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Zorluk: OrtaLinear Inequalities and Absolute Value

If xx is a real number that satisfies the inequality ∣3−2x∣≤7|3 - 2x| \le 7, what is the maximum possible value of 5−3x5 - 3x?

  1. A
    -10
  2. B
    -1
  3. C
    5
  4. 11Cevap
  5. E
    20

Cevap

The maximum possible value of 5−3x5 - 3x is 1111.
Solving ∣3−2x∣≤7|3 - 2x| \le 7 gives −7≤3−2x≤7-7 \le 3 - 2x \le 7. Subtracting 33 yields −10≤−2x≤4-10 \le -2x \le 4. Dividing by −2-2 and flipping the inequality signs gives −2≤x≤5-2 \le x \le 5. Because 5−3x5 - 3x has a negative coefficient for xx, it decreases as xx increases. Therefore, the maximum value of 5−3x5 - 3x occurs at the smallest value in the domain, x=−2x = -2. Substituting x=−2x = -2 into 5−3x5 - 3x gives 5−3(−2)=115 - 3(-2) = 11.

Adım Adım Çözüm

1
Rewrite the absolute value inequality ∣3−2x∣≤7|3 - 2x| \le 7 as a compound inequality.
−7≤3−2x≤7-7 \le 3 - 2x \le 7
An absolute value inequality of the form ∣u∣≤k|u| \le k (where k≥0k \ge 0) is equivalent to −k≤u≤k-k \le u \le k.
2
Isolate xx by subtracting 33 from all parts of the inequality and then dividing by −2-2.
−10≤−2x≤4  ⟹  5≥x≥−2-10 \le -2x \le 4 \implies 5 \ge x \ge -2, which is equivalent to −2≤x≤5-2 \le x \le 5
Dividing an inequality by a negative number reverses the direction of the inequality signs.
3
Determine which value of xx in the interval −2≤x≤5-2 \le x \le 5 maximizes the linear expression 5−3x5 - 3x.
The expression reaches its maximum at the minimum bound x=−2x = -2.
Since the coefficient of xx is negative (−3-3), smaller values of xx result in larger values for 5−3x5 - 3x.
4
Substitute x=−2x = -2 into 5−3x5 - 3x.
5−3(−2)=5+6=115 - 3(-2) = 5 + 6 = 11
Evaluating the linear expression at the lower bound yields its maximum value.

Anahtar Kavram

Solving linear absolute value inequalities and optimizing linear expressions over a closed interval.
Tahmini Süre:1m 30s
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Konular

Linear Inequalities and Absolute Value

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