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Math Help - Analysis Function

  1. #1
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    Analysis Function

    Is there a function such that, f is continuous on (1,5), reaches a maximum on (1,5), but does not reach a minimum on (1,5). Does such a function exist or is it impossible?
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  2. #2
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    Quote Originally Posted by thahachaina View Post
    Is there a function such that, f is continuous on (1,5), reaches a maximum on (1,5), but does not reach a minimum on (1,5). Does such a function exist or is it impossible?
    f both reaches a maximum on (1,5) and a minimum on (1,5), never heard of that => i guess it reaches a max on (1,5), but not a minimum on (1,5) is possible.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by thahachaina View Post
    Is there a function such that, f is continuous on (1,5), reaches a maximum on (1,5), but does not reach a minimum on (1,5). Does such a function exist or is it impossible?
    Hint, basically think of a function that when differentiated gives a function such that \exists{x_1}\in(1,5)\backepsilon{f(x_1)=0}

    This sounds hard, but it isn't. Think of trigonmetric function with an altered period for one.
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  4. #4
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    Thanks man. I kinda thought about it, but it looks to me like that function is not possible since it does not reach a maximum at (1,5). Is that right?
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  5. #5
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    Quote Originally Posted by thahachaina View Post
    Is there a function such that, f is continuous on (1,5), reaches a maximum on (1,5), but does not reach a minimum on (1,5). Does such a function exist or is it impossible?

    f(x)=-(x-1)(x-5)

    has a maximum at x=3 but has no minimum on the open interval (1,5) .

    (If it were a closed interval it would be a different matter)

    CB
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