Check the hypotheses of Rolle’s Theorem and the Mean Value Theorem and find a value of c that makes the appropriate conclusion true. Illustrate the conclusion with a graph. f(x) = x^2 + 1, [2,-2]

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- abb0t

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

- anonymous

so find the derivative of x^2 + 1?

- abb0t

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

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- anonymous

okay got you thanks

- abb0t

You don't need to check first two conditions since you know that your functions IS in fact continuous on the given interval and differentiable. So with that being said, best of luck.

- anonymous

@zepdrix can u help?

- anonymous

i did it but i think i got it wrong

- anonymous

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

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- anonymous

- experimentX

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

- experimentX

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

- anonymous

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

- experimentX

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

- experimentX

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

- anonymous

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

- anonymous

@zepdrix are u there?

- anonymous

@wio can u help

- anonymous

okay so its 0?

- experimentX

yes yes it is ..

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