Let `f(x)` be real valued continuous function on `R` defined as `f(x)=x^(2)e^(-|x|)` then `f(x)` is increasing in

Options

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

Explanation

To determine the intervals where the function \( f(x) = x^2 e^{-|x|} \) is increasing, we need to find its derivative \( f'(x) \) and analyze where this derivative is positive.

### Step 1: Define the function based on the absolute value
The function \( f(x) \) can be expressed differently depending on whether \( x \) is non-negative or negative:
- For \( x \geq 0 \): \( f(x) = x^2 e^{-x} \)
- For \( x < 0 \): \( f(x) = x^2 e^{x} \)

### Step 2: Differentiate \( f(x) \)
We will differentiate \( f(x) \) in both cases.

**Case 1: \( x \geq 0 \)**
Using the product rule:
\[ f'(x) = \frac{d}{dx}(x^2) \cdot e^{-x} + x^2 \cdot \frac{d}{dx}(e^{-x}) \]
\[ = 2x e^{-x} - x^2 e^{-x} \]
\[ = e^{-x}(2x - x^2) \]

**Case 2: \( x < 0 \)**
Again using the product rule:
\[ f'(x) = \frac{d}{dx}(x^2) \cdot e^{x} + x^2 \cdot \frac{d}{dx}(e^{x}) \]
\[ = 2x e^{x} + x^2 e^{x} \]
\[ = e^{x}(2x + x^2) \]

### Step 3: Find where \( f'(x) > 0 \)
We need to determine where \( f'(x) \) is positive in both cases.

**For \( x \geq 0 \):**
\[ f'(x) = e^{-x}(2x - x^2) > 0 \]
Since \( e^{-x} > 0 \) for all \( x \), we focus on the quadratic:
\[ 2x - x^2 > 0 \]
Factoring gives:
\[ -x(x - 2) > 0 \]
This inequality holds when \( 0 < x < 2 \).

**For \( x < 0 \):**
\[ f'(x) = e^{x}(2x + x^2) > 0 \]
Again, since \( e^{x} > 0 \) for all \( x \), we focus on:
\[ 2x + x^2 > 0 \]
Factoring gives:
\[ x(x + 2) > 0 \]
This inequality holds when \( x < -2 \).

### Step 4: Combine the intervals
From our analysis, we find:
- \( f(x) \) is increasing for \( x \in (0, 2) \)
- \( f(x) \) is increasing for \( x \in (-\infty, -2) \)

### Final Answer
Thus, the function \( f(x) \) is increasing in the intervals:
\[ (-\infty, -2) \cup (0, 2) \]