Find the range of the following functions: (where {.} and [.] represent fractional part and greatest integer part functions respectively)
`f(x)=(x+2)/(x^(2)-8x-4)`

Explanation

To find the range of the function \( f(x) = \frac{x + 2}{x^2 - 8x - 4} \), we will follow these steps:

### Step 1: Set the function equal to \( y \)
We start by rewriting the function in terms of \( y \):
\[ y = \frac{x + 2}{x^2 - 8x - 4} \]

### Step 2: Cross multiply to eliminate the fraction
To eliminate the fraction, we cross-multiply:
\[ y(x^2 - 8x - 4) = x + 2 \]
This simplifies to:
\[ yx^2 - 8yx - x - 4y - 2 = 0 \]

### Step 3: Rearrange the equation
Rearranging gives us a quadratic equation in \( x \):
\[ yx^2 - (8y + 1)x - (4y + 2) = 0 \]

### Step 4: Identify coefficients for the quadratic formula
In the quadratic equation \( Ax^2 + Bx + C = 0 \), we identify:
- \( A = y \)
- \( B = -(8y + 1) \)
- \( C = -(4y + 2) \)

### Step 5: Apply the discriminant condition
For \( x \) to have real solutions, the discriminant \( D \) must be non-negative:
\[ D = B^2 - 4AC \geq 0 \]
Substituting the values of \( A \), \( B \), and \( C \):
\[ D = (-(8y + 1))^2 - 4(y)(-(4y + 2)) \geq 0 \]
This simplifies to:
\[ (8y + 1)^2 + 4y(4y + 2) \geq 0 \]

### Step 6: Expand and simplify the discriminant
Expanding gives:
\[ 64y^2 + 16y + 1 + 16y^2 + 8y \geq 0 \]
Combining like terms results in:
\[ 80y^2 + 24y + 1 \geq 0 \]

### Step 7: Factor the quadratic expression
To factor \( 80y^2 + 24y + 1 \), we can use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{D}}{2a} \]
where \( a = 80 \), \( b = 24 \), and \( D = 24^2 - 4 \cdot 80 \cdot 1 = 576 - 320 = 256 \).

Calculating the roots:
\[ y = \frac{-24 \pm 16}{160} \]
This gives us two roots:
1. \( y_1 = \frac{-8}{160} = -\frac{1}{20} \)
2. \( y_2 = \frac{-40}{160} = -\frac{1}{4} \)

### Step 8: Analyze the intervals
The quadratic \( 80y^2 + 24y + 1 \) opens upwards (since \( a > 0 \)). The roots divide the number line into intervals. We check the sign of the quadratic in each interval:
- For \( y < -\frac{1}{4} \): Positive
- For \( -\frac{1}{4} < y < -\frac{1}{20} \): Negative
- For \( y > -\frac{1}{20} \): Positive

### Step 9: Determine the range
Thus, the range of \( f(x) \) is:
\[ (-\infty, -\frac{1}{4}] \cup [-\frac{1}{20}, \infty) \]

### Final Answer
The range of the function \( f(x) = \frac{x + 2}{x^2 - 8x - 4} \) is:
\[ (-\infty, -\frac{1}{4}] \cup [-\frac{1}{20}, \infty) \]