# Thread: simple function composite question

1. ## 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!

2. 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.

3. Originally Posted by croraf
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.

4. 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.

5. 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.