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Math Help - Proof involving continuity of two functions on the interval [a, b]?

  1. #1
    Member mfetch22's Avatar
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    Proof involving continuity of two functions on the interval [a, b]?

    I was wondering if somebody could check my proof for the following problem. I'm failry certain its correct (atleast correct in the general direction of the proof), I just wanted to make sure. Firstly, the proof makes use of the following theorems / properties:


    [1]

    If f is continuous on [a, b] (i.e. lim_{x \to a} f(x) = f(a) for all x \in [a, b]) and g is continuous on [a, b] then f - g is continuous on [a, b].

    [2]

    If f is continuous on [a, b], and f(a) < 0 < f(b), then f(x) = 0 for some x \in [a, b]

    Now heres the problem:


    Given that f and g are both continuous on [a, b], that f(a) < g(a), and that g(b) < f(b); then prove that f(x) = g(x) for some x \in [a, b]
    Now heres my proof. I know this problem will probably seem trivial and super easy to allot of people, but I just wanted to make sure I'm doing it correctly:


    Showing that:

    f(x) = g(x)

    For some x \in [a, b] is equivelant showing that:

    f(x) - g(x) = 0

    For some x \in [a, b]

    So, lets let:

    h(x) = f(x) - g(x)

    By the given information and property/theorem number [1], we know that h is continuous on [a, b]

    Now, since:

    f(a) < g(a)

    then:

    f(a) - g(a) = h(a) < 0

    And since:

    g(b) < f(b)

    then:

    0 < h(b) = f(b) - g(b)

    So we have:

    f(a) - g(a) < 0 < f(b) - g(b)

    Equivilantly...

    h(a) < 0 < h(b)

    Since we also know that h is continuous on [a, b], then by property/theorem [2] we are garunteed that:

    h(x) = 0

    For some x \in [a, b]

    Thus:

    f(x) - g(x) = 0

    For some x \in [a, b]

    Therefore we've shown we must have:

    f(x) = g(x)

    For some x \in [a, b]

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  2. #2
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    Looks good!
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