Find the maximum value of the function `x * e^(-x)`.

Answer

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

### Step 1: Differentiate the function
We start by finding the derivative \( f'(x) \) using the product rule. The product rule states that if you have two functions \( u \) and \( v \), then the derivative of their product is given by:
\[ (uv)' = u'v + uv' \]
Here, we let \( u = x \) and \( v = e^{-x} \).

Calculating the derivatives:
- \( u' = 1 \)
- \( v' = \frac{d}{dx}(e^{-x}) = -e^{-x} \)

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

### Step 2: Set the derivative 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 for any real \( x \), we only need to solve:
\[ 1 - x = 0 \implies x = 1 \]

### Step 3: Determine if it is a maximum
To confirm that this critical point is a maximum, we can use the second derivative test or analyze the sign of the first derivative around \( x = 1 \).

Calculating the second derivative \( f''(x) \):
\[ f'(x) = e^{-x}(1 - x) \]
Using the product rule again:
\[ f''(x) = \frac{d}{dx}(e^{-x})(1 - x) + e^{-x}\frac{d}{dx}(1 - x) \]
Calculating the derivatives:
- The derivative of \( e^{-x} \) is \( -e^{-x} \).
- The derivative of \( 1 - x \) is \( -1 \).

So,
\[ f''(x) = -e^{-x}(1 - x) + e^{-x}(-1) = -e^{-x}(1 - x + 1) = -e^{-x}(2 - x) \]

Now substituting \( x = 1 \):
\[ f''(1) = -e^{-1}(2 - 1) = -e^{-1} < 0 \]
Since \( f''(1) < 0 \), this indicates that \( x = 1 \) is a local maximum.

### Step 4: Find the maximum value
Now we find the maximum value by substituting \( x = 1 \) back into the original function:
\[ f(1) = 1 \cdot e^{-1} = \frac{1}{e} \]

### Conclusion
The maximum value of the function \( f(x) = x e^{-x} \) is:
\[ \boxed{\frac{1}{e}} \]