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

Explanation

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

### Step 1: Set up the equation
Let \( y = f(x) = \ln\left(\frac{\sqrt{8 - x^2}}{x - 2}\right) \).

### Step 2: Exponentiate both sides
Taking the exponential of both sides, we get:
\[ e^y = \frac{\sqrt{8 - x^2}}{x - 2} \]

### Step 3: Cross-multiply
Cross-multiplying gives:
\[ e^y (x - 2) = \sqrt{8 - x^2} \]

### Step 4: Square both sides
Squaring both sides to eliminate the square root yields:
\[ (e^y (x - 2))^2 = 8 - x^2 \]

### Step 5: Expand and rearrange
Expanding the left side:
\[ e^{2y} (x^2 - 4x + 4) = 8 - x^2 \]
Rearranging gives:
\[ (e^{2y} + 1)x^2 - 4e^{2y}x + (4e^{2y} - 8) = 0 \]

### Step 6: Identify coefficients
This is a quadratic equation in \( x \):
- \( a = e^{2y} + 1 \)
- \( b = -4e^{2y} \)
- \( c = 4e^{2y} - 8 \)

### Step 7: Apply the discriminant condition
For the quadratic to have real solutions, the discriminant must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Substituting the coefficients:
\[ (-4e^{2y})^2 - 4(e^{2y} + 1)(4e^{2y} - 8) \geq 0 \]
This simplifies to:
\[ 16e^{4y} - 4[(e^{2y} + 1)(4e^{2y} - 8)] \geq 0 \]

### Step 8: Simplify the discriminant
Expanding the second term:
\[ 16e^{4y} - 4(4e^{4y} - 8e^{2y} + 4e^{2y} - 8) \geq 0 \]
This simplifies to:
\[ 16e^{4y} - (16e^{4y} - 16e^{2y} + 32) \geq 0 \]
Which further simplifies to:
\[ 16e^{2y} - 32 \geq 0 \]

### Step 9: Solve the inequality
Solving \( 16e^{2y} \geq 32 \) gives:
\[ e^{2y} \geq 2 \implies 2y \geq \ln(2) \implies y \geq \frac{\ln(2)}{2} \]

### Step 10: Conclusion about the range
Since the quadratic discriminant must be non-negative for the function to have real outputs, we conclude that the range of \( f(x) \) is:
\[ \text{Range of } f(x) = \left[\frac{\ln(2)}{2}, \infty\right) \]