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Math Help - rolle's theorem

  1. #1
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    rolle's theorem

    for g(x)=x^3 + 3x^2 -2, show that there must ne one real root on [0,2]
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  2. #2
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    Quote Originally Posted by wisezeta View Post
    for g(x)=x^3 + 3x^2 -2, show that there must ne one real root on [0,2]
    try instead the Intermediate Value Theorem.
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    Quote Originally Posted by skeeter View Post
    try instead the Intermediate Value Theorem.
    can u do it using rolle's theorem?
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    Quote Originally Posted by airportman92 View Post
    can u do it using rolle's theorem?
    Rolle's theorem says that if a function is continuous on [a,b], differentiable on (a,b), and f(a) = f(b) , then there is at least one value c in the interval (a,b) where f'(c) = 0.

    I'm not saying it can't be done, but it would not be as simple as using the IVT.
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    Quote Originally Posted by skeeter View Post
    Rolle's theorem says that if a function is continuous on [a,b], differentiable on (a,b), and f(a) = f(b) , then there is at least one value c in the interval (a,b) where f'(c) = 0.

    I'm not saying it can't be done, but it would not be as simple as using the IVT.

    how would you do it using rolle's theorem though just for fun
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  6. #6
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    Quote Originally Posted by airportman92 View Post
    how would you do it using rolle's theorem though just for fun
    it would not be fun.
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  7. #7
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    Quote Originally Posted by skeeter View Post
    it would not be fun.
    but how do u do it hah
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  8. #8
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    Quote Originally Posted by airportman92 View Post
    but how do u do it hah
    I'm not sure.
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  9. #9
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    someone how do you use rolles theorem to do this one lol
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  10. #10
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    It's a useless detour (at least how I did it is), but since you want it so badly:

    Let g_c(x)= \frac{x^4}{4} + x^3 -2x +c=4x(x^3+4x^2-2)+c Let h(x)=x^3+4x^2-2 then h(0)<0 and h(2)>0 so by the intermediate value theorem there exists d\in (0,2) such that h(d)=0 and so g_c(0)=g_c(d)=c and by Rolle's theorem there exists y\in(0,d) such that g'_c (y)=0 but g'_c=g

    As skeeter said, not pretty at all.
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