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mathmate Group TitleBest ResponseYou've already chosen the best response.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 x1 or not.
 one year ago

abccindy Group TitleBest ResponseYou've already chosen the best response.0
the interval is [0,4]
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.1
... 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

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.1
Ok, you tell us: is \(x1\) continuous on \([a,b]\) and differentiable on \((a,b)\)?
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
Hint: dw:1358042258899:dw and try to answer questions by oldrin.
 one year ago

abccindy Group TitleBest ResponseYou've already chosen the best response.0
i got to that part but does the vertical asymptote have anything to do with it?
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
The main questions are: is f(x) continuous on (0,4), and is f(x) differentiable on [0,4]
 one year ago

abccindy Group TitleBest ResponseYou've already chosen the best response.0
yes to both
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
@geerky42 are we able to determine the derivative of f(x)=x1 at x=1?
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
@abccindy same question for you then: are we able to determine the derivative of f(x)=x1 at x=1?
 one year ago

abccindy Group TitleBest ResponseYou've already chosen the best response.0
no because it is 0
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
If it is zero, then the answer is yes, because zero is a number.
 one year ago

abccindy Group TitleBest ResponseYou've already chosen the best response.0
it is 11/ 11
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
Is the derivative really zero?
 one year ago

abccindy Group TitleBest ResponseYou've already chosen the best response.0
\[f'(x)=\frac{ x1 }{ x1 }\]
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
For the derivative to exist, the left and right limits of f(x) must be equal.
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
...exist and equal.
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.1
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

mathmate Group TitleBest ResponseYou've already chosen the best response.1
Would you propose the left limit, i.e. f'(1), and the right limt f'(1+).
 one year ago

abccindy Group TitleBest ResponseYou've already chosen the best response.0
Yes that is correct.
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
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

abccindy Group TitleBest ResponseYou've already chosen the best response.0
they are all undefined?
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
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 multivariable calculus in the near future.
 one year ago
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