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

Question

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).

2. 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?

3. 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.

Answer 1

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.

Answer 2

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 $r=0$.