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Thread: 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 $\displaystyle f:[0,1] \rightarrow \mathbb{R} $ is continous on $\displaystyle [0,1] $, $\displaystyle f(0) = 0 $ and $\displaystyle f(1)=1 $.
    Prove that there exists a $\displaystyle c \in (0,1) $ such that $\displaystyle f(c) = 1-c $.

    My solution:

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

    Then $\displaystyle \exists c \in (a,b) $ such that $\displaystyle 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 $\displaystyle g(x)=x+f(x)-1$ a continuous function? WHY?
    $\displaystyle g(0)=?~\&~g(1)=?$
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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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