7: Consider the lower bound
We also need to ensure that \( \ln(1 + x) \) is defined, which requires:

\[ 1 + x > 0 \implies x > -1 \]

### Step 8: Combine the conditions
Thus, combining both conditions, we find that:

\[ x \geq 0 \quad \text{and} \quad x > -1 \]

The intersection of these conditions gives us:

\[ x \in [0, \infty) \]

### Conclusion
The domain of the function \( f(x) = \sqrt{x - \ln(1 + x)} \) is:

\[ \text{Domain of } f = [0, \infty) \]