How to prove that a function can't have two roots in R 
If $a>0$, $b ∈ R$ and $n ∈ N-{0}$, prove that $f(x) = x^{2n+1}+ax+b$ can't have two real roots.

f(x) is continous in all R.
I thought on doing it by contradiction, say $f(x_0)=f(x_1)=0$ with $x_0<x_1$, we also have that $f'(x)=(2n+1)x^{2n}+a$
Can I do it with Rolle theorem? i.e.
$\exists x∈(x_0,x_1)/f'(x)=0$

You are in the right direction.
Note that $x^{2n}\ge 0$, $(2n+1)>0$, $a>0$, hence $$(2n+1)x^{2n}+a>0$$
