The difference of maximum and minimum values of `f(x) = x^(2) e^(-x)` is

Options

A. e
B. `1//e`
C. `1- (1)/ (e )`
D. `1 + (1)/(e )`

Answer

To find the difference between the maximum and minimum values of the function \( f(x) = x^2 e^{-x} \), we will follow these steps:

### Step 1: Differentiate the function
We need to find the critical points by differentiating \( f(x) \).

\[ f'(x) = \frac{d}{dx}(x^2 e^{-x}) \]

Using the product rule, where \( u = x^2 \) and \( v = e^{-x} \):

\[ f'(x) = u'v + uv' = (2x)e^{-x} + (x^2)(-e^{-x}) = e^{-x}(2x - x^2) \]

### Step 2: Set the derivative to zero
To find the critical points, we set \( f'(x) = 0 \):

\[ e^{-x}(2x - x^2) = 0 \]

Since \( e^{-x} \) is never zero, we have:

\[ 2x - x^2 = 0 \]

Factoring gives:

\[ x(2 - x) = 0 \]

Thus, the critical points are:

\[ x = 0 \quad \text{and} \quad x = 2 \]

### Step 3: Determine the nature of the critical points
We need to find the second derivative \( f''(x) \) to determine whether these points are maxima or minima.

First, we differentiate \( f'(x) \):

\[ f''(x) = \frac{d}{dx}(e^{-x}(2x - x^2)) \]

Using the product rule again:

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

Combining terms:

\[ f''(x) = e^{-x}(-2x + x^2 + 2 - 2x) = e^{-x}(x^2 - 4x + 2) \]

Now we evaluate \( f''(x) \) at the critical points.

1. **At \( x = 0 \)**:
\[ f''(0) = e^{0}(0^2 - 4 \cdot 0 + 2) = 2 \quad (\text{positive, so minimum}) \]

2. **At \( x = 2 \)**:
\[ f''(2) = e^{-2}(2^2 - 4 \cdot 2 + 2) = e^{-2}(4 - 8 + 2) = e^{-2}(-2) \quad (\text{negative, so maximum}) \]

### Step 4: Calculate the maximum and minimum values
Now we calculate the values of \( f(x) \) at these critical points:

1. **Maximum at \( x = 2 \)**:
\[ f(2) = 2^2 e^{-2} = 4 e^{-2} \]

2. **Minimum at \( x = 0 \)**:
\[ f(0) = 0^2 e^{0} = 0 \]

### Step 5: Find the difference
The difference between the maximum and minimum values is:

\[ \text{Difference} = f(2) - f(0) = 4 e^{-2} - 0 = 4 e^{-2} \]

### Final Answer
The difference of the maximum and minimum values of \( f(x) = x^2 e^{-x} \) is:

\[ \boxed{4 e^{-2}} \]