On which of the following intervals in the function `f(x)=2x^2-log|x|,xne0` increasing ?

Options

A. `(1/2,oo)`
B. `(-oo,-1//2) cup (0,1//2)`
C. `(-oo,-1//2)cup (1//2,oo)`
D. `(-1//2,0)cup(1//2,oo)`

Answer

To determine the intervals on which the function \( f(x) = 2x^2 - \log|x| \) (where \( x \neq 0 \)) is increasing, we will 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 \( f'(x) \).

\[ f'(x) = \frac{d}{dx}(2x^2) - \frac{d}{dx}(\log|x|) \]
\[ = 4x - \frac{1}{x} \]

### Step 2: Set the derivative greater than zero
Next, we set the derivative greater than zero to find the intervals where the function is increasing.

\[ 4x - \frac{1}{x} > 0 \]

### Step 3: Multiply through by \( x \) (considering \( x \neq 0 \))
To eliminate the fraction, we multiply through by \( x \). We need to consider the sign of \( x \) when doing this.

\[ 4x^2 - 1 > 0 \]

### Step 4: Factor the quadratic inequality
Now, we can factor the quadratic expression:

\[ (2x - 1)(2x + 1) > 0 \]

### Step 5: Identify critical points
The critical points from the factors are:

\[ 2x - 1 = 0 \Rightarrow x = \frac{1}{2} \]
\[ 2x + 1 = 0 \Rightarrow x = -\frac{1}{2} \]

### Step 6: Test intervals around the critical points
We will test the intervals determined by the critical points \( -\frac{1}{2} \) and \( \frac{1}{2} \):

1. **Interval \( (-\infty, -\frac{1}{2}) \)**: Choose \( x = -1 \)
\[ f'(-1) = 4(-1) - \frac{1}{-1} = -4 + 1 = -3 \quad (\text{negative}) \]

2. **Interval \( (-\frac{1}{2}, 0) \)**: Choose \( x = -\frac{1}{4} \)
\[ f'(-\frac{1}{4}) = 4(-\frac{1}{4}) - \frac{1}{-\frac{1}{4}} = -1 + 4 = 3 \quad (\text{positive}) \]

3. **Interval \( (0, \frac{1}{2}) \)**: Choose \( x = \frac{1}{4} \)
\[ f'(\frac{1}{4}) = 4(\frac{1}{4}) - \frac{1}{\frac{1}{4}} = 1 - 4 = -3 \quad (\text{negative}) \]

4. **Interval \( (\frac{1}{2}, \infty) \)**: Choose \( x = 1 \)
\[ f'(1) = 4(1) - \frac{1}{1} = 4 - 1 = 3 \quad (\text{positive}) \]

### Step 7: Summarize the intervals where \( f'(x) > 0 \)
From our tests, we find that \( f'(x) > 0 \) in the intervals:

- \( (-\frac{1}{2}, 0) \)
- \( (\frac{1}{2}, \infty) \)

### Final Result
Thus, the function \( f(x) = 2x^2 - \log|x| \) is increasing on the intervals:

\[ (-\frac{1}{2}, 0) \cup (\frac{1}{2}, \infty) \]