The function `f(x) = x^(2) e^(-x)` strictly increases on

Options

A. [0,2]
B. `[0, infty]`
C. `(- infty, 0] cup [ 2, infty)`
D. none of these

Answer

To determine the intervals on which the function \( f(x) = x^2 e^{-x} \) is strictly increasing, we will follow these steps:

### Step 1: Differentiate the function
We start by finding the derivative of \( f(x) \).

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

Using the product rule, we have:

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

Calculating the derivatives:

\[ \frac{d}{dx}(e^{-x}) = -e^{-x} \quad \text{and} \quad \frac{d}{dx}(x^2) = 2x \]

Thus, substituting these into the product rule gives:

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

This simplifies to:

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

### Step 2: Factor the derivative
We can factor out \( e^{-x} \):

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

This can be rewritten as:

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

### Step 3: Set the derivative to zero
To find the critical points, we set \( f'(x) = 0 \):

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

Since \( e^{-x} \) is never zero for any real \( x \), we focus on the other factors:

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

This gives us:

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

### Step 4: Analyze the sign of the derivative
Next, we will analyze the sign of \( f'(x) \) in the intervals determined by the critical points \( x = 0 \) and \( x = 2 \). The intervals to test are \( (-\infty, 0) \), \( (0, 2) \), and \( (2, \infty) \).

1. **Interval \( (-\infty, 0) \)**:
- Choose \( x = -1 \):
\[ f'(-1) = e^{1}(-1)(2 + 1) = e^{1}(-1)(3) < 0 \]
(Decreasing)

2. **Interval \( (0, 2) \)**:
- Choose \( x = 1 \):
\[ f'(1) = e^{-1}(1)(2 - 1) = e^{-1}(1)(1) > 0 \]
(Increasing)

3. **Interval \( (2, \infty) \)**:
- Choose \( x = 3 \):
\[ f'(3) = e^{-3}(3)(2 - 3) = e^{-3}(3)(-1) < 0 \]
(Decreasing)

### Conclusion
From the analysis, we find that the function \( f(x) = x^2 e^{-x} \) is strictly increasing on the interval \( (0, 2) \).

### Final Answer
The function \( f(x) = x^2 e^{-x} \) is strictly increasing on the interval \( (0, 2) \).