Question

Difficulty: HardLinear Inequalities and Absolute Value

For all real numbers xx satisfying the absolute value inequality ∣4x−12∣≤20|4x - 12| \le 20, the maximum possible value of the expression ∣2−3x∣|2 - 3x| is MM. What is the value of MM?

Answer: 22

Answer

The maximum possible value MM of the expression ∣2−3x∣|2 - 3x| on the domain −2≤x≤8-2 \le x \le 8 is 22.
Solving ∣4x−12∣≤20|4x - 12| \le 20 yields −20≤4x−12≤20-20 \le 4x - 12 \le 20, which simplifies to −2≤x≤8-2 \le x \le 8. Evaluating ∣2−3x∣|2 - 3x| over this interval gives a minimum of 00 (at x=2/3x = 2/3) and endpoint values of ∣2−3(−2)∣=8|2 - 3(-2)| = 8 and ∣2−3(8)∣=∣−22∣=22|2 - 3(8)| = |-22| = 22. Thus, the maximum value MM is 22.

Step-by-Step Solution

1
Unfold the given absolute value inequality into a compound linear inequality.
−20≤4x−12≤20-20 \le 4x - 12 \le 20
The inequality ∣u∣≤k|u| \le k for k≥0k \ge 0 is equivalent to −k≤u≤k-k \le u \le k.
2
Isolate the variable xx by adding 12 and dividing by 4.
−2≤x≤8-2 \le x \le 8
Adding 12 gives −8≤4x≤32-8 \le 4x \le 32. Dividing by positive 4 preserves inequality signs, yielding −2≤x≤8-2 \le x \le 8.
3
Evaluate the target expression ∣2−3x∣|2 - 3x| at the boundary points of the interval [−2,8][-2, 8].
For x=−2x = -2: ∣2−3(−2)∣=∣8∣=8|2 - 3(-2)| = |8| = 8. For x=8x = 8: ∣2−3(8)∣=∣−22∣=22|2 - 3(8)| = |-22| = 22.
The expression f(x)=∣2−3x∣f(x) = |2 - 3x| is convex and non-negative, reaching its local minimum of 0 at x=23x = \frac{2}{3}. Its maximum over a closed interval must occur at one of the endpoints.
4
Compare the evaluated values to find the maximum MM.
M=max⁡(8,22)=22M = \max(8, 22) = 22
Comparing 8 and 22 shows that 22 is the absolute maximum value achievable within the domain.

Key Concept

Solving linear absolute value inequalities to determine variable bounds and evaluating extreme values of absolute value expressions.
Estimated Time:2m 0s
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