For a cubic equation, prove that two critical points of the same sign imply one root For a cubic equation of form $x^3 + p x + q = 0$, where $p < 0$,
if the two critical points are of the same sign, it will have only one real root. This is easy to see from graph, but can you help me to find an analytic proof for this?
I tried something like the converse of IVP, but getting into logical fallacies every time. 
 A: We may assume that $f$ has no multiple roots as in that case one of the critical points would be zero and not have the same sign as the other.
Assume there are at least two real roots $x_1,x_2$ of $f(x)=x^2+ax^2+bx+c$. Then by polynomial division you find $f(x)=(x-x_1)(x-x_2)g(x)$, where $g$ is linear, i.e. $f$ has three (distinct!) real roots, wlog $x_1<x_2<x_3$. At each of these, $f$ switches signs (otherwise the root would be an extremum, hence a multiple root). By Rolle, each of the intervals $(x_1,x_2)$ and $(x_2,x_3)$ contains a critical point, and they have different signs because of the sign swirthc at $x_2$.
A: Let $f(x) = x^3 + p x + q$. Then $f'(x) = 3x^2 + p$. Let $a = -\sqrt{\frac{-p}{3}}$ and $b = \sqrt{\frac{-p}{3}}$, so $f$ is increasing when $x \le a$ or $x \ge b$. In the interval $a \le x \le b$ the function is decreasing.
Assume first that $f(a),f(b) \le 0$. Then because $f$ is (strictly) increasing to the left of $a$, we must have $f(x) < f(a) \le 0$ for all $x \in (-\infty,a)$. Similarly because $f$ is (strictly) decreasing between $a$ and $b$, we have $f(x) < f(a) \le 0$ for all $x \in (a,b]$. Finally because $f(x)$ is (strictly) increasing when $x \ge b$, it takes on the value $0$ exactly once in the interval $(b,\infty)$.
The case $f(a),f(b) \ge 0$ can be done in an analogous way.
