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

Question

mathmate · Answer

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 · Answer

the interval is [0,4]

anonymous · Answer

... you haven't given us enough information. It must be continuous over $[a,b]$ and differentiable over $(a,b)$, i.e. $0\notin(a,b)$.

anonymous · Answer

Ok, you tell us: is $|x-1|$ continuous on $[a,b]$ and differentiable on $(a,b)$?

anonymous · Answer

attachment

mathmate · Answer

Hint:
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and try to answer questions by oldrin.

anonymous · Answer

i got to that part but does the vertical asymptote have anything to do with it?

mathmate · Answer

The main questions are:
is f(x) continuous on (0,4), and
is f(x) differentiable on [0,4]

anonymous · Answer

yes to both

mathmate · Answer

@geerky42  are we able to determine the derivative of f(x)=|x-1| at x=1?

mathmate · Answer

@abccindy   same question for you then:
are we able to determine the derivative of f(x)=|x-1| at x=1?

anonymous · Answer

no because it is 0

mathmate · Answer

If it is zero, then the answer is yes, because zero is a number.

anonymous · Answer

it is 1-1/ |1-1|

mathmate · Answer

Is the derivative really zero?

anonymous · Answer

$$f'(x)=\frac{ x-1 }{ |x-1| }$$

mathmate · Answer

For the derivative to exist, the left and right limits of f(x) must be equal.

mathmate · Answer

...exist and equal.

anonymous · Answer

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 · Answer

Would you propose the left limit, i.e. f'(1-), and the right limt f'(1+).

anonymous · Answer

Yes that is correct.

mathmate · Answer

So if the derivative is undefined at x=1, what can we conclude if the MVT applies to the interval (0,4)?

anonymous · Answer

they are all undefined?

mathmate · Answer

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.