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Math Help - More calculus ideas and proofs.

  1. #1
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    More calculus ideas and proofs.

    1. For a continuous, differentiable function f(x), and an interval [a,b], supose that f(a) and f(b) have opposite signs and that f' does not equal 0. Show that f(x) is exactly one root between a and b.

    2. Show that for any numbers a and b, the inequality |sin(b) - sin(a)| is less than or equal to |a - b| is true.
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    1. Let's suppose it has more than one (it clearly has at least one by Bolzano), which would be, at least 2.

    x_1 and x_2 denote 2 of our roots

    We are supposing that
    x_1\in(a,b) and x_2\in(a,b) with x_1\neq{x_2} and f(x_1)=f(x_2)=0

    Then, by Rolle's Theorem, there exists c\in{(x_1,x_2)} such that: f'(c)=0

    But f'(x) can't be 0 in (a,b)

    Thus it is absurd to suppose it has more than 1 root
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    Quote Originally Posted by PaulRS View Post
    1. Let's suppose it has more than one (it clearly has at least one by Bolzano), which would be, at least 2.

    x_1 and x_2 denote 2 of our roots

    We are supposing that
    x_1\in(a,b) and x_2\in(a,b) with x_1\neq{x_2} and f(x_1)=f(x_2)=0

    Then, by Rolle's Theorem, there exists c\in{(x_1,x_2)} such that: f'(c)=0

    But f'(x) can't be 0 in (a,b)

    Thus it is absurd to suppose it has more than 1 root
    What do you mean by "Bolzano" is that how you say it in Ugurarian (that word probably does not exist ).
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    Quote Originally Posted by Prophet View Post
    2. Show that for any numbers a and b, the inequality |sin(b) - sin(a)| is less than or equal to |a - b| is true.
    Consider \sin x on [a,b] then by mean value there is x_0\in (a,b) so that \sin a - \sin b = \cos x_0 (a-b) so |\sin a - \sin b| = |\cos x_0||a-b| but |cos x_0|\leq 1 that means |\sin a - \sin b|\leq |a-b|.
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    Quote Originally Posted by ThePerfectHacker View Post
    What do you mean by "Bolzano"
    I was talking about Bolzano's Theorem

    Sorry, I am used to call it Bolzano directly
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    Quote Originally Posted by PaulRS View Post
    I was talking about Bolzano's Theorem

    Sorry, I am used to call it Bolzano directly
    There is something called the Bolzano-Weierstrass Theorem which is one of the most important theorems in Analysis.

    It says if x_n is a bounded sequence we can find a convergent sub-sequence.

    What you used: "If f(x) is continous on [a,b] and f(a)<0,f(b)>0 there exists a c\in(a,b) so that f(c)=0" is called Intermediate-Value theorem.

    (The intermediate-value theorem can be easily proven using Bolzano-Weierstrass theorem so maybe that is why you referred to it as Bolzano theorem. But if so, then how can you possibly leave out 'Weierstrass', he might get insulted).
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  7. #7
    Super Member PaulRS's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    What you used: "If f(x) is continous on [a,b] and f(a)<0,f(b)>0 there exists a c\in(a,b) so that f(c)=0"
    Well, that is what in Uruguay we commonly call Bolzano's Theorem.
    I thought that it was broadly called so.
    PlanetMath: Bolzano's theorem
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