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Composite function and continuerty

  1. Dec 7, 2005 #1
    Hello I have the following problem:

    Given two function f and g which are continuer on R, and some point c which belongs to R.

    I'm suppose to show that if f(c) = g(c), then h is continious on R.

    Isn't that the same as showing that that the composite function

    h(c) = g(f(c)) is continues on R???

    Best Regards,

    Fred
     
    Last edited: Dec 7, 2005
  2. jcsd
  3. Dec 7, 2005 #2
    You haven't told us what "h" is.
     
  4. Dec 7, 2005 #3
    h is definied as follows

    [tex]
    \begin{displaymath}
    h(c) = \left\{ \begin{array}{ll}
    f(c) \\
    g(c) \\
    \end{array} \right.
    \end{displaymath}
    [/tex]


    [tex]c \in \mathbb{R}[/tex] and f and g er continious on R.

    Then how do I show that if f(c) = g(c), then h(c) is continious on R?

    My own ideer is to show this if f and g are continious on R, then the composite function h is continious on R.

    Best Regards

    /Fred
     
    Last edited: Dec 7, 2005
  5. Dec 7, 2005 #4
    I'm afraid I have no idea what that means.
     
  6. Dec 7, 2005 #5
    Okay lets look it at another way,

    f and g: [tex][a,b] \rightarrow \mathbb{R}[/tex] are continious, [tex]c \in ]a,b[[/tex]. Next, let h be

    [tex]\begin{displaymath}h(x) = \left\{ \begin{array}{ll}f(x) \ \ \ \\g(x) \\\end{array} \right.\end{displaymath}[/tex]

    h is defined on [a,b]

    Now my assigment is the following:

    a) Let f and g be arbitrary functions. Show that if f(c) = g(c), then the function h is continious. I sure I need to use the epsilon-delta definition of continuerty, but if there is anybody out there who maybe can explain what I need to do by way of an example I would very much appriciate it :-)

    b) If [tex]f(c) \neq g(c)[/tex] the h is discontinious. Anybody who can direct me to a good example on how to show this?

    Sincerely and God bless You :-)

    Fred

    /Fred
     
    Last edited: Dec 7, 2005
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