Question

Domain:?

Answer 1

\[\LARGE \sqrt{|x-1|+|x-2|+|x-3|-6}\]

Answer 2

I want to do this with graphical method,would be appreciated

Answer 3

Domain={R}

Answer 4

no where near,absolutely wrong.

Answer 5

tell me what do you get for x=3?

Answer 6

and x=1 and x=2 too

Answer 7

http://www.wolframalpha.com/input/?i= |x-1|%2B|x-2|%2B|x-3|-6%3E0

Answer 8

yes that is the correct graph,i need some good explanation..

Answer 9

apparently it is \((-\infty, 0]\cup [2,\infty)\)

Answer 10

no again

Answer 11

tell me what do u get for x=3
like posted above

Answer 12

it will take a long time to do
you have to solve over different intervals

Answer 13

tell me over one interval

Answer 14

for |x-1|

Answer 15

if x<0 then why is the x-1 thing reversed,first doubt?

Answer 16

i am sorry my answer was wrong
it is \((-\infty, 0]\cup [4,\infty)\)

Answer 17

taking positive and negative value of piecewise function and solving the square root part greater than equal to 0 you get the following :

\[x \geq 0 \& x \geq 4\]

Answer 18

you have to break it up in to cases
if \(x>3\) then \(|x-1|=x-1,,|x-2|=x-2, |x-3|=x-3\)

Answer 19

so if \(x>3\) you are solving
\[x-1+x-2+x-3-6\geq 0\] in which case you get
\[3x-12\geq 0\] and so \(x\geq 4\)

Answer 20

So domain of the function is \[x \geq 4\]

Answer 21

If i have |x-1| then for x<0 it becomes x-1?why?

Answer 22

if \(x<1\) then \(|x-1|=1-x\)

Answer 23

i wrote that too but why :|

Answer 24

because if \(x<1\) then \(x-1<0\) and so \(|x-1|=-(x-1)=1-x\)

Answer 25

but modulus function makes it positive

Answer 26

yes, and if \(x-1\) is negative, then \(1-x\) is positive

Answer 27

why is the sign getting reversed? i thought | | will make it +ve no matter if x>0 or x<0

Answer 28

lets try it with numbers
suppose \(x=-2\)
then \(|-2-1|=|-3|=3=1-(-2)\)

Answer 29

seems legit but very non intuitive for me first time :P

Answer 30

the definition is this
\[ |x| = \left\{\begin{array}{rcc}
-x & \text{if} & x <0 \\
x & \text{if} & x\geq 0
\end{array}
\right.
\]

Answer 31

for kind of removing the mod right?

Answer 32

and so for example
\[ |x-1| = \left\{\begin{array}{rcc}
1-x & \text{if} & x <1 \\
x-1& \text{if} & x \geq 1
\end{array}
\right.
\]

Answer 33

yes, it seems like absolute value should be easy right? just "make it positive"
but in fact it is a very annoying piecewise function and hard to work with because of that fact

Answer 34

hmm..okay..clear enough..thanks for your time :")
why do i have 3 medals anyway :P

Answer 35

and the last thing,how do i get the points on the graph?
i am supposed to find dy/dx at all these intervals right?
If they turn out to be +ve then increasing slope..and vice versa..but what about the points?