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 ensure that all parts of the function are defined. This involves checking the conditions for the logarithm and the square root.

### Step 1: Conditions for the logarithm
The logarithm \( \log(2-x) \) is defined only when its argument is positive, and it cannot be equal to zero. Therefore, we have two conditions:

1. **Condition 1**: \( 2 - x > 0 \)
- This simplifies to \( x < 2 \).

2. **Condition 2**: \( \log(2-x) \neq 0 \)
- This means \( 2 - x \neq 1 \), which simplifies to \( x \neq 1 \).

### Step 2: Conditions for the square root
The square root \( \sqrt{x+1} \) is defined when its argument is non-negative:

3. **Condition 3**: \( x + 1 \geq 0 \)
- This simplifies to \( x \geq -1 \).

### Step 3: Combine the conditions
Now we need to combine all the conditions we found:

- From Condition 1: \( x < 2 \)
- From Condition 2: \( x \neq 1 \)
- From Condition 3: \( x \geq -1 \)

### Step 4: Determine the intersection
The valid values of \( x \) must satisfy all three conditions.

- The range from Condition 3 gives us \( x \geq -1 \).
- Condition 1 restricts \( x \) to be less than 2.
- Condition 2 excludes \( x = 1 \).

Thus, we can express the domain as:
- From \( -1 \) to \( 2 \), excluding \( 1 \).

In interval notation, the domain of the function is:
\[ [-1, 1) \cup (1, 2) \]

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