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In my calculus book it mentions that increasing functions maintain inequality relations and that's the reason you can apply $\exp$ and $\ln$ to two sides of an inequality to solve them. Is there some general classification for the types of functions that maintain inequality? For instance are they all 1 to 1 or have some other property in common?

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More is true: the functions which maintain inequalities are exactly the nondecreasing ones. – Did Jul 9 '12 at 9:00
So the only functions which maintain inequality are the class of increasing functions? – Robert S. Barnes Jul 9 '12 at 9:10
"Nondecreasing" is a little bit more general than "increasing"... – J. M. Jul 9 '12 at 9:26
Yes. A function maintains strong inequalities ($<$) if and only if it is increasing and weak inequalities ($\le$) if and only if it is non-decreasing. This is, in a way, the definition of non-decreasing. – yohBS Jul 9 '12 at 9:27
up vote 3 down vote accepted

In addition to m. k.'s answer, there is one very valuable criterion: If a function is continuous, it is monotonic if and only if it is injective (this is a consequence of the Intermediate Value Theorem). Therefore, a continuous function $f: [a,b] \to \mathbb R$ will maintain a strict inequality if and only if $f(a) < f(b)$ and it is injective.

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Is that supposed to read for any $a,b\in [a,b]$ as in "maintains a strict inequality iff $\forall a,b\in [a,b] f(a)<f(b) and f is injective." or are you just testing the endpoints? – Robert S. Barnes Jul 16 '12 at 7:09
I'm testing the endpoints. The argument runs as follows: f maintains a strict inequality iff is is strictly monotonously increasing (SMI). SMI functions are trivially injective, and if a continuous function with $f(a) < f(b)$ is not SMI, there are two points $x,y \in [a,b]$, $x<y$, such that $f(x) \ge f(y)$. If $f(x) = f(y)$, $f$ is not injective, so assume $f(x) > f(y)$. Now, we have to consider various cases, but for each case, we can apply the intermediate value theorem to find a point $z \in [a,b]$ such $z \neq z'$ for $z' \in \{ a,b,x,y\}$ and $f(z) = f(z')$. – Johannes Kloos Jul 16 '12 at 8:31

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