The function f(x) =` xe^(1-x)`

Options

A. strictly increases in the interval (1/2, 2)
B. increases in the interval `(0, oo)`
C. decreases in the interval (0, 2)
D. strictly decreases in the interval `(1, oo)`

Answer

To determine whether the function \( f(x) = x e^{1-x} \) is strictly increasing or decreasing, we need to find its derivative \( f'(x) \) and analyze its sign.

### Step 1: Differentiate the function
We will use the product rule to differentiate \( f(x) \). The product rule states that if you have two functions \( u(x) \) and \( v(x) \), then the derivative \( (uv)' = u'v + uv' \).

Let:
- \( u = x \) and \( v = e^{1-x} \)

Now, we find the derivatives:
- \( u' = 1 \)
- To find \( v' \), we will use the chain rule. The derivative of \( e^{g(x)} \) is \( e^{g(x)} g'(x) \), where \( g(x) = 1 - x \).
- Thus, \( g'(x) = -1 \)
- Therefore, \( v' = e^{1-x} \cdot (-1) = -e^{1-x} \)

Now applying the product rule:
\[ f'(x) = u'v + uv' = 1 \cdot e^{1-x} + x \cdot (-e^{1-x}) = e^{1-x} - x e^{1-x} \]
Factoring out \( e^{1-x} \):
\[ f'(x) = e^{1-x}(1 - x) \]

### Step 2: Analyze the sign of the derivative
To determine where \( f(x) \) is increasing or decreasing, we need to analyze the sign of \( f'(x) \).

1. **Find where \( f'(x) = 0 \)**:
\[ e^{1-x}(1 - x) = 0 \]
Since \( e^{1-x} \) is never zero, we set \( 1 - x = 0 \):
\[ x = 1 \]

2. **Test intervals around \( x = 1 \)**:
- For \( x < 1 \) (e.g., \( x = 0 \)):
\[ f'(0) = e^{1-0}(1 - 0) = e > 0 \quad \text{(function is increasing)} \]
- For \( x > 1 \) (e.g., \( x = 2 \)):
\[ f'(2) = e^{1-2}(1 - 2) = e^{-1}(-1) < 0 \quad \text{(function is decreasing)} \]

### Conclusion
- The function \( f(x) \) is strictly increasing for \( x < 1 \) and strictly decreasing for \( x > 1 \).
- Therefore, \( f(x) \) has a maximum at \( x = 1 \).

### Final Answer
- The function \( f(x) = x e^{1-x} \) is strictly increasing on the interval \( (-\infty, 1) \) and strictly decreasing on the interval \( (1, \infty) \).