DLS
  • DLS
Domain:?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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DLS
  • DLS
\[\LARGE \sqrt{|x-1|+|x-2|+|x-3|-6}\]
DLS
  • DLS
I want to do this with graphical method,would be appreciated
anonymous
  • anonymous
Domain={R}

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DLS
  • DLS
no where near,absolutely wrong.
DLS
  • DLS
tell me what do you get for x=3?
DLS
  • DLS
and x=1 and x=2 too
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=|x-1|%2B|x-2|%2B|x-3|-6%3E0
DLS
  • DLS
yes that is the correct graph,i need some good explanation..
anonymous
  • anonymous
apparently it is \((-\infty, 0]\cup [2,\infty)\)
DLS
  • DLS
no again
DLS
  • DLS
tell me what do u get for x=3 like posted above
anonymous
  • anonymous
it will take a long time to do you have to solve over different intervals
DLS
  • DLS
tell me over one interval
DLS
  • DLS
for |x-1|
DLS
  • DLS
if x<0 then why is the x-1 thing reversed,first doubt?
anonymous
  • anonymous
i am sorry my answer was wrong it is \((-\infty, 0]\cup [4,\infty)\)
anonymous
  • anonymous
taking positive and negative value of piecewise function and solving the square root part greater than equal to 0 you get the following : \[x \geq 0 \& x \geq 4\]
anonymous
  • anonymous
you have to break it up in to cases if \(x>3\) then \(|x-1|=x-1,,|x-2|=x-2, |x-3|=x-3\)
anonymous
  • anonymous
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 \(x\geq 4\)
anonymous
  • anonymous
So domain of the function is \[x \geq 4\]
DLS
  • DLS
If i have |x-1| then for x<0 it becomes x-1?why?
anonymous
  • anonymous
if \(x<1\) then \(|x-1|=1-x\)
DLS
  • DLS
i wrote that too but why :|
anonymous
  • anonymous
because if \(x<1\) then \(x-1<0\) and so \(|x-1|=-(x-1)=1-x\)
DLS
  • DLS
but modulus function makes it positive
anonymous
  • anonymous
yes, and if \(x-1\) is negative, then \(1-x\) is positive
DLS
  • DLS
why is the sign getting reversed? i thought | | will make it +ve no matter if x>0 or x<0
anonymous
  • anonymous
lets try it with numbers suppose \(x=-2\) then \(|-2-1|=|-3|=3=1-(-2)\)
DLS
  • DLS
seems legit but very non intuitive for me first time :P
anonymous
  • anonymous
the definition is this \[ |x| = \left\{\begin{array}{rcc} -x & \text{if} & x <0 \\ x & \text{if} & x\geq 0 \end{array} \right. \]
DLS
  • DLS
for kind of removing the mod right?
anonymous
  • anonymous
and so for example \[ |x-1| = \left\{\begin{array}{rcc} 1-x & \text{if} & x <1 \\ x-1& \text{if} & x \geq 1 \end{array} \right. \]
anonymous
  • anonymous
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
DLS
  • DLS
hmm..okay..clear enough..thanks for your time :") why do i have 3 medals anyway :P
DLS
  • DLS
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?

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