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Show that max(f,g) is continious?is my answer wrong?

  1. Jun 25, 2013 #1
    1. The problem statement, all variables and given/known data
    let f and g are continuous functions show that max(f,g) is continious


    3. The attempt at a solution
    the answer i gave is this
    maxf(g)=h(x)
    h(x)=f(x)==>f(x)-g(x)>0
    h(x)=g(x)==>f(x)-g(x)≤0
    thus h is continious

    now in the text book they defined max with the normal definition
    max=lf(x)-g(x)l+f(x)+g(x)/2
    is my answer wrong?
     
    Last edited: Jun 25, 2013
  2. jcsd
  3. Jun 25, 2013 #2

    micromass

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    Why does this imply that h is continuous?
     
  4. Jun 25, 2013 #3
    I thought it's obvious
    if f(x)>g(x) then h(x) = f(x) f continious then h continious
    if f(x)<=g(x) then h(x)=g(x) g gotinious thus h is continious
    Am i doing something wrong here?(spivak's calculus questions are quiet hard )
    Edit : i wasn't supposed to study cases?
     
  5. Jun 25, 2013 #4

    Mark44

    Staff: Mentor

    "Continious" is not a word - the one you mean is continuous.
    Presumably you mean max(f, g) = h(x)
    The above assume that f > g for all x in the domain or that g ≤ f for all x in the domain. What happens if one function is greater than the other in some places, but not in others?
    For absolute value, use |, not l (lower-case "ell"). Also, what gets divided by 2? As you wrote it, only g(x) is divided by 2. If that's not what you meant, use parentheses.
     
  6. Jun 25, 2013 #5

    micromass

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    This looks like the following proof:

    Consider ##f(x) = 1## for all ##x##, ##g(x) = -1## for all ##x## and take

    [tex]h(x) = \left\{\begin{array}{l} 1~\text{if}~x<0\\ -1~\text{if}~x\geq 0\end{array}\right.[/tex]

    If ##h(x)<0##, then ##h(x) = g(x)## and ##g## is continuous.
    If ##h(x)>0##, then ##h(x) = f(x)## and ##f## is continuous.
    Thus ##h## is continuous.

    This is essentially what you did. But the conclusion that ##h## is continuous is false of course.
     
  7. Jun 25, 2013 #6
    i was missing this part thank you
    micromass made me laugh a bit at my "proof" :).
    thanks guys .
     
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