Find domain of the function `f(x)=1/(log_10 (1-x)) +sqrt(x+2)`

Options

A. `[-2,1) `
B. `[-2,oo)`
C. `(-oo,1)`
D. `[-2,0) cup (0,1)`

Explanation

To find the domain of the function \( f(x) = \frac{1}{\log_{10}(1-x)} + \sqrt{x+2} \), we need to determine the values of \( x \) for which both components of the function are defined.

### Step 1: Analyze the first component \( g(x) = \frac{1}{\log_{10}(1-x)} \)

1. **Logarithm Condition**: The logarithm is defined only for positive arguments. Therefore, we need:
\[ 1 - x > 0 \implies x < 1 \]

2. **Denominator Condition**: The logarithm cannot be zero because it is in the denominator. Thus, we need:
\[ \log_{10}(1-x) \neq 0 \implies 1 - x \neq 1 \implies x \neq 0 \]

Combining these conditions, we find that the domain of \( g(x) \) is:
\[ (-\infty, 1) \setminus \{0\} \]

### Step 2: Analyze the second component \( h(x) = \sqrt{x+2} \)

1. **Square Root Condition**: The expression inside the square root must be non-negative:
\[ x + 2 \geq 0 \implies x \geq -2 \]

Thus, the domain of \( h(x) \) is:
\[ [-2, \infty) \]

### Step 3: Find the intersection of the domains

Now, we need to find the intersection of the domains of \( g(x) \) and \( h(x) \):
- Domain of \( g(x) \): \( (-\infty, 1) \setminus \{0\} \)
- Domain of \( h(x) \): \( [-2, \infty) \)

The intersection can be found as follows:
1. The interval \( (-\infty, 1) \) intersects with \( [-2, \infty) \) as \( [-2, 1) \).
2. We must exclude \( 0 \) from this interval.

Thus, the intersection is:
\[ [-2, 0) \cup (0, 1) \]

### Conclusion

The domain of the function \( f(x) \) is:
\[ [-2, 0) \cup (0, 1) \]