Using Rolle’s Theorem prove that the cubic equation $x^3+ax+b$ has at most one root

Question

A brief explanation of Rolle’s Theorem would be appreciated! I understand that it is a special case of the mean value theorem but I am not sure how to apply it to this question.

Edit: $a>0$

Answer 1

Let $f(x)=x^3+ax+b$. If $f(x)=0$ had two distinct solutions $x$ and $x'$, with $x<x'$, then you would have $f'(y)=0$ for some $y\in(x,x')$. But that cannot happen, since $f'(y)=3y^2+a>0$.

Answer 2

Rolle's Theorem says:

If a function $f(x)$ is

1. Continuous in the interval $\left[ a, b \right]$

2. Differentiable in the interval $\left( a, b \right)$

3. And if $f(a)=f(b)$,

then there exists one point c between a and b for which $f'(c)=0$

We can interpret this geometrically: The point c lying between a and b is where the tangent to the graph of $f(x)$ is parallel to th $x$ axis