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Math Help - Continuity Proof

  1. #1
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    Continuity Proof

    Sorry reposted:

    Let f(x) and g(x) be fcns de fned on an interval I and let h(x) = max(f(x); g(x)) and j(x) = min(f(x); g(x)). h(x) and
    j(x) are continuous on I. Show f(x) and g(x) are continuous on I or give a counterexample.
    Last edited by seams192; May 13th 2010 at 10:43 AM. Reason: reposted
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by seams192 View Post
    Let f(x) and g(x) be functions defi ned on an open interval I and let h(x) = max{f(x); g(x)} and j(x) = min{f(x); g(x)}. Suppose that h(x) and
    j(x) are continuous on I. Either show that f(x) and g(x) are also continuous on I or else give a counterexample.


    Any ideas on how to structure this proof using contradiction?

    Thanks!
    How much do you know about analysis?
    Hint:

    Spoiler:


    \left(f\vee g\right)^{-1}((a,\infty))=\left\{x\in I:f(x)>a\right\}\cup\left\{x\in I:g(x)>a\right\}
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  3. #3
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    I know analysis at an introduction course level. I know about delta-epsilon, continuity definitions, IVT, etc.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by seams192 View Post
    I know analysis at an introduction course level. I know about delta-epsilon, continuity definitions, IVT, etc.
    Hmm. Alright, so I guess the easiest way to say it then would be that \max\{f,g\}=\frac{f+g+|f-g|}{2}, can you show that's continuous by using some of your known theorems?
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  5. #5
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    Well, we know that the function h(x) is continuous, so the max(f,g) would be continuous. I know that there are rules for algebraic combinations of continuous functions being continuous, so I'm wondering if the max function can be worked backwards to prove f and g continuous that way. Is that where you were going with this?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by seams192 View Post
    Well, we know that the function h(x) is continuous, so the max(f,g) would be continuous. I know that there are rules for algebraic combinations of continuous functions being continuous, so I'm wondering if the max function can be worked backwards to prove f and g continuous that way. Is that where you were going with this?
    Oh, wow! I'm sorry, I completely misread the question. Please forgive me, I have to go. I hope another member can help you. Once again, I'm sorry
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    Quote Originally Posted by seams192 View Post
    Let f(x) and g(x) be functions defi ned on an open interval I and let h(x) = max{f(x); g(x)} and j(x) = min{f(x); g(x)}. Suppose that h(x) and
    j(x) are continuous on I. Either show that f(x) and g(x) are also continuous on I or else give a counterexample.


    Any ideas on how to structure this proof using contradiction?
    A better idea would be to look for a counterexample, perhaps by taking f to be a function that only takes the values 0 and 1, with a jump discontinuity at some point. Then think about how to construct g so as to make h and j continuous.
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  8. #8
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    Quote Originally Posted by seams192 View Post
    Let f(x) and g(x) be functions defi ned on an open interval I and let h(x) = max{f(x); g(x)} and j(x) = min{f(x); g(x)}. Suppose that h(x) and
    j(x) are continuous on I. Either show that f(x) and g(x) are also continuous on I or else give a counterexample.


    Any ideas on how to structure this proof using contradiction?

    Thanks!
    take two parts, first suposse than x (element of domain) is point internal, and demostrate than is continiuos

    second suposse that x satisfy f(x)=g(x) and probe that is continius.
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  9. #9
    Member mabruka's Avatar
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    Or simply take these two easy functions on [0,1]


    f(x)=1 if 0\leq x<\frac{1}{2}
    -1 if     \frac{1}{2} \leq x \leq 1


    g(x)= -1 if   0\leq x <\frac{1}{2}
    1 if    \frac{1}{2} \leq x \leq1


    both are discontinuous but max and min functions h,j are constant hence continuous
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  10. #10
    Senior Member Pinkk's Avatar
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    Uh, don't delete the question just because it has been answered.
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  11. #11
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    sorry pinkk...I put it back up
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