## DLS 3 years ago Domain:?

1. DLS

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

2. DLS

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

3. anonymous

Domain={R}

4. DLS

no where near,absolutely wrong.

5. DLS

tell me what do you get for x=3?

6. DLS

and x=1 and x=2 too

7. anonymous

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

8. DLS

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

9. anonymous

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

10. DLS

no again

11. DLS

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

12. anonymous

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

13. DLS

tell me over one interval

14. DLS

for |x-1|

15. DLS

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

16. anonymous

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

17. anonymous

taking positive and negative value of piecewise function and solving the square root part greater than equal to 0 you get the following : $x \geq 0 \& x \geq 4$

18. anonymous

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

19. 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 $$x\geq 4$$

20. anonymous

So domain of the function is $x \geq 4$

21. DLS

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

22. anonymous

if $$x<1$$ then $$|x-1|=1-x$$

23. DLS

i wrote that too but why :|

24. anonymous

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

25. DLS

but modulus function makes it positive

26. anonymous

yes, and if $$x-1$$ is negative, then $$1-x$$ is positive

27. DLS

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

28. anonymous

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

29. DLS

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

30. anonymous

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

31. DLS

for kind of removing the mod right?

32. anonymous

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

33. 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

34. DLS

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

35. 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?