Let `f(x) = log (1-x) +sqrt(x^(2)-1). `Then dom (f )=?

Options

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

Explanation

To find the domain of the function \( f(x) = \log(1-x) + \sqrt{x^2 - 1} \), we need to determine the conditions under which both components of the function are defined.

### Step 1: Analyze the logarithmic part \( \log(1-x) \)

The logarithmic function \( \log(1-x) \) is defined when its argument is positive:
\[ 1 - x > 0 \]
This simplifies to:
\[ x < 1 \]

### Step 2: Analyze the square root part \( \sqrt{x^2 - 1} \)

The square root function \( \sqrt{x^2 - 1} \) is defined when its argument is non-negative:
\[ x^2 - 1 \geq 0 \]
This can be factored as:
\[ (x - 1)(x + 1) \geq 0 \]
To solve this inequality, we find the critical points:
- \( x = -1 \)
- \( x = 1 \)

Now we test the intervals determined by these critical points:
1. **Interval \( (-\infty, -1) \)**: Choose \( x = -2 \):
\[ (-2 - 1)(-2 + 1) = (-3)(-1) = 3 \quad (\text{positive}) \]
2. **Interval \( (-1, 1) \)**: Choose \( x = 0 \):
\[ (0 - 1)(0 + 1) = (-1)(1) = -1 \quad (\text{negative}) \]
3. **Interval \( (1, \infty) \)**: Choose \( x = 2 \):
\[ (2 - 1)(2 + 1) = (1)(3) = 3 \quad (\text{positive}) \]

From this analysis, the solution to the inequality \( (x - 1)(x + 1) \geq 0 \) is:
\[ x \leq -1 \quad \text{or} \quad x \geq 1 \]

### Step 3: Combine the conditions

Now we combine the conditions from both parts:
1. From \( \log(1-x) \): \( x < 1 \)
2. From \( \sqrt{x^2 - 1} \): \( x \leq -1 \) or \( x \geq 1 \)

The intersection of these conditions is:
- For \( x \leq -1 \): This satisfies \( x < 1 \).
- For \( x \geq 1 \): This does not satisfy \( x < 1 \).

Thus, the only valid condition is:
\[ x \leq -1 \]

### Conclusion

The domain of the function \( f(x) \) is:
\[ \text{dom}(f) = (-\infty, -1] \]