The function f, defined by `f(x) = (x^2)/(2) + "ln" x - 2 cos x` increase for `x epsilon`

Options

A. `R^-`
B. `R^+`
C. `R-{0}`
D. `[1,oo)`

Explanation

To determine the intervals where the function \( f(x) = \frac{x^2}{2} + \ln x - 2 \cos x \) is increasing, we need to follow these steps:

### Step 1: Find the first derivative of the function
We start by differentiating \( f(x) \) with respect to \( x \):

\[ f'(x) = \frac{d}{dx}\left(\frac{x^2}{2}\right) + \frac{d}{dx}(\ln x) - \frac{d}{dx}(2 \cos x) \]

Calculating each term:
- The derivative of \( \frac{x^2}{2} \) is \( x \).
- The derivative of \( \ln x \) is \( \frac{1}{x} \).
- The derivative of \( -2 \cos x \) is \( 2 \sin x \).

Thus, we have:

\[ f'(x) = x + \frac{1}{x} + 2 \sin x \]

### Step 2: Set the first derivative greater than zero
To find where the function is increasing, we need to solve the inequality:

\[ f'(x) > 0 \]

This translates to:

\[ x + \frac{1}{x} + 2 \sin x > 0 \]

### Step 3: Analyze the inequality
We can rearrange the inequality to:

\[ x + \frac{1}{x} + 2 \sin x \geq 0 \]

### Step 4: Consider the cases for \( x \)
1. **Case 1**: When \( x > 0 \)

Since \( x > 0 \), both \( x \) and \( \frac{1}{x} \) are positive. Thus, \( x + \frac{1}{x} > 0 \). The term \( 2 \sin x \) can vary between -2 and 2. Therefore, we need to ensure that:

\[ x + \frac{1}{x} + 2 \sin x > 0 \]

This will always hold true for \( x > 0 \) since \( x + \frac{1}{x} \) is always positive and dominates the behavior of the function.

2. **Case 2**: When \( x < 0 \)

Here, \( x + \frac{1}{x} \) can be negative, and \( 2 \sin x \) will also vary. However, since we are looking for increasing intervals, we will focus on \( x > 0 \).

### Step 5: Conclusion
From the analysis, we conclude that the function \( f(x) \) is increasing for:

\[ x > 0 \]

### Final Answer
The function \( f(x) \) is increasing for \( x \in (0, \infty) \).

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