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abccindy

  • 2 years ago

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

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  1. mathmate
    • 2 years ago
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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.

  2. abccindy
    • 2 years ago
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    the interval is [0,4]

  3. oldrin.bataku
    • 2 years ago
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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. \(0\notin(a,b)\).

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

  5. abccindy
    • 2 years ago
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  6. mathmate
    • 2 years ago
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    Hint: |dw:1358042258899:dw| and try to answer questions by oldrin.

  7. abccindy
    • 2 years ago
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    i got to that part but does the vertical asymptote have anything to do with it?

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

  9. abccindy
    • 2 years ago
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    yes to both

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

  11. mathmate
    • 2 years ago
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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?

  12. abccindy
    • 2 years ago
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    no because it is 0

  13. mathmate
    • 2 years ago
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    If it is zero, then the answer is yes, because zero is a number.

  14. abccindy
    • 2 years ago
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    it is 1-1/ |1-1|

  15. mathmate
    • 2 years ago
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    Is the derivative really zero?

  16. abccindy
    • 2 years ago
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    \[f'(x)=\frac{ x-1 }{ |x-1| }\]

  17. mathmate
    • 2 years ago
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    For the derivative to exist, the left and right limits of f(x) must be equal.

  18. mathmate
    • 2 years ago
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    ...exist and equal.

  19. oldrin.bataku
    • 2 years ago
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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.

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

  21. abccindy
    • 2 years ago
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    Yes that is correct.

  22. mathmate
    • 2 years ago
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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)?

  23. abccindy
    • 2 years ago
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    they are all undefined?

  24. mathmate
    • 2 years ago
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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.

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