The interval in which `y=x^2""e^(-x)` is increasing is (A) `(-oo,""""oo)` (B) ( 2, 0) (C) `(2,""""oo)` (D) (0, 2)

Options

A. `(-infty, infty)`
B. `(-2, 0)`
C. `(2, infty)`
D. `(0, 2)`

Explanation

To determine the interval in which the function \( y = x^2 e^{-x} \) is increasing, we need to follow these steps:

### Step 1: Find the derivative of the function
To find where the function is increasing, we first need to compute its derivative \( y' \).

Given:
\[ y = x^2 e^{-x} \]

Using the product rule, we have:
\[ y' = \frac{d}{dx}(x^2) \cdot e^{-x} + x^2 \cdot \frac{d}{dx}(e^{-x}) \]

Calculating the derivatives:
\[ \frac{d}{dx}(x^2) = 2x \]
\[ \frac{d}{dx}(e^{-x}) = -e^{-x} \]

Thus,
\[ y' = 2x e^{-x} + x^2 (-e^{-x}) = 2x e^{-x} - x^2 e^{-x} \]

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

### Step 3: Set the derivative greater than zero
For the function to be increasing, we need:
\[ y' > 0 \]
This leads to:
\[ e^{-x} (2x - x^2) > 0 \]

Since \( e^{-x} > 0 \) for all \( x \), we only need to solve:
\[ 2x - x^2 > 0 \]

### Step 4: Solve the inequality
Rearranging gives:
\[ -x^2 + 2x > 0 \]
or
\[ x(2 - x) > 0 \]

### Step 5: Find the critical points
The critical points occur when:
\[ x = 0 \quad \text{or} \quad x = 2 \]

### Step 6: Test intervals
We will test the intervals determined by the critical points \( ( -\infty, 0 ) \), \( ( 0, 2 ) \), and \( ( 2, \infty ) \):

1. **Interval \( (-\infty, 0) \)**: Choose \( x = -1 \):
\[ y' = (-1)(2 - (-1)) = -1(3) < 0 \quad \text{(decreasing)} \]

2. **Interval \( (0, 2) \)**: Choose \( x = 1 \):
\[ y' = (1)(2 - 1) = 1(1) > 0 \quad \text{(increasing)} \]

3. **Interval \( (2, \infty) \)**: Choose \( x = 3 \):
\[ y' = (3)(2 - 3) = 3(-1) < 0 \quad \text{(decreasing)} \]

### Conclusion
The function \( y = x^2 e^{-x} \) is increasing in the interval \( (0, 2) \).

Thus, the answer is:
**(D) \( (0, 2) \)**