Range of the function `f(x)=(x^(2)-2)/(2x-3)` is

Options

A. `(-oo,1] uu[2,oo)`
B. `[1,2]`
C. `(-oo,0)uu[2,oo)`
D. `(0,2]`

Answer

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

### Step 1: Set the function equal to \( y \)
Let \( y = \frac{x^2 - 2}{2x - 3} \).

### Step 2: Cross-multiply to eliminate the fraction
Rearranging gives:
\[ y(2x - 3) = x^2 - 2 \]
This simplifies to:
\[ 2xy - 3y = x^2 - 2 \]

### Step 3: Rearrange the equation into standard quadratic form
Rearranging gives:
\[ x^2 - 2xy + (3y - 2) = 0 \]
This is a quadratic equation in \( x \).

### Step 4: Identify coefficients for the quadratic equation
In the quadratic equation \( ax^2 + bx + c = 0 \):
- \( a = 1 \)
- \( b = -2y \)
- \( c = 3y - 2 \)

### Step 5: Apply the discriminant condition
For \( x \) to have real solutions, the discriminant must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Substituting the coefficients:
\[ (-2y)^2 - 4(1)(3y - 2) \geq 0 \]
This simplifies to:
\[ 4y^2 - 12y + 8 \geq 0 \]

### Step 6: Simplify the inequality
Dividing the entire inequality by 4 gives:
\[ y^2 - 3y + 2 \geq 0 \]

### Step 7: Factor the quadratic
Factoring the quadratic:
\[ (y - 2)(y - 1) \geq 0 \]

### Step 8: Determine the intervals
To find the intervals where this inequality holds, we identify the roots:
- \( y = 1 \)
- \( y = 2 \)

Using a number line, we test intervals:
1. For \( y < 1 \): Choose \( y = 0 \) → \( (0 - 2)(0 - 1) = 2 > 0 \) (True)
2. For \( 1 < y < 2 \): Choose \( y = 1.5 \) → \( (1.5 - 2)(1.5 - 1) = -0.25 < 0 \) (False)
3. For \( y > 2 \): Choose \( y = 3 \) → \( (3 - 2)(3 - 1) = 2 > 0 \) (True)

### Step 9: Write the solution
The solution to the inequality is:
\[ y \in (-\infty, 1] \cup [2, \infty) \]

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