Examkin

Soru

Zorluk: OrtaLinear Inequalities and Absolute Value

If xx is a real number that satisfies ∣3−2x∣≤9|3 - 2x| \le 9, what is the maximum possible value of ∣x−5∣|x - 5|?

Cevap: 8

Cevap

The maximum possible value of ∣x−5∣|x - 5| is 8.
Solving ∣3−2x∣≤9|3 - 2x| \le 9 yields −9≤3−2x≤9-9 \le 3 - 2x \le 9. Subtracting 3 gives −12≤−2x≤6-12 \le -2x \le 6. Dividing by −2-2 and reversing inequality signs results in −3≤x≤6-3 \le x \le 6. To maximize ∣x−5∣|x - 5|, find the point in [−3,6][-3, 6] that is farthest from 5. Evaluating at the endpoints gives ∣−3−5∣=8|-3 - 5| = 8 and ∣6−5∣=1|6 - 5| = 1. The maximum possible value is 8.

Adım Adım Çözüm

1
Solve the absolute value inequality ∣3−2x∣≤9|3 - 2x| \le 9.
−9≤3−2x≤9-9 \le 3 - 2x \le 9
By definition of absolute value, ∣u∣≤k|u| \le k means −k≤u≤k-k \le u \le k for k≥0k \ge 0.
2
Isolate −2x-2x by subtracting 3 from each part.
−12≤−2x≤6-12 \le -2x \le 6
Subtracting a constant from all parts preserves the inequality direction.
3
Isolate xx by dividing each part by −2-2.
−3≤x≤6-3 \le x \le 6
Dividing an inequality by a negative number flips the inequality signs.
4
Determine the maximum value of ∣x−5∣|x - 5| over the interval [−3,6][-3, 6].
Maximum occurs at x=−3x = -3, where ∣−3−5∣=8|-3 - 5| = 8.
The expression ∣x−5∣|x - 5| represents the distance from xx to 5 on the number line. The endpoint x=−3x = -3 is furthest from 5.

Anahtar Kavram

Linear inequalities and absolute value as distance on a number line
Bu soruyu puanla

Konular

Linear Inequalities and Absolute Value

Bu konudan daha fazla soru