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BlingBlong

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

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  1. Hermella
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    WHOA 2 LAZY 2 ANSWER BOO HOO

    • 2 years ago
  2. BlingBlong
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    x^(2)-1 if x^(2) -1 >=0 and -x^(2) + 1 x^(2) -1 <0

    • 2 years ago
  3. Hermella
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    FIGURE IT OUTIT IS YOUR HW

    • 2 years ago
  4. Callum29
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    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?

    • 2 years ago
  5. Hermella
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    MAKE ME

    • 2 years ago
  6. BlingBlong
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    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.

    • 2 years ago
  7. myininaya
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    @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.

    • 2 years ago
  8. myininaya
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    So how would you just graph x^2-1? this is a parabola right with y-intercept -1 and x-intercept -1 and 1

    • 2 years ago
  9. myininaya
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    |dw:1326156295808:dw|

    • 2 years ago
  10. Callum29
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    -x^(2) + 1 if x < 1 violates -x^(2) + 1 if x^(2) -1 <0. Take x = 1/2

    • 2 years ago
  11. myininaya
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    now what does -x^2+1 look like (note we will worry about the domains in a sec)

    • 2 years ago
  12. myininaya
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    |dw:1326156367729:dw|

    • 2 years ago
  13. myininaya
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    now the first graph was for x>=1 so we need to play with's domain

    • 2 years ago
  14. myininaya
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    |dw:1326156439379:dw|

    • 2 years ago
  15. myininaya
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    now for the 2nd part we have x<1

    • 2 years ago
  16. myininaya
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    |dw:1326156472771:dw|

    • 2 years ago
  17. myininaya
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    something like that

    • 2 years ago
  18. BlingBlong
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    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

    • 2 years ago
  19. myininaya
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    wait something is wrong with my graph its not all above the x-axis

    • 2 years ago
  20. myininaya
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    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?

    • 2 years ago
  21. BlingBlong
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    yup

    • 2 years ago
  22. BlingBlong
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    don't understand what is incorrect with your graph it seems correct to me

    • 2 years ago
  23. myininaya
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    \[(x-1)(x+1) \ge 0 \] 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)

    • 2 years ago
  24. myininaya
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    there we go

    • 2 years ago
  25. myininaya
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    i should have done all the work earlier i just skipped this algebra part

    • 2 years ago
  26. myininaya
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    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)

    • 2 years ago
  27. myininaya
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    now back to our parabolas

    • 2 years ago
  28. BlingBlong
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    so (1,-1) does not exist in the domain?

    • 2 years ago
  29. myininaya
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    so x^2-1 we had |dw:1326157039943:dw| but we didn't have it for (-1,1)

    • 2 years ago
  30. myininaya
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    so we actually have |dw:1326157076390:dw|

    • 2 years ago
  31. myininaya
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    that is for the |x^2-1|=x^2-1 for the intervals (-inf,-1),(1,inf)

    • 2 years ago
  32. BlingBlong
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    why is (-1, 1) excluded from the domain I don't understand

    • 2 years ago
  33. myininaya
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    now for the middle part (-1,1) we have |x^2-1|=-(x^2-1)

    • 2 years ago
  34. myininaya
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    |dw:1326157146780:dw|

    • 2 years ago
  35. myininaya
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    |x^2-1|=x^2-1 for x^2-1>0 right?

    • 2 years ago
  36. myininaya
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    now did you solve x^2-1>0?

    • 2 years ago
  37. myininaya
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    you will get that |x^2-1|=x^2-1 is only true for the intervals (-inf,-1),(1,inf)

    • 2 years ago
  38. myininaya
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    now for |x^2-1|=-(x^2-1) is only true for the intervals (-1,1)

    • 2 years ago
  39. BlingBlong
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    oh, because it is the only area of the graph that will make the graph less than zero

    • 2 years ago
  40. myininaya
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    yes x^2-1<0 on the interval (-1,1)

    • 2 years ago
  41. BlingBlong
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    so when solving these it is important to make a chart and plug in numbers between the intervals

    • 2 years ago
  42. myininaya
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    yes most definitely

    • 2 years ago
  43. BlingBlong
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    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

    • 2 years ago
  44. BlingBlong
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    after considering the graph of the various rules of the function

    • 2 years ago
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