consider the function f(x) =`1(1+(1)/(x))^(x)`
The domain of f(x) is

Options

A. `(-1,0)cup(0,oo)`
B. R-{0}
C. `(-oo,-1)cup(0,oo)cup(0,oo)`
D. (0,oo)`

Explanation

To find the domain of the function \( f(x) = 1 \cdot \left( 1 + \frac{1}{x} \right)^x \), we need to ensure that the expression inside the function is defined and valid.

### Step-by-step Solution:

1. **Identify the function**:
We start with the function given:
\[ f(x) = \left( 1 + \frac{1}{x} \right)^x \]

2. **Determine the conditions for the function to be defined**:
The term \( \frac{1}{x} \) implies that \( x \) cannot be zero, as division by zero is undefined. Therefore, we have:
\[ x \neq 0 \]

3. **Analyze the term \( 1 + \frac{1}{x} \)**:
Next, we need to ensure that \( 1 + \frac{1}{x} > 0 \). This will help us understand the behavior of the function:
\[ 1 + \frac{1}{x} > 0 \]
Rearranging this gives:
\[ \frac{1}{x} > -1 \]
This can be rewritten as:
\[ 1 > -x \quad \text{or} \quad x > -1 \]

4. **Combine the conditions**:
From the conditions derived, we have:
- \( x \neq 0 \)
- \( x > -1 \)

The critical points to consider are \( x = -1 \) and \( x = 0 \).

5. **Test intervals**:
We will test the intervals determined by these critical points:
- For \( x < -1 \): Choose \( x = -2 \):
\[ 1 + \frac{1}{-2} = 1 - 0.5 = 0.5 > 0 \quad \text{(valid)} \]
- For \( -1 < x < 0 \): Choose \( x = -0.5 \):
\[ 1 + \frac{1}{-0.5} = 1 - 2 = -1 < 0 \quad \text{(invalid)} \]
- For \( x > 0 \): Choose \( x = 1 \):
\[ 1 + \frac{1}{1} = 2 > 0 \quad \text{(valid)} \]

6. **Conclusion**:
The valid intervals where the function is defined are:
- From \( -\infty \) to \( -1 \) (excluding -1)
- From \( 0 \) to \( +\infty \) (excluding 0)

Thus, the domain of the function \( f(x) \) is:
\[ (-\infty, -1) \cup (0, +\infty) \]