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

Options

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

Answer

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 the derivative \( \frac{dy}{dx} \).

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

Differentiating with respect to \( x \):

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

### Step 2: Set the derivative greater than zero
For the function to be increasing, the derivative must be greater than zero:

\[ 2x + e^{-x} > 0 \]

### Step 3: Analyze the derivative
Since \( e^{-x} \) is always positive for all real \( x \), we can analyze the inequality:

\[ 2x + e^{-x} > 0 \]

This will always hold true because \( e^{-x} \) is positive and \( 2x \) can be negative or positive depending on the value of \( x \).

### Step 4: Find critical points
To find critical points, we can set the derivative equal to zero:

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

However, since \( e^{-x} \) is always positive, \( 2x \) cannot equal \(-e^{-x}\) for any real \( x \). Therefore, there are no critical points where the derivative is zero.

### Step 5: Determine the intervals
Since \( e^{-x} \) is always positive, \( 2x + e^{-x} > 0 \) for all \( x \). Thus, the function \( y = x^2 - e^{-x} \) is increasing for all values of \( x \).

### Conclusion
The interval in which \( y = x^2 - e^{-x} \) is increasing is:

\[ (-\infty, \infty) \]