The function `f(x) = x^(2)/e^(x)` is monotonically increasing if

Options

A. Only `x lt 0`
B. Only `x gt 2`
C. `0 lt x lt 2`
D. `x in (-infty, 0) cup (2, infty)`

Answer

To determine when the function \( f(x) = \frac{x^2}{e^x} \) is monotonically increasing, we need to find the derivative of the function and analyze its sign.

### Step-by-Step Solution:

1. **Find the Derivative**:
We start by applying the quotient rule to find \( f'(x) \). The quotient rule states that if you have a function \( f(x) = \frac{g(x)}{h(x)} \), then:
\[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} \]
Here, \( g(x) = x^2 \) and \( h(x) = e^x \).

- Calculate \( g'(x) \):
\[ g'(x) = 2x \]
- Calculate \( h'(x) \):
\[ h'(x) = e^x \]

Now applying the quotient rule:
\[ f'(x) = \frac{(2x)(e^x) - (x^2)(e^x)}{(e^x)^2} \]

Simplifying the numerator:
\[ f'(x) = \frac{e^x(2x - x^2)}{e^{2x}} = \frac{2x - x^2}{e^x} \]

2. **Determine When \( f'(x) > 0 \)**:
For the function to be monotonically increasing, we need:
\[ f'(x) > 0 \implies \frac{2x - x^2}{e^x} > 0 \]
Since \( e^x > 0 \) for all \( x \), we only need to consider the numerator:
\[ 2x - x^2 > 0 \]

3. **Solve the Inequality**:
Rearranging gives:
\[ x(2 - x) > 0 \]
This product is positive when both factors are either positive or both are negative.

- The roots of the equation \( x(2 - x) = 0 \) are \( x = 0 \) and \( x = 2 \).
- The intervals to test are \( (-\infty, 0) \), \( (0, 2) \), and \( (2, \infty) \).

Testing these intervals:
- For \( x < 0 \) (e.g., \( x = -1 \)): \( -1(2 - (-1)) = -1(3) < 0 \) (not increasing)
- For \( 0 < x < 2 \) (e.g., \( x = 1 \)): \( 1(2 - 1) = 1(1) > 0 \) (increasing)
- For \( x > 2 \) (e.g., \( x = 3 \)): \( 3(2 - 3) = 3(-1) < 0 \) (not increasing)

Therefore, \( f'(x) > 0 \) when:
\[ 0 < x < 2 \]

### Conclusion:
The function \( f(x) = \frac{x^2}{e^x} \) is monotonically increasing for \( 0 < x < 2 \).