Find the domain of each of the following functions: `f(x)=1/(log_(10)(1-x))+sqrt(x+2)`

Explanation

To find the domain of the function \( f(x) = \frac{1}{\log_{10}(1-x)} + \sqrt{x+2} \), we need to analyze the conditions under which each component of the function is defined.

### Step 1: Analyze the square root function
The term \( \sqrt{x + 2} \) requires that the expression inside the square root is non-negative:
\[ x + 2 \geq 0 \]
This simplifies to:
\[ x \geq -2 \]
Thus, the first condition for the domain is:
\[ x \in [-2, \infty) \]

### Step 2: Analyze the logarithmic function
Next, we consider the term \( \log_{10}(1 - x) \). The logarithm is defined only for positive arguments:
\[ 1 - x > 0 \]
This simplifies to:
\[ x < 1 \]
Thus, the second condition for the domain is:
\[ x \in (-\infty, 1) \]

### Step 3: Ensure the logarithm does not equal zero
Since \( f(x) \) has a term \( \frac{1}{\log_{10}(1-x)} \), we need to ensure that the logarithm is not equal to zero, as this would make the function undefined:
\[ \log_{10}(1 - x) \neq 0 \]
This occurs when:
\[ 1 - x \neq 1 \quad \Rightarrow \quad x \neq 0 \]

### Step 4: Combine the conditions
Now we need to combine all the conditions:
1. From the square root: \( x \in [-2, \infty) \)
2. From the logarithm: \( x \in (-\infty, 1) \)
3. From the logarithm not being zero: \( x \neq 0 \)

The intersection of these intervals gives us:
- The interval from the square root is \( [-2, \infty) \).
- The interval from the logarithm is \( (-\infty, 1) \).
- Excluding \( x = 0 \).

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

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