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by vikram4689 » Sat Jun 04, 2011 4:57 am
General Rule of | |:
|x| is +ve for x>0
|x| is -ve for x<0

f(1.5)= |x| + |x-2|
= x + [-(x-2)]
= 1.5 + [-0.5)]
=>f(1.5)=1
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by Frankenstein » Sat Jun 04, 2011 5:10 am
vikram4689 wrote:General Rule of | |:
|x| is +ve for x>0
|x| is -ve for x<0

f(1.5)= |x| + |x-2|
= x + [-(x-2)]
= 1.5 + [-0.5)]
=>f(1.5)=1
Hi,
|x| is never negative.
f(1.5) = |1.5|+|1.5-2| = 1.5+0.5 = 2.
In fact, for any value of x satisfying 0 ≤ x ≤ 2, f(x) = |x| + |x - 2| =2
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by Ozlemg » Sat Jun 04, 2011 5:16 am
I agree with Frank!

Because 0 ≤ x ≤ 2,

f(x) = |x| + |x - 2| =2
First part of the equation (red) is (+)ve and second part of (blue) it is (-)ve
so x-x+2=2
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by vikram4689 » Sat Jun 04, 2011 7:33 am
Sorry pals I DO NOT agree on this ( though i made a silly mistake on a -ve sign but concept is correct)

General Rule of | |: Google MODULUS FUNCTION GRAPH, it is symm. about y-axis
|x| is +ve for x>0
|x| is -ve for x<0

f(1.5)= |x| + |x-2|
= x + [-(x-2)]
= 1.5 + [-(-0.5)] mistook 1.5-2 as 0.5 , correcting it to -0.5
=2
=>f(1.5)=2
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by cans » Sat Jun 04, 2011 9:17 am
If f(x) = |x| + |x - 2| then find the value of F(1.5) for 0 ≤ x ≤ 2
f(1.5) = |1.5| + |1.5-2| = 1.5 + |-.5| = 1.5 + .5 =2
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by Anurag@Gurome » Sat Jun 04, 2011 9:38 am
vikram4689 wrote:General Rule of | |: Google MODULUS FUNCTION GRAPH, it is symm. about y-axis
Yes, the modulus graph is symmetrical about y-axis but that does not mean
vikram4689 wrote:|x| is +ve for x>0
|x| is -ve for x<0
|x| is always non-negative for any value of x.

By definition of |x|,
  • |x| = x for x ≥ 0
    |x| = -x for x < 0
Hence, if x is positive, |x| = x = +ve
and if x is negative, |x| = -x = -(-ve) = +ve
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by Frankenstein » Sat Jun 04, 2011 9:39 am
vikram4689 wrote:Sorry pals I DO NOT agree on this ( though i made a silly mistake on a -ve sign but concept is correct)

General Rule of | |: Google MODULUS FUNCTION GRAPH, it is symm. about y-axis
|x| is +ve for x>0
|x| is -ve for x<0
Hi,
You might be having a good picture of modulus but what you have written is still wrong.
Can you give an example for |x| being -ve. By definition modulus is non-negative.
It would be correct to say
|x| is x for x>0
|x| is -x for x<0 --> As x is -ve, -x will be positive.
I think this is what you have intended to write.
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by venmic » Sun Jun 05, 2011 2:03 pm
why did you convert the modulus to [-(x-2)] negaive ???????
i disagree
vikram4689 wrote:General Rule of | |:
|x| is +ve for x>0
|x| is -ve for x<0

f(1.5)= |x| + |x-2|
= x + [-(x-2)]
= 1.5 + [-0.5)]
=>f(1.5)=1
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by vikram4689 » Sun Jun 05, 2011 6:29 pm
Actually i confused the whole scenario ;).... what i intended was |x| is -ve for x<0 and when that value of x is substituted in -x than it becomes +ve.

For |x-2|, it is -ve for x<2 i.e.-(x-2) and when i substitute x=1.5 it becomes +ve

Apologies for creating unintended confusion.
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by TuanNguyen87 » Sun Jun 05, 2011 8:48 pm
I think we just need to replace x by 1.5 and because 1.5 > 0. Thus,

f(1.5)= |1.5| + |1.5 - 2|= 1.5 + (2 - 1.5)= 2 :)
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by aftableo2006 » Sun Jun 05, 2011 11:14 pm
f(x)=|x|+|x-2|
for x>2 f(x)=x+x-2=2x-2
for x<0 f(x)=-x-x+2=2-2x
for 0<x>2 f(x)=x-x+2=2
s0 f(1.5)=2
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by [email protected] » Thu Jun 09, 2011 7:51 pm
OA is 2
IT IS TIME TO BEAT THE GMAT

LEARNING, APPLICATION AND TIMING IS THE FACT OF GMAT AND LIFE AS WELL... KEEP PLAYING!!!

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by MBA.Aspirant » Thu Jun 09, 2011 7:57 pm
f(1.5) = |1.5| + |1.5-2| = 1.5 + 0.5 = 2
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