calculus
posted by Jamie .
Let f(x) = (x+1)/(x1). Show that there are no vlue of c such that
f(2)f(0) =f'(c)(20). 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 ((x1)(x+1))/(x1)^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)/20) = 4/2 = 2
The derivative is
f'(x) = 2/(x1)^2
For that to equal 2, you must have
2 = 2/(x1)^2
(x1)^2 = 1
That is not possible since (x1)^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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