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Can we say (for sure) that "the function is increasing” to mean that the first derivative is positive?
Whenever $f'$ (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.
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 $$ f'_\ell(x)=\sum\limits_{q<x}a_q\qquad\text{and}\qquad f'_r(x)=\sum\limits_{q\leqslant x}a_q. $$ 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 $f'_\ell$ and $f'_r$ 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 $(g_q)'_\ell(x)=[x>q]$ and $(g_q)'_r(x)=[x\geqslant q]$.
No. Consider function $f(x)=x^3$. It is increasing on $\mathbb R$ but $f'(0)=0$.
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 $$ f(x) = \begin{cases} x & \text{if }x<0, \\ 2x & \text{if }x\ge 0. \end{cases} $$ The derivative is undefined at $x=0$. The function is everywhere increasing.
