$f$ is strictly increasing and $g$ is decreasing. How to find whether $f\circ g$ and $g\circ f$ are increasing, decreasing, strictly increasing or strictly decreasing?

This is what I did,

$f \circ g=f(g(x))$

If we take $x_1 < x_2$,

$f(x_1) < f(x_2) $ and $\ g(x_1) \ge g(x_2) $

Assuming $f(g(x_1)) < f(g(x_2))$, then,

$g(x_1)<g(x_2) \implies x_2<x_1$

This is a contradiction, therefore our assumption is wrong. After this, what should I assume to prove this? If I assume $f(g(x_1)) \ge f(g(x_2)) \ $, a problem occurs since $f$ is strictly increasing and the assumption has an equality possibility.

Or is there a more effective method than this? Can it be applied to prove the same for $g\circ f$?


It is not necessary to produce a contradiction.

When $x<y$ then $g(x)\geq g(y)$ and therefore $f\bigl(g(x)\bigr)\geq f\bigl(g(y)\bigr)$. This already proves that $f\circ g$ is decreasing. You cannot hope for more: It could be that $g(x)=g(y)$, so that we obtain an instance of $x<y$ and $f\circ g(x)=f\circ g(y)$.

Similarly for $g\circ f$.


You have shown that $f\circ g(x_1)<f\circ g(x_2)\implies x_2<x_1$ - that is, $f\circ g$ is decreasing.

Yes, the same method shows that $g\circ f$ is also decreasing.

  • $\begingroup$ I forgot to mention(editted the question) , I need to find if the functions are increasing, decreasing, strictly increasing or strictly decreasing. $\endgroup$ – S.Dan May 29 '14 at 5:02

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