The true set of real values of `x` for which the function `f(x)=xlnx-x+1` is positive is

Options

A. `(1,oo)`
B. `(1//e,oo)`
C. `[e,oo)`
D. `(0,1)cup(1,oo)`

Explanation

To determine the set of real values of \( x \) for which the function \( f(x) = x \ln x - x + 1 \) is positive, we will follow these steps:

### Step 1: Set up the inequality
We need to find when \( f(x) > 0 \):
\[ f(x) = x \ln x - x + 1 > 0 \]

### Step 2: Differentiate the function
To analyze the function, we first find its derivative:
\[ f'(x) = \frac{d}{dx}(x \ln x) - \frac{d}{dx}(x) + \frac{d}{dx}(1) \]
Using the product rule for the term \( x \ln x \):
\[ f'(x) = \ln x + x \cdot \frac{1}{x} - 1 + 0 = \ln x + 1 - 1 = \ln x \]

### Step 3: Analyze the derivative
We need to find where \( f'(x) > 0 \):
\[ \ln x > 0 \]
This implies:
\[ x > e^0 = 1 \]

### Step 4: Determine the intervals
Since \( f'(x) > 0 \) for \( x > 1 \), the function \( f(x) \) is increasing in the interval \( (1, \infty) \).

### Step 5: Evaluate the function at the boundary
Next, we check the value of \( f(x) \) at \( x = 1 \):
\[ f(1) = 1 \ln 1 - 1 + 1 = 0 \]
Since \( f(1) = 0 \) and \( f(x) \) is increasing for \( x > 1 \), we conclude that \( f(x) > 0 \) for \( x > 1 \).

### Conclusion
The true set of real values of \( x \) for which \( f(x) > 0 \) is:
\[ \boxed{(1, \infty)} \]