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u0860867

  • one year ago

Urgent help needed Pls Help

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  1. u0860867
    • one year ago
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  2. allank
    • one year ago
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    Do u understand the concept of continuity?

  3. u0860867
    • one year ago
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    Not completely @ allank

  4. allank
    • one year ago
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    Okay. Lemme walk u through it..

  5. u0860867
    • one year ago
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    @allank

  6. allank
    • one year ago
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    If we have the graph of a function, it is continuous if it is differentiable at all points i.e. it has no breaks, sharp corners, etc.

  7. Euler271
    • one year ago
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    if lim of f(x) as x approaches a from the left = the limit as x approaches a from the right, the limit exists. if the limit exists at a, then the function is continuous at a. [square brackets] mean that the number is included in the interval (round brackets) mean that the point is not included in the interval.

  8. aditya96
    • one year ago
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    [-4,2),(-2,2),[2,4),(4,6),(6,8)

  9. allank
    • one year ago
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    Carry on @Euler271

  10. u0860867
    • one year ago
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    @allank so when u say sharp corners do you also mean by turning points where it slowly rises and then falls

  11. allank
    • one year ago
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    No, the function is differentiable there. |dw:1369549739726:dw|

  12. u0860867
    • one year ago
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    @allank so from this graph how do we determine where the graph is continous??? Sorry for hassling u again and again

  13. u0860867
    • one year ago
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    @Euler271 do u think you could help me solve this question as well

  14. allank
    • one year ago
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    No prob. So avoid intervals where we breaks like: |dw:1369550343320:dw|

  15. allank
    • one year ago
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    And vertical asymptotes like: |dw:1369550390687:dw|

  16. RaphaelFilgueiras
    • one year ago
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    |dw:1369550623413:dw|

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