Question

The function \( f(x) = \frac{x^2 - 4x + 3}{x^2 - 5x + 6} \) is defined for all real values of \( x \) except:

• A. 2 and 3
• B. 3 and 4
• C. 2 and 6
• D. 3 and 6

Hint:

\begin{itemize} \item A rational function $f(x) = P(x)/Q(x)$ is undefined where its denominator $Q(x)=0$. \item Set the denominator $x^2-5x+6$ to zero and solve for $x$. \item Factor $x^2-5x+6 = (x-2)(x-3)$. \item The roots are $x=2$ and $x=3$. These are the values where the function is not defined. \item Even if a factor cancels (like $(x-3)$ here), the original function is still undefined at that point due to the initial division by zero. \end{itemize}

Answer 1

The Correct Option is A

A rational function \( f(x) = \frac{P(x)}{Q(x)} \) is undefined when its denominator \( Q(x) \) is equal to zero. In this case, \( f(x) = \frac{x^2 - 4x + 3}{x^2 - 5x + 6} \). The denominator is \( Q(x) = x^2 - 5x + 6 \). We need to find the values of \( x \) for which \( x^2 - 5x + 6 = 0 \). Factorize the quadratic equation: \[ x^2 - 5x + 6 = (x - 2)(x - 3) \] So, \( (x - 2)(x - 3) = 0 \). This gives \( x - 2 = 0 \) or \( x - 3 = 0 \). Therefore, \( x = 2 \) or \( x = 3 \). The function is undefined at \( x = 2 \) and \( x = 3 \). Now, check the numerator at these points to determine the type of discontinuity: Numerator \( P(x) = x^2 - 4x + 3 = (x - 1)(x - 3) \). At \( x = 2 \): \[ P(2) = 2^2 - 4 \cdot 2 + 3 = 4 - 8 + 3 = -1 \] \[ Q(2) = 0 \] So, \( f(2) = \frac{-1}{0} \), which is undefined (vertical asymptote). At \( x = 3 \): \[ P(3) = 3^2 - 4 \cdot 3 + 3 = 9 - 12 + 3 = 0 \] \[ Q(3) = 0 \] So, \( f(3) = \frac{0}{0} \), which is indeterminate (removable discontinuity, or "hole"). For \( x \neq 3 \), \[ f(x) = \frac{(x - 1)(x - 3)}{(x - 2)(x - 3)} = \frac{x - 1}{x - 2} \] However, the original function is still undefined at \( x = 3 \) due to division by zero in the original expression.

The function is defined for all real values of \( x \) except where the denominator is zero, i.e., \( x = 2 \) and \( x = 3 \). \[ \boxed{\text{2 and 3}} \]