The set of real values of x that satisfy the | vedclass

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(B) Given the inequality: ${\log _{1/2}}({x^2} - 6x + 12) \ge - 2$. Since the base of the logarithm is $1/2$ (which is between $0$ and $1$),the inequa

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$[2, 4]$

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(B) Given the inequality: ${\log _{1/2}}({x^2} - 6x + 12) \ge - 2$. Since the base of the logarithm is $1/2$ (which is between $0$ and $1$),the inequality sign reverses when we remove the logarithm: ${x^2} - 6x + 12 \le {(1/2)^{-2}}$. ${x^2} - 6x + 12 \le 4$. ${x^2} - 6x + 8 \le 0$. Factoring the quadratic: $(x - 2)(x - 4) \le 0$. The solution to this inequality is $x \in [2, 4]$. We must also ensure the argument of the logarithm is positive: ${x^2} - 6x + 12 > 0$. The discriminant $D = {(-6)^2} - 4(1)(12) = 36 - 48 = -12 Since the leading coefficient is positive and $D Thus,the solution set is $[2, 4]$.

The set of real values of $x$ that satisfy the inequality ${\log _{1/2}}({x^2} - 6x + 12) \ge - 2$ is:

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