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.
- 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).
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?
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.