find the interval(s) in which the function `f(x)= (4x^(2)+1)/(x)`, is decreasing.

Answer

To find the interval(s) in which the function \( f(x) = \frac{4x^2 + 1}{x} \) is decreasing, we will follow these steps:

### Step 1: Rewrite the function
We can simplify the function for easier differentiation:
\[ f(x) = \frac{4x^2 + 1}{x} = 4x + \frac{1}{x} \]

### Step 2: Differentiate the function
Next, we need to find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(4x) + \frac{d}{dx}\left(\frac{1}{x}\right) \]
Using the power rule and the derivative of \( \frac{1}{x} \):
\[ f'(x) = 4 - \frac{1}{x^2} \]

### Step 3: Set the derivative less than zero
To find where the function is decreasing, we set the derivative less than zero:
\[ 4 - \frac{1}{x^2} < 0 \]

### Step 4: Solve the inequality
Rearranging the inequality gives:
\[ 4 < \frac{1}{x^2} \]
Taking the reciprocal (and remembering to flip the inequality) gives:
\[ \frac{1}{4} > x^2 \]
This implies:
\[ x^2 < \frac{1}{4} \]

### Step 5: Take the square root
Taking the square root of both sides gives:
\[ |x| < \frac{1}{2} \]
This means:
\[ -\frac{1}{2} < x < \frac{1}{2} \]

### Step 6: Conclusion
Thus, the function \( f(x) \) is decreasing in the interval:
\[ \left(-\frac{1}{2}, \frac{1}{2}\right) \]