## BlingBlong Group Title How would I graph an expression like |x^(2)-1| I know I would first convert it to a piece wise function where: 2 years ago 2 years ago

1. Hermella Group Title

WHOA 2 LAZY 2 ANSWER BOO HOO

2. BlingBlong Group Title

x^(2)-1 if x^(2) -1 >=0 and -x^(2) + 1 x^(2) -1 <0

3. Hermella Group Title

FIGURE IT OUTIT IS YOUR HW

4. Callum29 Group Title

Isn't it just the graph of x^2 - 1 but with the part which is below the x-axis inverted so that it is above the x-axis? You will lose differentiability at x=-1 and x=1 but continuity is preserved so there is no need to define it piece-wise?

5. Hermella Group Title

MAKE ME

6. BlingBlong Group Title

So x^(2)-1 if x^(2) -1 >=0 and -x^(2) + 1 x^(2) -1 <0 x^(2) - 1 if x >= 1 and -x^(2) + 1 if x < 1 Thus I can just input values into both sides, anyone able to provide me with a more difficult piece wise function. I know it is elementary I just want to refresh my memory.

7. myininaya Group Title

@Hermella please don't yell at people to just figure it out on their own. They come on this site to get help. If you don't want to help, then just ignore their question.

8. myininaya Group Title

So how would you just graph x^2-1? this is a parabola right with y-intercept -1 and x-intercept -1 and 1

9. myininaya Group Title

|dw:1326156295808:dw|

10. Callum29 Group Title

-x^(2) + 1 if x < 1 violates -x^(2) + 1 if x^(2) -1 <0. Take x = 1/2

11. myininaya Group Title

now what does -x^2+1 look like (note we will worry about the domains in a sec)

12. myininaya Group Title

|dw:1326156367729:dw|

13. myininaya Group Title

now the first graph was for x>=1 so we need to play with's domain

14. myininaya Group Title

|dw:1326156439379:dw|

15. myininaya Group Title

now for the 2nd part we have x<1

16. myininaya Group Title

|dw:1326156472771:dw|

17. myininaya Group Title

something like that

18. BlingBlong Group Title

Oh thanks for going through this step by step, it is clear to me now. Could you give me an example of a piece wise function that is not continuous if not I understand

19. myininaya Group Title

wait something is wrong with my graph its not all above the x-axis

20. myininaya Group Title

ok so we had $|x^2-1|=(x^2-1) \text{ if } x^2-1 \ge 0$ $|x^2-1|=-(x^2-1) \text{ if } x^2-1<0$ right?

21. BlingBlong Group Title

yup

22. BlingBlong Group Title

don't understand what is incorrect with your graph it seems correct to me

23. myininaya Group Title

$(x-1)(x+1) \ge 0$ this is 0 when x=1 and x=-1 ---|----|--- -1 1 now we need to test where (x-1)(x+1) is greater than 0 so we try the first interval (-inf,-1) we choose a number like -2 (-2-1)(-2+1)=(-3)(-1)=3>0 so on the interval (-inf,-1)=> (x-1)(x+1)>0 what about the interval (-1,1) we choose a number like 0 (0-1)(0+1)=-1(1)=-1<0 so on this interval we have that (x-1)(x+1)<0 now last interval (1,inf) we choose a number to try like 2 (2-1)(2+1)=1(3)=3>0 so for the interval (1,inf)=> (x-1)(x+1)>0 so for this |x^2-1|=x^2-1 happens on (-inf,-1) and (1,inf) now lets check out the other equality |x^2-1|=-(x^2-1)=-x^2+1 this is for x^2-1<0 (x-1)(x+1)<0 btw (-1,1)

24. myininaya Group Title

there we go

25. myininaya Group Title

i should have done all the work earlier i just skipped this algebra part

26. myininaya Group Title

so do you understand |x^2-1|=x^2-1 on the intervals (-inf,-1), (1,inf) and |x^2-1|=-(x^2-1) on the interval (-1,1)

27. myininaya Group Title

now back to our parabolas

28. BlingBlong Group Title

so (1,-1) does not exist in the domain?

29. myininaya Group Title

so x^2-1 we had |dw:1326157039943:dw| but we didn't have it for (-1,1)

30. myininaya Group Title

so we actually have |dw:1326157076390:dw|

31. myininaya Group Title

that is for the |x^2-1|=x^2-1 for the intervals (-inf,-1),(1,inf)

32. BlingBlong Group Title

why is (-1, 1) excluded from the domain I don't understand

33. myininaya Group Title

now for the middle part (-1,1) we have |x^2-1|=-(x^2-1)

34. myininaya Group Title

|dw:1326157146780:dw|

35. myininaya Group Title

|x^2-1|=x^2-1 for x^2-1>0 right?

36. myininaya Group Title

now did you solve x^2-1>0?

37. myininaya Group Title

you will get that |x^2-1|=x^2-1 is only true for the intervals (-inf,-1),(1,inf)

38. myininaya Group Title

now for |x^2-1|=-(x^2-1) is only true for the intervals (-1,1)

39. BlingBlong Group Title

oh, because it is the only area of the graph that will make the graph less than zero

40. myininaya Group Title

yes x^2-1<0 on the interval (-1,1)

41. BlingBlong Group Title

so when solving these it is important to make a chart and plug in numbers between the intervals

42. myininaya Group Title

yes most definitely

43. BlingBlong Group Title

Thanks I think it is a bit more clear in my head how to solve these now I need to focus on the output of the piece wise function and use that to guide me on how to graph the function

44. BlingBlong Group Title

after considering the graph of the various rules of the function