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

Explanation

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

### Step 1: Rewrite the function
We start by letting \( y = f(x) \):
\[ y = \frac{x^2 - 2x + 4}{x^2 + 2x + 4} \]

### Step 2: Cross-multiply
Cross-multiplying gives us:
\[ y(x^2 + 2x + 4) = x^2 - 2x + 4 \]
Expanding this, we get:
\[ yx^2 + 2yx + 4y = x^2 - 2x + 4 \]

### Step 3: Rearranging the equation
Rearranging the equation to bring all terms to one side:
\[ (y - 1)x^2 + (2y + 2)x + (4y - 4) = 0 \]

### Step 4: Identify coefficients
The coefficients of the quadratic equation are:
- \( A = y - 1 \)
- \( B = 2y + 2 \)
- \( C = 4y - 4 \)

### Step 5: Condition for real roots
For \( x \) to have real roots, the discriminant must be non-negative:
\[ B^2 - 4AC \geq 0 \]
Substituting the coefficients:
\[ (2y + 2)^2 - 4(y - 1)(4y - 4) \geq 0 \]

### Step 6: Simplifying the discriminant
Calculating the discriminant:
\[ (2y + 2)^2 = 4y^2 + 8y + 4 \]
\[ 4(y - 1)(4y - 4) = 4(4y^2 - 4y - 4y + 4) = 16y^2 - 16y \]
Now we have:
\[ 4y^2 + 8y + 4 - (16y^2 - 16y) \geq 0 \]
This simplifies to:
\[ 4y^2 + 8y + 4 - 16y^2 + 16y \geq 0 \]
\[ -12y^2 + 24y + 4 \geq 0 \]
Dividing by -4 (and reversing the inequality):
\[ 3y^2 - 6y - 1 \leq 0 \]

### Step 7: Finding the roots
Using the quadratic formula to find the roots of \( 3y^2 - 6y - 1 = 0 \):
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \]
Calculating the discriminant:
\[ 36 + 12 = 48 \]
Thus, the roots are:
\[ y = \frac{6 \pm \sqrt{48}}{6} = \frac{6 \pm 4\sqrt{3}}{6} = 1 \pm \frac{2\sqrt{3}}{3} \]

### Step 8: Interval for the range
The roots are:
\[ y_1 = 1 - \frac{2\sqrt{3}}{3}, \quad y_2 = 1 + \frac{2\sqrt{3}}{3} \]
The quadratic \( 3y^2 - 6y - 1 \) opens upwards, so the range of \( y \) is:
\[ y \in \left[1 - \frac{2\sqrt{3}}{3}, 1 + \frac{2\sqrt{3}}{3}\right] \]

### Final Answer
The range of the function \( f(x) \) is:
\[ \boxed{\left[1 - \frac{2\sqrt{3}}{3}, 1 + \frac{2\sqrt{3}}{3}\right]} \]