# Can we say (for sure) that "the function is increasing” to mean that the ﬁrst derivative is positive?

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?

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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 $g'(0) =0$. –  martini Oct 11 '11 at 16:29
Increasing functions don't even have to be continuous! –  barrycarter Oct 11 '11 at 16:41

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 $(g_q)'_r(x)=[x\geqslant q]$.

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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.

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No. Consider function $f(x)=x^3$. It is increasing on $\mathbb R$ but $f'(0)=0$.

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