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Zorluk: ZorLinear Inequalities and Absolute Value

If xx is a real number that satisfies both ∣3x−12∣≤9|3x - 12| \le 9 and ∣2−x∣≥4|2 - x| \ge 4, the maximum possible value of the expression 5−2x5 - 2x is MM and the minimum possible value is mm. What is the value of M−mM - m?

Cevap: 2

Cevap

The value of M−mM - m is 2.
Solving ∣3x−12∣≤9|3x - 12| \le 9 yields 1≤x≤71 \le x \le 7. Solving ∣2−x∣≥4|2 - x| \ge 4 yields x≤−2x \le -2 or x≥6x \ge 6. The set of xx-values satisfying both conditions is the intersection [6,7][6, 7]. Since 5−2x5 - 2x is a linear expression with a negative coefficient, its maximum value MM occurs at the smallest value of xx (x=6x = 6), giving M=5−2(6)=−7M = 5 - 2(6) = -7. Its minimum value mm occurs at the largest value of xx (x=7x = 7), giving m=5−2(7)=−9m = 5 - 2(7) = -9. Thus, M−m=−7−(−9)=2M - m = -7 - (-9) = 2.

Adım Adım Çözüm

1
Solve the inequality ∣3x−12∣≤9|3x - 12| \le 9
1≤x≤71 \le x \le 7
Expanding absolute value yields −9≤3x−12≤9-9 \le 3x - 12 \le 9. Adding 12 gives 3≤3x≤213 \le 3x \le 21, then dividing by 3 yields 1≤x≤71 \le x \le 7.
2
Solve the inequality ∣2−x∣≥4|2 - x| \ge 4
x≤−2x \le -2 or x≥6x \ge 6
Absolute value inequality ∣u∣≥a|u| \ge a splits into u≥au \ge a or u≤−au \le -a. Here 2−x≥4  ⟹  x≤−22 - x \ge 4 \implies x \le -2, and 2−x≤−4  ⟹  x≥62 - x \le -4 \implies x \ge 6.
3
Determine the overlapping domain for xx
6≤x≤76 \le x \le 7
Combining 1≤x≤71 \le x \le 7 with x≤−2x \le -2 or x≥6x \ge 6 leaves only the interval 6≤x≤76 \le x \le 7.
4
Evaluate maximum MM and minimum mm of 5−2x5 - 2x on 6≤x≤76 \le x \le 7
M=−7M = -7 and m=−9m = -9
Since −2x-2x decreases as xx increases, the maximum occurs at x=6x = 6 (M=5−12=−7M = 5 - 12 = -7) and the minimum occurs at x=7x = 7 (m=5−14=−9m = 5 - 14 = -9).
5
Calculate the difference M−mM - m
22
Subtracting mm from MM gives −7−(−9)=2-7 - (-9) = 2.

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Linear Inequalities and Absolute Value
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Linear Inequalities and Absolute Value

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