Flex for a monotone strictly increasing function

Question

First of all sorry if my question could be too trivial for the standards of this site but I am not so expert.

I am studying how perform the qualitative plots of functions of real variables. In particular I am interested in understand the relation in between the first derivative and the second derivative.

I mean: if I have a function that is always increasing, more specifically always strictly increasing, i.e $f'(x)>0$ $\forall x\in \mathbb{R}$, then is it possibile that there exists a point in which the function passes from a phase of "acceleration" to a phase of "deceleration", so from convexity to concavity?

$\textbf{So the question is:}$ is it possible that a function always strictly increasing has a flex point (where the second derivative is null)? I am trying to think about some examples but I can't...maybe my fault since I'm just getting started in studying all these stuffs.

EDIT: I have thought $x^3$ that is strictly increasing but has a flex point in $0$, am I right?

Answer 1

Consider some polynomial of the form $f(x) = a_0+a_1x+a_2x^2+a_3x^3+\dots$

Your first requirement is that $f(x)$ is always increasing. Thus $f'(x) = a_1 + 2a_2x+3a_3x^2+\dots>0\quad \forall x$.

The second requirement (which is easier to satisfy) is that, it has some point where $f''(x) =0$. Therefore $f''(x) = 2a_2+6a_3x + 12a_4x^2+\dots$ has some solution $f''(x) = 0$.

If you think about it, there are many ways to satisfy these two requirements.

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Two examples i'll give:

1. for $a>0$, $f(x) = a\cdot x^3$

$\qquad f'(x) = a\cdot 3x^2 >0$,$f''(x) = 6ax$ has a solution for $f''(x) =0$ at $x=0$

2. for $a_1,a_3,a_5,\dots>0$, $f(x) = a_1+a_3x^3+a_5x^5+\dots$,

$\qquad f'(x) = 3a_3x^2+5a_5x^4+7a_7x^6+\dots >0$,$f''(x) = 6a_3x+20a_5x^3+42a_7x^5+\dots$ has a solution for $f''(x) =0$ at $x=0$