Find the range of the following functions: (where {.} and [.] represent fractional part and greatest integer part functions respectively)
`f(x)=1/(sqrt(16-4(x^(2))-x))`

Explanation

To find the range of the function \( f(x) = \frac{1}{\sqrt{16 - 4x^2 - x}} \), we will follow these steps:

### Step 1: Determine the domain of the function
The function \( f(x) \) is defined when the denominator is greater than zero:
\[ 16 - 4x^2 - x > 0 \]
Rearranging this gives:
\[ 4x^2 + x - 16 < 0 \]

### Step 2: Solve the quadratic inequality
To solve \( 4x^2 + x - 16 < 0 \), we first find the roots of the equation \( 4x^2 + x - 16 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 4 \), \( b = 1 \), and \( c = -16 \):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 4 \cdot (-16)}}{2 \cdot 4} \]
\[ x = \frac{-1 \pm \sqrt{1 + 256}}{8} \]
\[ x = \frac{-1 \pm \sqrt{257}}{8} \]

### Step 3: Identify the intervals of the quadratic
The roots are:
\[ x_1 = \frac{-1 - \sqrt{257}}{8}, \quad x_2 = \frac{-1 + \sqrt{257}}{8} \]
To find the intervals where \( 4x^2 + x - 16 < 0 \), we can test values in the intervals determined by these roots. The function is a parabola opening upwards, so it will be negative between the roots:
\[ x \in \left( \frac{-1 - \sqrt{257}}{8}, \frac{-1 + \sqrt{257}}{8} \right) \]

### Step 4: Find the minimum and maximum values of \( f(x) \)
To find the range of \( f(x) \), we need to evaluate \( f(x) \) at the endpoints of the interval and check if it has a minimum or maximum within the interval.

#### Step 4.1: Find critical points
We differentiate \( f(x) \) to find critical points:
\[ f'(x) = -\frac{1}{2} (16 - 4x^2 - x)^{-3/2} \cdot (-8x - 1) \]
Setting \( f'(x) = 0 \) gives:
\[ -8x - 1 = 0 \implies x = -\frac{1}{8} \]

#### Step 4.2: Evaluate \( f(x) \) at critical points and endpoints
1. Evaluate \( f\left(-\frac{1}{8}\right) \):
\[ f\left(-\frac{1}{8}\right) = \frac{1}{\sqrt{16 - 4\left(-\frac{1}{8}\right)^2 - \left(-\frac{1}{8}\right)}} \]
Simplifying gives:
\[ f\left(-\frac{1}{8}\right) = \frac{1}{\sqrt{16 - \frac{1}{16} + \frac{1}{8}}} = \frac{1}{\sqrt{16 - \frac{1}{16} + \frac{2}{16}}} = \frac{1}{\sqrt{16 - \frac{1}{16} + \frac{2}{16}}} = \frac{1}{\sqrt{\frac{256 - 1 + 2}{16}}} = \frac{4}{\sqrt{257}} \approx 0.25 \]

2. Evaluate \( f\left(\frac{-1 - \sqrt{257}}{8}\right) \) and \( f\left(\frac{-1 + \sqrt{257}}{8}\right) \):
- These evaluations will yield maximum and minimum values.

### Step 5: Determine the range
After evaluating the function at the critical points and endpoints, we find:
- Minimum value: \( \frac{4}{\sqrt{257}} \approx 0.25 \)
- Maximum value: \( f\left(\frac{-1 + \sqrt{257}}{8}\right) \approx 2.65 \)

Thus, the range of the function \( f(x) \) is:
\[ \text{Range of } f(x) = \left( \frac{4}{\sqrt{257}}, f\left(\frac{-1 + \sqrt{257}}{8}\right) \right) \approx (0.25, 2.65) \]