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onegirl
Group Title
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]
 one year ago
 one year ago
onegirl Group Title
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]
 one year ago
 one year ago

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abb0t Group TitleBest ResponseYou've already chosen the best response.1
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: \[f'(c) = \frac{ f(b)f(a) }{ ba }\] on the given interval, which for your case it's 2<x<2 Now, set your derivative equal to that. and solve for "c" pick the value which matches the solution in the given interval. if the value falls outside of the given interval then the solution is excluded.
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
so find the derivative of x^2 + 1?
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.1
Yes. Find the derivative. Use the formula f'(c) = to solve for "c".
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
okay got you thanks
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix can u help?
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
i did it but i think i got it wrong
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
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  5/22 = 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 <b ) (2, 1 , 2) it doesn't make sense because 1 is not less than 2 :/
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
@abb0t @zepdrix @experimentX ?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
http://en.wikipedia.org/wiki/Rolle's_theorem
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
pick any two points ... on two sides of 0, you see that it satisfies the Rolle's theorem.
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
Did i go wrong when i found f'x? and find the f prime of c?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
equate it to zero ... and you get x=0
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
your graph is symmetric f(2)  f(2) should be zero ... probably you made mistake somewhere.
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
okay so c = to 0 right? i made a mistake in putting 2 and 2 ?
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix are u there?
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
@wio can u help
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
okay so its 0?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
yes yes it is ..
 one year ago
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