Determine the intervals in which the following functions are strictly increasing or strictly decreasing :
`f(x)=(4x^(2)+1)/(x).`

Answer

To determine the intervals in which the function \( f(x) = \frac{4x^2 + 1}{x} \) is strictly increasing or strictly decreasing, we will follow these steps:

### Step 1: Find the derivative of the function
To analyze the increasing and decreasing behavior of the function, we first need to find its derivative \( f'(x) \).

\[ f(x) = \frac{4x^2 + 1}{x} = 4x + \frac{1}{x} \]

Now, we differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx}(4x) + \frac{d}{dx}\left(\frac{1}{x}\right) = 4 - \frac{1}{x^2} \]

### Step 2: Determine where the derivative is positive or negative
The function \( f(x) \) is strictly increasing where \( f'(x) > 0 \) and strictly decreasing where \( f'(x) < 0 \).

Set the derivative greater than zero:

\[ 4 - \frac{1}{x^2} > 0 \]

Rearranging gives:

\[ 4 > \frac{1}{x^2} \]

This can be rewritten as:

\[ x^2 > \frac{1}{4} \]

Taking the square root of both sides, we find:

\[ |x| > \frac{1}{2} \]

This leads to two intervals:

\[ x < -\frac{1}{2} \quad \text{or} \quad x > \frac{1}{2} \]

### Step 3: Determine where the derivative is negative
Now, we find where the derivative is less than zero:

\[ 4 - \frac{1}{x^2} < 0 \]

Rearranging gives:

\[ 4 < \frac{1}{x^2} \]

This can be rewritten as:

\[ x^2 < \frac{1}{4} \]

Taking the square root of both sides, we find:

\[ |x| < \frac{1}{2} \]

This leads to the interval:

\[ -\frac{1}{2} < x < \frac{1}{2} \]

### Step 4: Summary of intervals
From the analysis, we conclude:

- The function \( f(x) \) is **strictly increasing** on the intervals \( (-\infty, -\frac{1}{2}) \) and \( (\frac{1}{2}, \infty) \).
- The function \( f(x) \) is **strictly decreasing** on the interval \( (-\frac{1}{2}, \frac{1}{2}) \).

### Final Answer
- **Increasing**: \( (-\infty, -\frac{1}{2}) \cup (\frac{1}{2}, \infty) \)
- **Decreasing**: \( (-\frac{1}{2}, \frac{1}{2}) \)