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Retired Moderator
Joined: 16 Apr 2020
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f(x) = |x 2| + |2.5 x| + |3.6 x|
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24 Sep 2021, 10:34
24 Sep 2021, 10:34
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Question Stats:
54% (01:48) correct
45% (01:11) wrong based on 84 sessions
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\(f(x) = |x – 2| + |2.5 – x| + |3.6 – x|\)
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
Minimum value of \(f(x)\) |
2.5 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
Re: f(x) = |x 2| + |2.5 x| + |3.6 x|
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24 Sep 2021, 14:27
24 Sep 2021, 14:27
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the absolute is minimized when its value is 0
setting up pair-wise function as below
\(min f(x)=|x–2|, x=2\)
\(min f(x)=|2.5-x|, x=2.5\)
\(min f(x)=|3.6-x|, x=3.6\)
we can notice that x is different, and we must choose one value for \(x\), because it's one and not pair-wise function
let us choose the middle value, \(x=2.5\), \(f(2.5)=|2.5–2|+|2.5–2.5|+|3.6–2.5|=1.6\)
answer is B, as Quantity A=1.6 < Quantity B=2.5
setting up pair-wise function as below
\(min f(x)=|x–2|, x=2\)
\(min f(x)=|2.5-x|, x=2.5\)
\(min f(x)=|3.6-x|, x=3.6\)
we can notice that x is different, and we must choose one value for \(x\), because it's one and not pair-wise function
let us choose the middle value, \(x=2.5\), \(f(2.5)=|2.5–2|+|2.5–2.5|+|3.6–2.5|=1.6\)
answer is B, as Quantity A=1.6 < Quantity B=2.5
KarunMendiratta wrote:
\(f(x) = |x – 2| + |2.5 – x| + |3.6 – x|\)
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
Minimum value of \(f(x)\) |
2.5 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
