Cubic Depressed Form ! What can we deduce form it?

Question

Cubic depressed form with equation $f(x) = x^3 + px + q$

The question is, when $p$ is positive, will the function have $3$ real roots ? or does it have to have $1$ real and $2$ complex roots?

My question is not the same as Maths cubic equation discriminant....

I need to know the possibilities of roots when p is positive.

Answer 1

We have $f'(x) = 3x^2+p \ge p > 0 $. Consequently, $f$ is strictly increasing.

• $f(x)=0$ has at most one real solution. Otherwise take two distinct real solution $x_1<x_2$. Since $f$ is strictly increasing we have $0=f(x_1)<f(x_2)=0$. Absurd.
• Since $$\lim_{x\to +\infty}f(x) = +\infty,\quad\text{and }\quad \lim_{x\to -\infty}f(x) = -\infty.$$ The function $f$ is continuous and $f(x)$ goes from $-\infty$ to $+\infty$ when $x$ goes from $-\infty$ to $+\infty$. By continuity, $f(x)$ has to take the value $0$ at least once.

Finally, there is exactly one solution to the equation $f(x)=0$ for $x\in \mathbb{R}$.