Question

solution of |x-1|+|x-2|+|x-3|>=6 is??

Answer 1

@mathmale please help me

Answer 2

adding absolute values means we the proper addition of "lines" within special intervals

Answer 3

im thinking my a and c are redundant at the moment and just the vertex portions are critical for finding changes in direction

Answer 4

i didnt get u :(

Answer 5

|x-1|+|x-2|+|x-3|
1 2 3 ... our critical points





1 2 3
<------------------------------------------->
-(x-1) +(x-1) +(x-1) +(x-1)
-(x-2) -(x-2) +(x-2) +(x-2)
-(x-3) -(x-3) -(x-3) +(x-3)


we have lines that add up in each interval like this

Answer 6

each interval can then be assessed for an overall view of what is happening

Answer 7

1 2 3
<------------------------------------------->
-(x-1) +(x-1) +(x-1) +(x-1)
-(x-2) -(x-2) +(x-2) +(x-2)
-(x-3) -(x-3) -(x-3) +(x-3)
----- ----- ----- -------
-3x+6 x+4 x+0 3x-6

Answer 8

cant v combine the eqn like...|x-1 + x-2 + x-3| ??
sorry m totally stucked wid this mod

Answer 9

is this correct??
3x-6 >=6

Answer 10

we cant combine them since they change at different parts of the domain

Answer 11

but yes, we can then subtract 6 from each intervals writeup and see if there is a value int he interval that gets us a zero

Answer 12

and 3x>=12
x>=4 ?

Answer 13

1 2 3
<------------------------------------------->
-3x+6 x+4 x+0 3x-6
-3x>=0 x-2>=0 x-6>=0 3x-12>=0

Answer 14

might have to dbl chk my additions for the intervals but thats the basic run of it

Answer 15

-3x>=0 x-2>=0 x-6>=0 3x-12>=0
x <= 0 x >=2 x>=6 x >=4
good iffy not in good

Answer 16

-x+4-6 for the second one :) forgot a -

-x-2 >= 0
-x >= 2

x <= -2 is not in the intervla