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mathmate
 2 years ago
Best ResponseYou've already chosen the best response.1The 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.

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

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

mathmate
 2 years ago
Best ResponseYou've already chosen the best response.1Hint: dw:1358042258899:dw and try to answer questions by oldrin.

abccindy
 2 years ago
Best ResponseYou've already chosen the best response.0i got to that part but does the vertical asymptote have anything to do with it?

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

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

mathmate
 2 years ago
Best 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?

mathmate
 2 years ago
Best ResponseYou've already chosen the best response.1If it is zero, then the answer is yes, because zero is a number.

mathmate
 2 years ago
Best ResponseYou've already chosen the best response.1Is the derivative really zero?

abccindy
 2 years ago
Best ResponseYou've already chosen the best response.0\[f'(x)=\frac{ x1 }{ x1 }\]

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

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.1The 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
 2 years ago
Best ResponseYou've already chosen the best response.1Would you propose the left limit, i.e. f'(1), and the right limt f'(1+).

mathmate
 2 years ago
Best ResponseYou've already chosen the best response.1So if the derivative is undefined at x=1, what can we conclude if the MVT applies to the interval (0,4)?

abccindy
 2 years ago
Best ResponseYou've already chosen the best response.0they are all undefined?

mathmate
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, 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.
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