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onegirl
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]
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<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.
so find the derivative of x^2 + 1?
Yes. Find the derivative. Use the formula f'(c) = to solve for "c".
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.
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 <b ) (2, 1 , -2) it doesn't make sense because 1 is not less than -2 :/
@abb0t @zepdrix @experimentX ?
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 ?