What do the complex roots in the derivative $f'(x)$ say about $f(x)$?

Question

For example, take the function $\frac13 x^3 + x + c$. Its derivative is $x^2 + 1$, which has no real roots, but two complex roots. Looking at their graphs, it seems that non-real roots produce no horizontal tangent at all. So I'm inclined to say that there's no implication at all, but maybe maybe there are features of $f(x)$ that I have not noticed.

Answer 1

You need to give up on the idea of focusing on just the real-valued graph and think instead of the polynomial as a function with complex number inputs and outputs: a function $f:\mathbf C \to \mathbf C$. When $f$ has degree $n$, $f$ is locally an $n$-to-$1$ mapping except at the numbers $f(r)$ where $f’(r) = 0$.

For example, let $f(z) = (1/3)z^3+z$. Then $f’(z) = 0$ at $z = \pm i$, with $f(i) = (1/3)(-i) + i = (2/3)i$ and $f(-i) = (1/3)i - i = (-2/3)i$. So for each $w$ in $\mathbf C$ other than $(\pm 2/3)i$, the equation $f(z) = w$ is $3$-to-$1$ near $w$: each number near $w$ is an $f$-value $3$ times. But this is false for the equations $f(z) = (2/3)i$ and $f(z) = (-2/3)i$, which each have just $2$ solutions. For instance, $$ (1/3)z^3 + z = (2/3)i $$ is equivalent to $$ 0 = (1/3)z^3 + z -(2/3)i = (1/3)(z-i)^2(z+2i), $$ so the only solutions of $f(z) = (2/3)i$ are $i$ and $-2i$.

This is a very important phenomenon in more advanced math, where it is related to ramification points or branch points of nonconstant holomorphic mappings between Riemann surfaces.