The interval in which the function f given by `f(x) = xe^(-x)` is strictly increasing, is:

Options

A. `(-infty, infty)`
B. (0,1)
C. `(-infty, 1)`
D. `(1,infty)`

Answer

To determine the interval in which the function \( f(x) = xe^{-x} \) is strictly increasing, we need to follow these steps:

### Step 1: Find the derivative of the function
To find where the function is increasing, we first need to compute the derivative \( f'(x) \).

Using the product rule, we differentiate \( f(x) = x \cdot e^{-x} \):
\[ f'(x) = \frac{d}{dx}(x) \cdot e^{-x} + x \cdot \frac{d}{dx}(e^{-x}) \]
\[ f'(x) = 1 \cdot e^{-x} + x \cdot (-e^{-x}) \]
\[ f'(x) = e^{-x} - xe^{-x} \]
\[ f'(x) = e^{-x}(1 - x) \]

### Step 2: Determine where the derivative is positive
The function \( f(x) \) is increasing where \( f'(x) > 0 \):
\[ e^{-x}(1 - x) > 0 \]
Since \( e^{-x} > 0 \) for all \( x \), we can focus on the inequality:
\[ 1 - x > 0 \]
This simplifies to:
\[ x < 1 \]

### Step 3: Identify the interval
Since \( x < 1 \) indicates that the function is increasing for all values of \( x \) less than 1, we can express this as:
\[ (-\infty, 1) \]

### Conclusion
Thus, the interval in which the function \( f(x) = xe^{-x} \) is strictly increasing is:
\[ \boxed{(-\infty, 1)} \]