# Injectivity of $f_r : \mathbb{R} \to \mathbb{R}$ Defined as $f_r(x) = x^3 + rx + 1$: Inflection Points and Injectivity

I have a function $f_r : \mathbb{R} \to \mathbb{R}$ defined as $f_r(x) = x^3 + rx + 1$.

Provided $r > 0$, the derivative will always be strictly positive, meaning that the function will be injective (one-to-one).

I am told the following:

At $r = 0$, we have $f_r = x^3 + 1$, and this only has a turning point at $x = 0$, but the second derivative is also $0$ at this point.

1. I am struggling to understand the point this statement is trying to make. It says that "at $r = 0$, we have $f_r = x^3 + 1$, and this only has a turning point at $x = 0$", but then it says "but the second derivative is also $0$ at this point"; the way it is phrased seems to imply that it is trying to make some contrasting point between these two statements, but I am not understanding what this is?

Since the second derivative is $0$ at the point, it means that it is an inflection point. But this doesn't mean that the function is not injective at this point, right? For instance, $f(x) = x^3$ has a turning point at $x = 0$, but its second derivative is also $0$ at this point, and $f(x) = x^3$ is injective (one-to-one).

1. The function would still be injective for $r = 0$, right? Just because the derivative is positive at all points except $x = 0$ does not necessarily mean that the function is not injective at this point?

2. Finally, I think my problem is that I'm struggling to understand what the presence of inflection points mean for injectivity. What DO the presence of inflection points mean for injectivity?

I would greatly appreciate it if people could please take the time to clarify this.

We can show that $f_r$ is injective without the derivative. Note that $$f_r(x)-f_r(y)=(x-y)(x^2+xy+y^2+r).$$ Hence $f_r(x)=f_r(y)$ if and only if $x=y$ because for $(x,y)\not=(0,0)$ $$\underbrace{x^2+xy+y^2}_{> 0}+r>r\geq 0.$$

P.S. Injectivity is not influenced by the appearance of an inflection point! If $f'(x)\geq0$ for $x\in (a,b)$ and $f'(x)=0$ at a finite number of points in $(a,b)$ then $f$ is strictly increasing by the Mean Value Theorem.

• Thanks for the response, but I was more-so looking for something that directly addresses my 3 points above. Enlightening approach, nonetheless. Commented Mar 29, 2018 at 13:15
• Injectivity is not influenced by the appearance of the inflection point! If $f'\geq0$ in $(a,b)$ and $f'=0$ at a finite number of points in $(a,b)$ then $f$ is strictly increasing by the MVT. Commented Mar 29, 2018 at 13:20
• Ahh, ok. So, just to be absolutely sure, I'll repeat my question to Mundron Schmidt: if a function is injective along some interval, must that interval necessarily also include the inflection point? After all, as you say, the inflection point is simply a boundary. Commented Mar 29, 2018 at 13:27
• No, $x^2$ is strictly increasing in $(0,+\infty)$ but it has no inflection point. Commented Mar 29, 2018 at 13:32

You are totally right that $f_0$ is still injective. The inflection point doesn't infect the injectivity of the function.

There is no contrast to the injectivity between $f_r$ for $r>0$ and $f_0$. But the rise of an inflection point can be seen as some kind of boundary for the injectivity. In this example, the injectivity breaks for $r<0$. But this break is not sudden because there appeared an inflection point for $r=0$.

• Thanks for the response. So, if a function is injective along some interval, must that interval necessarily also include the inflection point? After all, as you say, the inflection point is simply a boundary. Commented Mar 29, 2018 at 13:20
• I don't understand your question because I talked about a family of functions, where the inflection point marks the boundary of the injective functions. But you talk about one injective function on some interval.One specific injective function doesn't need to have an inflection point. And you can also have a family of injective functions without inflection points. Commented Mar 29, 2018 at 13:28
• I mean, if $r = 0$, then the derivative $f' > 0$ at all points except at $x = 0$, since it is an inflection point. So, generally speaking, if a function has a strictly positive or negative derivative at all points except at the inflection points, must this necessarily mean that the inflection points don't "break" injectivity? For instance, we could imagine a function with $f' > 0$ everywhere except at multiple inflection points where $f' = 0$ and $f'' = 0$. Commented Mar 29, 2018 at 13:33