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BlingBlong

  • 2 years ago

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

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  1. Hermella
    • 2 years ago
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    WHOA 2 LAZY 2 ANSWER BOO HOO

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

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

  4. Callum29
    • 2 years ago
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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?

  5. Hermella
    • 2 years ago
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    MAKE ME

  6. BlingBlong
    • 2 years ago
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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.

  7. myininaya
    • 2 years ago
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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.

  8. myininaya
    • 2 years ago
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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

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

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

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

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

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

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

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

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

  17. myininaya
    • 2 years ago
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    something like that

  18. BlingBlong
    • 2 years ago
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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

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

  20. myininaya
    • 2 years ago
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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?

  21. BlingBlong
    • 2 years ago
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    yup

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

  23. myininaya
    • 2 years ago
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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)

  24. myininaya
    • 2 years ago
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    there we go

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

  26. myininaya
    • 2 years ago
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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)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  42. myininaya
    • 2 years ago
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    yes most definitely

  43. BlingBlong
    • 2 years ago
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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

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

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