Title: Can we say (for sure) that "the function is increasing” to mean that the ﬁrst derivative is positive?
Tags: calculus;intuition
Question: Can we  say (for sure) that "the function is increasing” to mean that the ﬁrst derivative is positive?
Whenever $f&#39;$ (the first derivative) is positive the function is increasing,but does that imply  if a function is increasing the first derivative must be positive?
Please explain with an example.
Comment: Increasing functions don't even have to be continuous!

Comment: No. First, increasing functions need not to be everywhere differentiable (for example $f\colon \mathbb R\to \mathbb R$ given by $f(x) = x$ for $x \ge 0$ and $f(x) = 2x$ for $x \le 0$), second: If $f$ is differentiable, you only need to have $f' \ge 0$, not $f' > 0$, for example: $g(x) = x^3$ has $g'(0) =0$.

Here is the accepted answer: No. Consider a dense countable subset $Q$ of $\mathbb R$ and a family $(a_q)_{q\in Q}$ of positive real numbers such that $\sum\limits_{q\in Q}a_q(1+|q|)$ converges. For every $x$, let $(x)^+=\max\{x,0\}$ denote the positive part of $x$.
Then, the function $f$ defined by
f(x)=\sum\limits_{q\in Q}a_q(x-q)^+
is well defined for every real number $x$. The left and right derivatives of $f$ exist everywhere, with
f&#39;_\ell(x)=\sum\limits_{q<x}a_q\qquad\text{and}\qquad
f&#39;_r(x)=\sum\limits_{q\leqslant x}a_q.
Thus, $f$ is strictly increasing and strictly convex, differentiable at every point not in $Q$, and not differentiable at every point in $Q$.
To prove the existence of $f&#39;_\ell$ and $f&#39;_r$ at every point, one can come back to the definitions of the left and right derivatives as the limits, if these exist, of $\pm(f(x\pm h)-f(x))/h$ when $h\to0^+$.
Or one can use directly the fact that each function $g_q$ defined by $g_q(x)=(x-q)^+$ has left and right derivatives $(g_q)&#39;_\ell(x)=[x&gt;q]$ and $(g_q)&#39;_r(x)=[x\geqslant q]$.

Here is another answer: No. Consider function $f(x)=x^3$. It is increasing on $\mathbb R$ but $f&#39;(0)=0$.

Here is another answer: Someone mentioned $f(x)=x^3$, for which the derivative at one point is $0$ but the function is everywhere increasing.
It can also happen that the derivative at one point is undefined and the function is everywhere increasing.  For example, let
f(x) = \begin{cases} x &amp; \text{if }x<0, \\ 2x &amp; \text{if }x\ge 0. \end{cases}
The derivative is undefined at $x=0$.  The function is everywhere increasing.

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