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Well, to satisfy rolle's theorem, you must satisfy the 3 conditions (you can find them in your book - if not, I can write them down for you if you don't have access to a book).
Now to find the value of "c" using mean value theorem, you must start by finding the derivative of the function
I'll leave you to find it's derivative as it is quite simple for this function.
Basically, it must follow the first two conditions of rolle's theorem to apply the formula
Then, plug it in to the formula: \[f'(c) = \frac{ f(b)-f(a) }{ b-a }\]
on the given interval, which for your case it's -2

so find the derivative of x^2 + 1?

Yes. Find the derivative.
Use the formula f'(c) =
to solve for "c".

okay got you thanks

i did it but i think i got it wrong

f'(c) = f(-2) - f(2)/ -2 - 2 so f(-2) -2^2 + 1 = -3, f(2) 2^2 + 1 = 5.
So the slope will be 2 because -3 - 5/-2-2 = -8/-4 which is 2.
So To c : f'(x) = 2x, f'(c) = 2c = 2
2c = 2 so c will equal one (1) but when i checked (a< c **
**

http://en.wikipedia.org/wiki/Rolle's_theorem

pick any two points ... on two sides of 0, you see that it satisfies the Rolle's theorem.

Did i go wrong when i found f'x? and find the f prime of c?

equate it to zero ... and you get x=0

your graph is symmetric
f(-2) - f(2) should be zero ... probably you made mistake somewhere.

okay so c = to 0 right? i made a mistake in putting -2 and 2 ?

okay so its 0?

yes yes it is ..