# Why does a monotonic function always have a positive rate of change?

Based on my book

The rate of change of $$f(u)$$ as $$u$$ changes can be measured by looking at the change in $$f$$ between two values of $$u$$, divided by the change in $$u$$: $$\frac{Δf}{Δu} = \frac{f(u_2)-f(u_1)}{u_2-u_1}$$For a monotonic transformation, $$f(u_2)-f(u_1)$$ always has the same sign as $$u_2-u_1$$. Thus a monotonic function always has a positive rate of change. This means that the graph of a monotonic function will always have a positive slope, as depicted in Figure 4.1A.

Now what I don't understand is if the monotonic function is as simple as $$f(u) = u +1$$ (Changing it slightly from the non-monotonic function) how come it comes it will always have a positive rate of change compared to the non-monotonic function?

• 1. What does "normal function" mean in this context. 2. Do you know about the Cantor function? Commented Feb 3, 2019 at 14:18
• @kimchilover Thank you very much for your quick reply, by normal function I mean the non monotonic one (so in my case $f(u) = u$) Commented Feb 3, 2019 at 14:19
• I still really don't know what you are asking. That everything is called $f$ is no help. Commented Feb 3, 2019 at 15:02
• Unless I'm blatantly missing something, this is an odd definition of monotonic. Usually a function that only decreases would be considered monotonic too. But there might be debate about whether it's allowed to have sections where it stays constant. Commented Feb 3, 2019 at 15:25
• I wouldn't trust that book very much if I were you, since that excerpt is quite misleading. The slope isn't $\Delta f/\Delta u$, it's the limit of that thing as $\Delta u \to 0$. And even if $f$ is strictly increasing, there may be points where the slope is zero, like for $f(u)=u^3$ at $u=0$. Commented Feb 3, 2019 at 15:49