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Math Help - simple function composite question

  1. #1
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    simple function composite question

    f and  g are functions from R to R

    1) If f \circ g is a monotonically increasing function, does that mean that f is monotonically increasing function?

    2) If f \circ g and f are monotonically increasing functions, does that mean that g is a monotonically increasing function

    I think for the first one f must be a monotonically increasing funtion. It doesn't matter what g is since it is in the domain of f. We can define f however we want to but if f \circ g is a monotonically increasing function, then definitely f is too.

    The second one I think is false for the reasons I just gave.

    Am I correct?

    Thanks!
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  2. #2
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    I think the first is false and the second true.

    Consider f = -x, g = -x then (fog)(x) = -(-x) = x which is monotonically increasing, and f and g are monotonically decreasing.
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  3. #3
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    Quote Originally Posted by croraf View Post
    I think the first is false and the second true.

    Consider f = -x, g = -x then (fog)(x) = -(-x) = x which is monotonically increasing, and f and g are monotonically decreasing.
    Thanks! I'm always backwards (:

    So now I have to prove why the second claim is true. Those are usually annoying.
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  4. #4
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    The proofs for this kind of questions go like this:

    If f is m.i. and g is m.i. then for x1>x2 follows g(x1)>g(x2) and therefore (fog)(x1) = f(g(x1)) > f(g(x2)) = (fog)(x2), the composition is monotonically increasing.

    If f is m.d. and g is m.d. then for x1>x2 follows g(x1)<g(x2) and therefore (fog)(x1) = f(g(x1)) > f(g(x2)) = (fog)(x2), the composition is monotonically increasing.

    Similarly for the two other combinations (m.d. m.i. and m.i. m.d.); you can find that the composition is m.d.

    So fog is monotonically increasing only if (f is m.i and g is m.i.) or (f is m.d. and g is m.d.). From this your answers come from.
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  5. #5
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    You can write it more handsome the answers.

    For the first you found the contradiction.
    For the second (to explain what I said before):
    If fog is m.i. than for x1>x2 you have (fog)(x1)>(fog)(x2) or f(g(x1))>f(g(x2)),
    and because f is m.i. you have g(x1) > g(x2) (the argument of the left f must be greater from the argument of the right f),
    and because x1 > x2 g is also m.i.
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