The domain of difinition of the function
`y=1/(log_10(1-x))+sqrt(x+2)`

Options

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

Answer

To find the domain of the function \( y = \frac{1}{\log_{10}(1-x)} + \sqrt{x+2} \), we need to ensure that both components of the function are defined.

### Step 1: Analyze the logarithmic function
The term \( \log_{10}(1-x) \) is defined only when its argument is positive:
\[ 1 - x > 0 \implies x < 1 \]
Additionally, the logarithm cannot be zero, as it would make the denominator undefined:
\[ \log_{10}(1-x) \neq 0 \implies 1 - x \neq 1 \implies x \neq 0 \]
So, from the logarithmic part, we have:
\[ x < 1 \quad \text{and} \quad x \neq 0 \]

### Step 2: Analyze the square root function
The term \( \sqrt{x+2} \) is defined when its argument is non-negative:
\[ x + 2 \geq 0 \implies x \geq -2 \]

### Step 3: Combine the conditions
Now we combine the conditions from both parts:
1. From the logarithmic function: \( x < 1 \) and \( x \neq 0 \)
2. From the square root function: \( x \geq -2 \)

### Step 4: Determine the overall domain
Combining these inequalities, we find:
- The lower bound is \( x \geq -2 \).
- The upper bound is \( x < 1 \), excluding \( x = 0 \).

Thus, the domain of the function can be expressed in interval notation as:
\[ [-2, 0) \cup (0, 1) \]

### Final Answer
The domain of the function \( y = \frac{1}{\log_{10}(1-x)} + \sqrt{x+2} \) is:
\[ [-2, 0) \cup (0, 1) \]

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