Range of the function `f(x)=(x-2)/(x^2-4x+3)` is

Options

A. `(-oo,0)`
B. `R`
C. `(0,oo)`
D. `R-{0}`

Explanation

To find the range of the function \( f(x) = \frac{x - 2}{x^2 - 4x + 3} \), we will follow these steps:

### Step 1: Set the function equal to \( y \)
Let \( f(x) = y \):
\[ y = \frac{x - 2}{x^2 - 4x + 3} \]

### Step 2: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ y(x^2 - 4x + 3) = x - 2 \]
This can be rearranged to:
\[ yx^2 - 4yx + 3y - x + 2 = 0 \]

### Step 3: Rearrange the equation
Rearranging the equation, we have:
\[ yx^2 - (4y + 1)x + (3y + 2) = 0 \]
This is a quadratic equation in \( x \).

### Step 4: Apply the condition for real roots
For the quadratic equation to have real roots, the discriminant \( D \) must be greater than or equal to zero:
\[ D = b^2 - 4ac \]
Here, \( a = y \), \( b = -(4y + 1) \), and \( c = 3y + 2 \). Thus, the discriminant is:
\[ D = (-(4y + 1))^2 - 4(y)(3y + 2) \]

### Step 5: Simplify the discriminant
Calculating the discriminant:
\[ D = (4y + 1)^2 - 4y(3y + 2) \]
Expanding this:
\[ D = 16y^2 + 8y + 1 - (12y^2 + 8y) \]
Simplifying further:
\[ D = 16y^2 + 8y + 1 - 12y^2 - 8y = 4y^2 + 1 \]

### Step 6: Set the discriminant greater than or equal to zero
Now we need to find when:
\[ 4y^2 + 1 \geq 0 \]
Since \( 4y^2 \) is always non-negative and \( 1 \) is positive, \( 4y^2 + 1 \) is always greater than zero for all real values of \( y \).

### Conclusion: Determine the range
Since the discriminant is always non-negative, the quadratic equation has real roots for all values of \( y \). Therefore, the range of the function \( f(x) \) is:
\[ \text{Range of } f(x) = \text{All real numbers} \]

### Final Answer
The range of the function \( f(x) = \frac{x - 2}{x^2 - 4x + 3} \) is all real numbers.

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