Function

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

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

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

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

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

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.

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

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.

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

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

OA is 2

f(1.5) = |1.5| + |1.5-2| = 1.5 + 0.5 = 2