Is there the continuous real root of the cubic equation and is there a closed formula to present it?

Question

For any cubic equation, $ax^{3}+bx^{2}+cx+d=0$, we know there is always a real root if $a,b,c,d$ are all real. Suppose that $a,b,c,d$ are continuous and real function with respect of $i\in \mathbb{R}$, formulated as $a(i),b(i),c(i),d(i)$, is there the continuous real root, $x(i)$, of the cubic equation? If there is, I wonder whether there is a closed formula to present it.

Answer 1

No, this is not possible and Mark Bennet's argument in the comments is correct. Suppose that the roots begin at the positions $(-1 - i, -1 + i, 1)$, so that our continuous real root must be $1$. Next have the complex roots travel towards the real line until they meet there, forming a double root at $-1$. Our continuous real root must still be $1$. Then have the double root separate into two real roots, the middle one of which travels from $-1$ to $1$. Our continuous real root must still be $1$. Now we have a double root at $1$ which can separate into a pair of complex roots, so that we end up with the roots $(-1, 1 + i, 1 - i)$. At the point at which our double root at $1$ travels away from the real line the only real root is $-1$ but our continuous real root is stuck at $1$.

However, the possibility of the roots colliding is the only obstacle, and away from the discriminant locus, where all the roots must be distinct, this is possible. The complement of the discriminant locus has two connected components; in one of them there's a unique real root which varies continuously (even smoothly as Mor A. says in the comments), and in another one there are three distinct real roots, all three of which vary continuously (again, even smoothly as Mor A. says in the comments).