anonymous
  • anonymous
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]
Mathematics
chestercat
  • chestercat
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abb0t
  • 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
  • anonymous
so find the derivative of x^2 + 1?
abb0t
  • abb0t
Yes. Find the derivative. Use the formula f'(c) = to solve for "c".

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anonymous
  • anonymous
okay got you thanks
abb0t
  • 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
  • anonymous
@zepdrix can u help?
anonymous
  • anonymous
i did it but i think i got it wrong
anonymous
  • 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
anonymous
  • anonymous
@abb0t @zepdrix @experimentX ?
experimentX
  • experimentX
http://en.wikipedia.org/wiki/Rolle's_theorem
experimentX
  • experimentX
pick any two points ... on two sides of 0, you see that it satisfies the Rolle's theorem.
anonymous
  • anonymous
Did i go wrong when i found f'x? and find the f prime of c?
experimentX
  • experimentX
equate it to zero ... and you get x=0
experimentX
  • experimentX
your graph is symmetric f(-2) - f(2) should be zero ... probably you made mistake somewhere.
anonymous
  • anonymous
okay so c = to 0 right? i made a mistake in putting -2 and 2 ?
anonymous
  • anonymous
@zepdrix are u there?
anonymous
  • anonymous
@wio can u help
anonymous
  • anonymous
okay so its 0?
experimentX
  • experimentX
yes yes it is ..

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