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.
and are functions from to
1) If is a monotonically increasing function, does that mean that is monotonically increasing function?
2) If and are monotonically increasing functions, does that mean that is a monotonically increasing function
I think for the first one must be a monotonically increasing funtion. It doesn't matter what is since it is in the domain of . We can define however we want to but if is a monotonically increasing function, then definitely is too.
The second one I think is false for the reasons I just gave.
Am I correct?
Thanks!
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.
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.