If f(x) = frac{2x-3}{(x-2)(x-3)} is a real-va | vedclass

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(D) Let $y = \frac{2x-3}{x^2-5x+6}$.For $x$ to be real, the discriminant $D$ of the quadratic equation in $x$ must be non-negative.$y(x^2-5x+6) = 2x-3

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-$2$

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(D) Let $y = \frac{2x-3}{x^2-5x+6}$.For $x$ to be real, the discriminant $D$ of the quadratic equation in $x$ must be non-negative.$y(x^2-5x+6) = 2x-3$$yx^2 - (5y+2)x + (6y+3) = 0$For $x \in \mathbb{R}$, $D = b^2 - 4ac \geq 0$.$(5y+2)^2 - 4y(6y+3) \geq 0$$25y^2 + 20y + 4 - 24y^2 - 12y \geq 0$$y^2 + 8y + 4 \geq 0$.The roots of $y^2 + 8y + 4 = 0$ are $y = \frac{-8 \pm \sqrt{64-16}}{2} = -4 \pm 2\sqrt{3}$.Thus, $y \in (-\infty, -4-2\sqrt{3}] \cup [-4+2\sqrt{3}, \infty)$.The values not taken by $f(x)$ lie in the interval $(-4-2\sqrt{3}, -4+2\sqrt{3})$.Since $-4-2\sqrt{3} \approx -7.46$ and $-4+2\sqrt{3} \approx -0.53$, the value $1$ is in the range, $2$ is in the range, $-10$ is in the range, but $-2$ is $NOT$ in the range of $f(x)$.

If $f(x) = \frac{2x-3}{(x-2)(x-3)}$ is a real-valued function, then the value that $f(x)$ does not take is:

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