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

Explanation

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

### Step 1: Analyze the logarithmic part
The logarithmic function \( \log(2-x) \) is defined only when its argument is positive, and it cannot be equal to zero since it is in the denominator of the fraction.

1. **Condition for the logarithm to be defined:**
\[ 2 - x > 0 \implies x < 2 \]

2. **Condition for the logarithm to be non-zero:**
\[ \log(2 - x) \neq 0 \implies 2 - x \neq 1 \implies x \neq 1 \]

### Step 2: Analyze the square root part
The square root function \( \sqrt{x+1} \) is defined when the expression inside the square root is non-negative.

3. **Condition for the square root to be defined:**
\[ x + 1 \geq 0 \implies x \geq -1 \]

### Step 3: Combine the conditions
Now we have three conditions to satisfy:
1. \( x < 2 \)
2. \( x \neq 1 \)
3. \( x \geq -1 \)

### Step 4: Determine the domain
To find the domain, we combine these conditions:

- From \( x \geq -1 \) and \( x < 2 \), we can say \( -1 \leq x < 2 \).
- However, since \( x \neq 1 \), we need to exclude 1 from our interval.

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

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

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