# simple function composite question

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• Oct 28th 2010, 12:34 PM
jayshizwiz
simple function composite question
\$\displaystyle f\$ and \$\displaystyle g\$ are functions from \$\displaystyle R\$ to \$\displaystyle R\$

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

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

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

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

Am I correct?

Thanks!
• Oct 28th 2010, 12:47 PM
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.
• Oct 28th 2010, 12:58 PM
jayshizwiz
Quote:

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.
• Oct 28th 2010, 01:13 PM
croraf
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.
• Oct 28th 2010, 01:28 PM
croraf
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.