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Math Help - Easy Analysis IVT

  1. #1
    Senior Member slevvio's Avatar
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    Easy Analysis IVT

    Help!

    Q: 1) State the intermediate value theorem
    2) Suppose  f:[0,1] \rightarrow \mathbb{R} is continous on  [0,1] ,  f(0) = 0 and  f(1)=1 .
    Prove that there exists a  c \in (0,1) such that  f(c) = 1-c .

    My solution:

    A: 1) Suppose  f:[a,b] \rightarrow \mathbb{R} is continuous on  [a,b] with  f(a) < 0 < f(b) .

    Then  \exists c \in (a,b)  such that  f(c) = 0 .

    2) ????????????????

    Any help would be appreciated thanks
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  2. #2
    MHF Contributor

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    HINT for #2.
    Is g(x)=x+f(x)-1 a continuous function? WHY?
    g(0)=?~\&~g(1)=?
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  3. #3
    Senior Member slevvio's Avatar
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    g is continous on [0,1] because its limit as x tends to c is g(c).

    thanks !
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