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#1 2007-01-01 05:24:15

Prakash Panneer
Member
Registered: 2006-06-01
Posts: 110

Mean Value Theorem

Please help me with that.....



Give an example of a function on a closed interval which satisfies the conclusion of the Mean Value Theorem, but not the hypothesis.

Thanks in Advance up


Letter, number, arts and science
of living kinds, both are the eyes.

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#2 2007-01-01 08:08:38

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: Mean Value Theorem

do you mean the mean value theorem for derivatives or the mean value theorem for integrals?


A logarithm is just a misspelled algorithm.

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#3 2007-01-01 08:35:39

Prakash Panneer
Member
Registered: 2006-06-01
Posts: 110

Re: Mean Value Theorem

Mean value theorem for derivatives.


Letter, number, arts and science
of living kinds, both are the eyes.

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#4 2007-01-01 11:53:08

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: Mean Value Theorem

hope i don't screw the precise definition up, but as far as i remember, the mean value states that if f(x) is differentiable and continuous on an open interval (a,  b) then there is some number c betweeen a and b (a <= c <= b) such that f'(c) = (f(a) - f(b) )/(a - b)

Usually all they want is to find two points on the graph of the function, find the slope of the line that cuts through both of them, and find a point thats horizontally between those two points where the derivitive is equal to that same slope.

Here's how you would go about answering the problem:

let f(x) = x^2  then f'(x) = 2x, lets look at the open interval (1, 3)  so the two points on the graph are (1, f(1)) and (3, f(3)) or in otherwords, (1,1) and (3, 9) the slope of the line that passes through these points is (9 - 1)/(3 - 1) which is 4. Now the mean value theorem states that somewhere between 1 and 3 (which are the endpoints of the interval) there bust be value c such that f'(c) = 4 (which is the slope of the line through the endpoints of the interval) so now we just need to show that there is a value of x such that f'(x) = 4 and is between 1 and 3. If f'(x) = 4 so 2x = 4 so x = 2. 2 is between 1 and 3 so thats how you demonstrate the mean value theorem.


A logarithm is just a misspelled algorithm.

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#5 2007-01-01 12:05:57

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: Mean Value Theorem

Heres how you might write down your answer:

let f(x) = x^2 on the interval (1, 3)

by the mean value there is some value c between 1 and 3 such that f'(c) = (f(3) - f(1))/(3 - 1) = 4.

f'(c) = 2c, if 2c = 4 then c = 2.  1 <= 2 <= 3. This confirms the mean value theorem.


A logarithm is just a misspelled algorithm.

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