# calculus

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Let f(x) = (x+1)/(x-1). Show that there are no vlue of c such that
f(2)-f(0) =f'(c)(2-0). Why does this not contradict the Mean Value Theorem?

I plugged 2 and 0 into the original problem and got 3 and -1 . Then I found
the derivative to be ((x-1)-(x+1))/(x-1)^2. Whould would I do next? I am
confused at this part.

The mean value theorem says, in this example, that there IS some value of c in the interval such that f'(c) = f(2)-f(0)/2-0) = 4/2 = 2

The derivative is
f'(x) = -2/(x-1)^2

For that to equal 2, you must have
2 = -2/(x-1)^2
(x-1)^2 = -1
That is not possible since (x-1)^2 must always be positive.

The reason the Mean Value Theorm seems to be violated here is that the function f(x) is not continuous in the inveral x = 0 to 2. The Mean Value Theorem does not apply in such situations.

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