anonymous
  • anonymous
does the mean value theorem apply to |x-1|?
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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mathmate
  • mathmate
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.
anonymous
  • anonymous
the interval is [0,4]
anonymous
  • anonymous
... you haven't given us enough information. It must be continuous over \([a,b]\) and differentiable over \((a,b)\), i.e. \(0\notin(a,b)\).

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More answers

anonymous
  • anonymous
Ok, you tell us: is \(|x-1|\) continuous on \([a,b]\) and differentiable on \((a,b)\)?
anonymous
  • anonymous
2 Attachments
mathmate
  • mathmate
Hint: |dw:1358042258899:dw| and try to answer questions by oldrin.
anonymous
  • anonymous
i got to that part but does the vertical asymptote have anything to do with it?
mathmate
  • mathmate
The main questions are: is f(x) continuous on (0,4), and is f(x) differentiable on [0,4]
anonymous
  • anonymous
yes to both
mathmate
  • mathmate
@geerky42 are we able to determine the derivative of f(x)=|x-1| at x=1?
mathmate
  • mathmate
@abccindy same question for you then: are we able to determine the derivative of f(x)=|x-1| at x=1?
anonymous
  • anonymous
no because it is 0
mathmate
  • mathmate
If it is zero, then the answer is yes, because zero is a number.
anonymous
  • anonymous
it is 1-1/ |1-1|
mathmate
  • mathmate
Is the derivative really zero?
anonymous
  • anonymous
\[f'(x)=\frac{ x-1 }{ |x-1| }\]
mathmate
  • mathmate
For the derivative to exist, the left and right limits of f(x) must be equal.
mathmate
  • mathmate
...exist and equal.
anonymous
  • anonymous
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.
mathmate
  • mathmate
Would you propose the left limit, i.e. f'(1-), and the right limt f'(1+).
anonymous
  • anonymous
Yes that is correct.
mathmate
  • mathmate
So if the derivative is undefined at x=1, what can we conclude if the MVT applies to the interval (0,4)?
anonymous
  • anonymous
they are all undefined?
mathmate
  • mathmate
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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