Find the range of `f(x)=(x^(2)-2)/(x^(2)-3` :

Options

A. `(1, oo)`
B. `(-oo, 2//3)`
C. `(-oo, 2//3)`
D. `(-oo, 2//3]uu(1, oo)`

Answer

To find the range of the function \( f(x) = \frac{x^2 - 2}{x^2 - 3} \), we will follow these steps:

### Step 1: Set the function equal to \( y \)
We start by setting the function equal to \( y \):
\[ y = \frac{x^2 - 2}{x^2 - 3} \]

### Step 2: Cross-multiply to eliminate the fraction
To eliminate the fraction, we cross-multiply:
\[ y(x^2 - 3) = x^2 - 2 \]
This simplifies to:
\[ yx^2 - 3y = x^2 - 2 \]

### Step 3: Rearrange the equation
Rearranging the equation gives us:
\[ yx^2 - x^2 = 3y - 2 \]
Factoring out \( x^2 \) from the left side:
\[ x^2(y - 1) = 3y - 2 \]

### Step 4: Solve for \( x^2 \)
Now, we can express \( x^2 \) in terms of \( y \):
\[ x^2 = \frac{3y - 2}{y - 1} \]

### Step 5: Analyze the conditions for \( x^2 \)
Since \( x^2 \) must be non-negative (i.e., \( x^2 \geq 0 \)), we set the condition:
\[ \frac{3y - 2}{y - 1} \geq 0 \]

### Step 6: Determine the critical points
To find the values of \( y \) that satisfy this inequality, we identify the critical points by setting the numerator and denominator to zero:
1. **Numerator**: \( 3y - 2 = 0 \) gives \( y = \frac{2}{3} \)
2. **Denominator**: \( y - 1 = 0 \) gives \( y = 1 \)

### Step 7: Test intervals around the critical points
We will test intervals determined by the critical points \( y = \frac{2}{3} \) and \( y = 1 \):
- For \( y < \frac{2}{3} \): Choose \( y = 0 \)
\[ \frac{3(0) - 2}{0 - 1} = \frac{-2}{-1} = 2 \quad (\text{positive}) \]
- For \( \frac{2}{3} < y < 1 \): Choose \( y = 0.8 \)
\[ \frac{3(0.8) - 2}{0.8 - 1} = \frac{2.4 - 2}{-0.2} = \frac{0.4}{-0.2} = -2 \quad (\text{negative}) \]
- For \( y > 1 \): Choose \( y = 2 \)
\[ \frac{3(2) - 2}{2 - 1} = \frac{6 - 2}{1} = 4 \quad (\text{positive}) \]

### Step 8: Conclusion on intervals
From our tests, we find:
- The function is non-negative for \( y < \frac{2}{3} \) and \( y > 1 \).
- The function is undefined at \( y = 1 \).

### Step 9: Write the range
Thus, the range of the function \( f(x) \) is:
\[ (-\infty, \frac{2}{3}] \cup (1, \infty) \]