The function `f(x)=x e^(1-x)` stricly

Options

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

Answer

To determine whether the function \( f(x) = x e^{1-x} \) is increasing or decreasing, we will follow these steps:

### Step 1: Rewrite the Function
We can rewrite the function as:
\[ f(x) = \frac{x}{e^x} \cdot e \]
This form will help us in differentiating the function easily.

### Step 2: Find the Derivative
To find the nature of the function, we need to compute its derivative \( f'(x) \). We will use the quotient rule for differentiation, which states:
\[ \frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Here, let \( u = x \) and \( v = e^x \).

Calculating the derivatives:
- \( \frac{du}{dx} = 1 \)
- \( \frac{dv}{dx} = e^x \)

Now applying the quotient rule:
\[ f'(x) = \frac{e^x \cdot 1 - x \cdot e^x}{(e^x)^2} \]
This simplifies to:
\[ f'(x) = \frac{e^x - x e^x}{e^{2x}} = \frac{e^x(1 - x)}{e^{2x}} = \frac{1 - x}{e^x} \]

### Step 3: Analyze the Sign of the Derivative
Now, we need to determine where \( f'(x) \) is positive or negative. The expression \( \frac{1 - x}{e^x} \) is positive when \( 1 - x > 0 \) and negative when \( 1 - x < 0 \).

- \( 1 - x > 0 \) implies \( x < 1 \)
- \( 1 - x < 0 \) implies \( x > 1 \)

Since \( e^x \) is always positive for all \( x \), the sign of \( f'(x) \) is determined solely by \( 1 - x \).

### Step 4: Conclusion on Increasing/Decreasing Intervals
From our analysis:
- \( f'(x) > 0 \) when \( x < 1 \) (function is increasing)
- \( f'(x) < 0 \) when \( x > 1 \) (function is decreasing)

At \( x = 1 \), \( f'(1) = 0 \), indicating a critical point.

### Final Answer
Thus, the function \( f(x) = x e^{1-x} \) is:
- **Increasing** on the interval \( (-\infty, 1) \)
- **Decreasing** on the interval \( (1, \infty) \)