Here's the question you clicked on:
abccindy
does the mean value theorem apply to |x-1|?
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.
... you haven't given us enough information. It must be continuous over \([a,b]\) and differentiable over \((a,b)\), i.e. \(0\notin(a,b)\).
Ok, you tell us: is \(|x-1|\) continuous on \([a,b]\) and differentiable on \((a,b)\)?
Hint: |dw:1358042258899:dw| and try to answer questions by oldrin.
i got to that part but does the vertical asymptote have anything to do with it?
The main questions are: is f(x) continuous on (0,4), and is f(x) differentiable on [0,4]
@geerky42 are we able to determine the derivative of f(x)=|x-1| at x=1?
@abccindy same question for you then: are we able to determine the derivative of f(x)=|x-1| at x=1?
If it is zero, then the answer is yes, because zero is a number.
Is the derivative really zero?
\[f'(x)=\frac{ x-1 }{ |x-1| }\]
For the derivative to exist, the left and right limits of f(x) must be equal.
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.
Would you propose the left limit, i.e. f'(1-), and the right limt f'(1+).
So if the derivative is undefined at x=1, what can we conclude if the MVT applies to the interval (0,4)?
they are all undefined?
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.