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Zorluk: OrtaReal Numbers, Number Line, and Absolute Value

If xx and yy are real numbers such that ∣x−4∣≤3|x - 4| \le 3 and ∣y+2∣≤5|y + 2| \le 5, what is the maximum possible value of ∣x−y∣|x - y|?

Cevap: 14

Cevap

The maximum possible value of ∣x−y∣|x - y| is 14.
Solving ∣x−4∣≤3|x - 4| \le 3 gives the closed interval [1,7][1, 7] for xx. Solving ∣y+2∣≤5|y + 2| \le 5 gives the closed interval [−7,3][-7, 3] for yy. The maximum possible value of ∣x−y∣|x - y| is the maximum distance between a point in [1,7][1, 7] and a point in [−7,3][-7, 3], which is 7−(−7)=147 - (-7) = 14.

Adım Adım Çözüm

1
Determine the range of possible values for xx.
1≤x≤71 \le x \le 7
The inequality ∣x−4∣≤3|x - 4| \le 3 represents all numbers within distance 3 of 4 on the number line.
2
Determine the range of possible values for yy.
−7≤y≤3-7 \le y \le 3
The inequality ∣y+2∣≤5|y + 2| \le 5 represents all numbers within distance 5 of -2 on the number line.
3
Find the maximum distance between any point xx in [1,7][1, 7] and any point yy in [−7,3][-7, 3].
14
The maximum absolute difference ∣x−y∣|x - y| occurs between the upper endpoint of the xx-interval (77) and the lower endpoint of the yy-interval (−7-7).

Anahtar Kavram

Absolute value inequalities as distance intervals on the real number line and maximizing differences between bounded variables
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