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abccindy Group Title

does the mean value theorem apply to |x-1|?

  • one year ago
  • one year ago

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  1. mathmate Group Title
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    The mean value theorem applies to a specified closed interval, and it must be differentiable within the (open) interval. This should help you put restriction on whether the MVT applies to |x-1| or not.

    • one year ago
  2. abccindy Group Title
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    the interval is [0,4]

    • one year ago
  3. oldrin.bataku Group Title
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    ... you haven't given us enough information. It must be continuous over \([a,b]\) and differentiable over \((a,b)\), i.e. \(0\notin(a,b)\).

    • one year ago
  4. oldrin.bataku Group Title
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    Ok, you tell us: is \(|x-1|\) continuous on \([a,b]\) and differentiable on \((a,b)\)?

    • one year ago
  5. abccindy Group Title
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    • one year ago
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  6. mathmate Group Title
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    Hint: |dw:1358042258899:dw| and try to answer questions by oldrin.

    • one year ago
  7. abccindy Group Title
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    i got to that part but does the vertical asymptote have anything to do with it?

    • one year ago
  8. mathmate Group Title
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    The main questions are: is f(x) continuous on (0,4), and is f(x) differentiable on [0,4]

    • one year ago
  9. abccindy Group Title
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    yes to both

    • one year ago
  10. mathmate Group Title
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    @geerky42 are we able to determine the derivative of f(x)=|x-1| at x=1?

    • one year ago
  11. mathmate Group Title
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    @abccindy same question for you then: are we able to determine the derivative of f(x)=|x-1| at x=1?

    • one year ago
  12. abccindy Group Title
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    no because it is 0

    • one year ago
  13. mathmate Group Title
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    If it is zero, then the answer is yes, because zero is a number.

    • one year ago
  14. abccindy Group Title
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    it is 1-1/ |1-1|

    • one year ago
  15. mathmate Group Title
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    Is the derivative really zero?

    • one year ago
  16. abccindy Group Title
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    \[f'(x)=\frac{ x-1 }{ |x-1| }\]

    • one year ago
  17. mathmate Group Title
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    For the derivative to exist, the left and right limits of f(x) must be equal.

    • one year ago
  18. mathmate Group Title
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    ...exist and equal.

    • one year ago
  19. oldrin.bataku Group Title
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    The derivative is undefined... c'mon people! There are two different limits of for the slope at x=1 depending on whether you approach from the left or right.

    • one year ago
  20. mathmate Group Title
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    Would you propose the left limit, i.e. f'(1-), and the right limt f'(1+).

    • one year ago
  21. abccindy Group Title
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    Yes that is correct.

    • one year ago
  22. mathmate Group Title
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    So if the derivative is undefined at x=1, what can we conclude if the MVT applies to the interval (0,4)?

    • one year ago
  23. abccindy Group Title
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    they are all undefined?

    • one year ago
  24. mathmate Group Title
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    Yes, the derivative at x=1 is undefined because f'(1-)=-1, f'(1+)=+1, so f'(1-)\(\ne\)f'(1+), by definition of the derivative, it is undefined at x=1. This concept is even more important when you work with multi-variable calculus in the near future.

    • one year ago
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