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

Options

A. neither a maximum nor a minimum at x= 1
B. a minimum at x=1
C. a maximum at x= 1
D. a maximum at `x = -1`

Answer

To find the maximum and minimum values of the function \( f(x) = x e^{-x} \), we will follow these steps:

### Step 1: Find the derivative of the function
We need to differentiate \( f(x) \) with respect to \( x \).

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

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

### Step 3: Determine the nature of the critical point
We will use the first derivative test to determine whether this critical point is a maximum or minimum. We will check the sign of \( f'(x) \) around \( x = 1 \).

- For \( x < 1 \) (e.g., \( x = 0 \)):
\[ f'(0) = e^{0}(1 - 0) = 1 > 0 \quad \text{(increasing)} \]

- For \( x > 1 \) (e.g., \( x = 2 \)):
\[ f'(2) = e^{-2}(1 - 2) = e^{-2}(-1) < 0 \quad \text{(decreasing)} \]

### Step 4: Conclusion about the critical point
Since \( f'(x) \) changes from positive to negative at \( x = 1 \), this indicates that \( x = 1 \) is a local maximum.

### Step 5: Find the maximum value
Now we will calculate the maximum value of \( f(x) \) at \( x = 1 \):
\[ f(1) = 1 \cdot e^{-1} = \frac{1}{e} \]

### Final Result
The function \( f(x) = x e^{-x} \) has a maximum at \( x = 1 \) with a maximum value of \( \frac{1}{e} \).

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