On which of the following intervals is the function `f(x)=2x^2 - log |x| , x ne 0` increasing ?

Options

A. `(1/2,oo)`
B. `(-oo ,-1/2 ) uu (1/2 ,oo)`
C. ` (-oo , -1/2 ) uu (0,1/2)`
D. ` (-1/2 ,0)uu (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 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 \( f'(x) \).

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

Using the power rule and the derivative of the logarithm, we get:

\[ f'(x) = 4x - \frac{1}{x} \]

### Step 2: Set the derivative greater than or equal to zero
The function is increasing where \( f'(x) \geq 0 \):

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

### Step 3: Solve the inequality
To solve the inequality, we can rearrange it:

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

This can be factored as:

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

### Step 4: Find the critical points
Setting each factor to zero gives us the critical points:

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

### Step 5: Analyze the intervals
The critical points divide the number line into intervals. We will test the sign of \( f'(x) \) in each interval:

1. \( (-\infty, -\frac{1}{2}) \)
2. \( (-\frac{1}{2}, 0) \)
3. \( (0, \frac{1}{2}) \)
4. \( (\frac{1}{2}, \infty) \)

### Step 6: Test each interval
- **Interval 1: \( (-\infty, -\frac{1}{2}) \)**
Choose \( x = -1 \):
\[ f'(-1) = 4(-1) - \frac{1}{-1} = -4 + 1 = -3 \quad (\text{Not increasing}) \]

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

- **Interval 3: \( (0, \frac{1}{2}) \)**
Choose \( x = \frac{1}{4} \):
\[ f'(\frac{1}{4}) = 4(\frac{1}{4}) - \frac{1}{\frac{1}{4}} = 1 - 4 = -3 \quad (\text{Not increasing}) \]

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

### Step 7: Conclusion
The function \( f(x) \) is increasing on the intervals \( (-\frac{1}{2}, 0) \) and \( (\frac{1}{2}, \infty) \).

### Final Answer
The intervals on which the function is increasing are:
- \( (-\frac{1}{2}, 0) \)
- \( (\frac{1}{2}, \infty) \)