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Math Help - Continuous function problem

  1. #1
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    Continuous function problem

    Q: Suppose that f: [0,2] -> [0,4] is continuous. Show that there is an x \epsilon[0,2] such that f(x) = 2x.

    Proof: All I know is use the intermediate value theorem...
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  2. #2
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    What happens if f is nowhere differentiable?
    Are you saying that f is a surjection (onto)?
    There has got to be more to this question than you posted!
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  3. #3
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    That is the whole problem. But there is a hint here provided by my professor:

    Hint: Consider the function g: [0,2] -> R defined by g(x) = f(x) - 2x and use the intermediate value theorem.

    Now, again, we have not learn about using derivatives in proof yet.

    Thank you!
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  4. #4
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    Define a function on the interval [0,2] as,
    g(x)=f(x)-2x
    Since, f(x),2x are countinous on [0,2] so too is g(x)=f(x)-2x continous on [0,2].
    Now,
    g(0)=f(0)-0=f(0)
    g(2)=f(2)-4

    By definition f(0)\geq 0

    If f(0)=0 then the proof is complete.
    Because x=0\in [0,1] is one such point. So there is no harm in assuming that f(0)>0

    By definition f(2)\leq 4

    If f(2)=4 then the proof is complete.
    Because x=2\in [0,2] is one such point. So there is no harm in assuming that f(2)<4

    Thus, g(0)=f(0)>0 and g(2)=f(2)-4<0.

    Since g(x) is continous on [0,2] there is a point c\in [0,2] such that g(c)=0 by the intermediate value theorem.
    Thus, g(c)=f(c)-2c=0. Thus, f(c)=2c for some point in [0,2].
    Q.E.D.
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