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What is the precise difference between strictly increasing and increasing functions?? I see these terms being thrown around a lot My guess is that strictly increasing mean that derivative is only greater than 0 and in case of just increasing derivative can be greater than or equal or 0?

Schwarz Kugelblitz
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    Something not noted in the answers is that, even in the differentiable case, it is not the case that a function is strictly increasing iff its derivative is positive everywhere. Indeed, the function $f(x)=x^3$ is strictly increasing, but $f'(0)=0$. – Wojowu Aug 25 '19 at 16:59
  • The definitions also imply that $f(x) = 0$ is both a (non-strictly) increasing and decreasing flat-line function. – Jamie McGuigan Sep 11 '22 at 22:43

3 Answers3

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You're basically correct (update: as stated in this answer and question comment by Wojowu, I forgot to add that strictly increasing differentiable functions do not require the derivative to always be positive), except the concept of "strictly increasing" and "increasing" functions also applies to non-differentiable functions. In particular, for all $x \gt y$, where $x$ and $y$ are in the function domain, strictly increasing means $f(x) \gt f(y)$ while just increasing means $f(x) \ge f(y)$.

John Omielan
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    To give a specific example, a constant function is increasing but not strictly increasing. –  Aug 25 '19 at 05:08
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Strictly increasing means that $f(x)> f(y)$ for $x>y$. While increasing means that $f(x)\geq f(y)$ for $x>y$.

Example:

$f:\mathbb{R}\to \mathbb{R}, x\mapsto x$ is strictly increasing.

And the function $f:\mathbb{R}\to\mathbb{R}, x\mapsto \lfloor x\rfloor$ is increasing.

$\lfloor x\rfloor$ notes the flooring-function: https://en.wikipedia.org/wiki/Floor_and_ceiling_functions#Definition_and_properties

And looks like this:

enter image description here

Note that this function stays constant on certain intervals, but increases over time.

Cornman
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Note that your inituition about derivatives is only $3$ quarters correct: If the function $f$ under consideration is differentiable in the considered interval open $I$, then

$$ f \text{ is increasing in } I \Longleftrightarrow f'(x) \ge 0\; \forall x \in I.$$

OTOH, under the same conditions only one of the implications holds for strictly increasing functions:

$$ f \text{ is strictly increasing in } I \Longleftarrow f'(x) > 0\; \forall x \in I.$$

The other implication is wrong, for example $f(x)=x^3$ is strictly increasing in $I=(-1,1)$, while the derivative is zero at $x=0$.

Ingix
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