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anonymous
 3 years ago
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]
anonymous
 3 years ago
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
 3 years ago
Best ResponseYou've already chosen the best response.1Well, 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so find the derivative of x^2 + 1?

abb0t
 3 years ago
Best ResponseYou've already chosen the best response.1Yes. Find the derivative. Use the formula f'(c) = to solve for "c".

abb0t
 3 years ago
Best ResponseYou've already chosen the best response.1You 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
 3 years ago
Best ResponseYou've already chosen the best response.0i did it but i think i got it wrong

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0f'(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 :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@abb0t @zepdrix @experimentX ?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Did i go wrong when i found f'x? and find the f prime of c?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0equate it to zero ... and you get x=0

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay so c = to 0 right? i made a mistake in putting 2 and 2 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@zepdrix are u there?
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