If f(x) = frac{x + 2}{x^2 - 3x + 1}, then the | vedclass

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(A) rational function $f(x) = \frac{P(x)}{Q(x)}$ is undefined when the denominator $Q(x) = 0$.Set the denominator equal to zero: $x^2 - 3x + 1 = 0$.Us

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$x = \frac{3 + \sqrt{5}}{2}, x = \frac{3 - \sqrt{5}}{2}$

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(A) rational function $f(x) = \frac{P(x)}{Q(x)}$ is undefined when the denominator $Q(x) = 0$.Set the denominator equal to zero: $x^2 - 3x + 1 = 0$.Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1, b = -3, c = 1$.$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(1)}}{2(1)}$$x = \frac{3 \pm \sqrt{9 - 4}}{2}$$x = \frac{3 \pm \sqrt{5}}{2}$Thus, $f(x)$ is not defined for $x = \frac{3 + \sqrt{5}}{2}$ and $x = \frac{3 - \sqrt{5}}{2}$.

If $f(x) = \frac{x + 2}{x^2 - 3x + 1}$, then the values of $x$ for which $f(x)$ is not defined are

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