The greatest value of the funxtion `f(x)=xe^(-x) " in " [0,oo]` is

Options

A. 0
B. `1//e`
C. `-e`
D. e

Answer

To find the greatest value of the function \( f(x) = x e^{-x} \) in the interval \([0, \infty)\), we will follow these steps:

### Step 1: Find the derivative of the function
To locate the maximum value, we first need to find the derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(x e^{-x}) \]
Using the product rule, we have:
\[ f'(x) = e^{-x} + x \frac{d}{dx}(e^{-x}) = e^{-x} - x e^{-x} \]
Thus, we can simplify this to:
\[ f'(x) = e^{-x}(1 - x) \]

### Step 2: Set the derivative equal to zero
To find the critical points, we set the derivative equal to zero:
\[ e^{-x}(1 - x) = 0 \]
Since \( e^{-x} \) is never zero, we can set the other factor to zero:
\[ 1 - x = 0 \implies x = 1 \]

### Step 3: Determine if it is a maximum using the second derivative test
Next, we find the second derivative \( f''(x) \):
\[ f''(x) = \frac{d}{dx}(e^{-x}(1 - x)) \]
Using the product rule again:
\[ f''(x) = \frac{d}{dx}(e^{-x}) \cdot (1 - x) + e^{-x} \cdot \frac{d}{dx}(1 - x) \]
Calculating this gives:
\[ f''(x) = -e^{-x}(1 - x) - e^{-x} \]
Simplifying:
\[ f''(x) = -e^{-x}(1 - x + 1) = -e^{-x}(2 - x) \]
Now, we evaluate the second derivative at \( x = 1 \):
\[ f''(1) = -e^{-1}(2 - 1) = -e^{-1} \]
Since \( f''(1) < 0 \), this indicates that \( x = 1 \) is a point of local maximum.

### Step 4: Calculate the maximum value
Now, we find the maximum value of the function at \( x = 1 \):
\[ f(1) = 1 \cdot e^{-1} = e^{-1} \]
This can also be expressed as:
\[ f(1) = \frac{1}{e} \]

### Conclusion
Thus, the greatest value of the function \( f(x) = x e^{-x} \) in the interval \([0, \infty)\) is:
\[ \frac{1}{e} \]