For all real numbers x satisfying the absolute value inequality |4x - 12| \le 20, the maxi… — GRE General Test

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

Answer: 22

Answer

The maximum possible value

M

of the expression

|2 - 3x|

on the domain

-2 \le x \le 8

is 22.

Solving

|4x - 12| \le 20

yields

-20 \le 4x - 12 \le 20

, which simplifies to

-2 \le x \le 8

. Evaluating

|2 - 3x|

over this interval gives a minimum of

0

(at

x = 2/3

) and endpoint values of

|2 - 3(-2)| = 8

and

|2 - 3(8)| = |-22| = 22

. Thus, the maximum value

M

is 22.

Step-by-Step Solution

Unfold the given absolute value inequality into a compound linear inequality.

-20 \le 4x - 12 \le 20

The inequality

|u| \le k

for

k \ge 0

is equivalent to

-k \le u \le k

Isolate the variable

x

by adding 12 and dividing by 4.

-2 \le x \le 8

Adding 12 gives

-8 \le 4x \le 32

. Dividing by positive 4 preserves inequality signs, yielding

-2 \le x \le 8

Evaluate the target expression

|2 - 3x|

at the boundary points of the interval

[-2, 8]

For

x = -2

|2 - 3(-2)| = |8| = 8

. For

x = 8

|2 - 3(8)| = |-22| = 22

The expression

f(x) = |2 - 3x|

is convex and non-negative, reaching its local minimum of 0 at

x = \frac{2}{3}

. Its maximum over a closed interval must occur at one of the endpoints.

Compare the evaluated values to find the maximum

M

M = \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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