The interval, in which `y=2x^(2)e^(-2x)` is increasing is :

Options

A. `(-oo, oo)`
B. `(-1, 0)`
C. `(1, oo)`
D. `(0, 1).`

Answer

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

### Step 1: Find the derivative of the function
We start with the function:
\[ y = f(x) = 2x^2 e^{-2x} \]
To find where the function is increasing, we need to compute the derivative \( f'(x) \).

Using the product rule, where \( u = 2x^2 \) and \( v = e^{-2x} \):
\[ f'(x) = u'v + uv' \]
Calculating \( u' \) and \( v' \):
- \( u' = \frac{d}{dx}(2x^2) = 4x \)
- \( v' = \frac{d}{dx}(e^{-2x}) = -2e^{-2x} \)

Now substituting these into the product rule:
\[ f'(x) = (4x)e^{-2x} + (2x^2)(-2e^{-2x}) \]
Simplifying this:
\[ f'(x) = 4x e^{-2x} - 4x^2 e^{-2x} \]
Factoring out the common terms:
\[ f'(x) = 4x e^{-2x}(1 - x) \]

### Step 2: Set the derivative greater than zero
To find the intervals where the function is increasing, we set the derivative greater than zero:
\[ 4x e^{-2x}(1 - x) > 0 \]
Since \( e^{-2x} \) is always positive for all \( x \), we can simplify our inequality to:
\[ 4x(1 - x) > 0 \]

### Step 3: Solve the inequality
Now we need to solve \( 4x(1 - x) > 0 \). This can be broken down into two factors:
1. \( 4x > 0 \) implies \( x > 0 \)
2. \( 1 - x > 0 \) implies \( x < 1 \)

Thus, the solution to the inequality is:
\[ 0 < x < 1 \]

### Step 4: Conclusion
The function \( y = 2x^2 e^{-2x} \) is increasing in the interval:
\[ (0, 1) \]