How to prove $f(x) = 4x^{3} + 4x - 6$ has exactly one real root? 
How can I show that $f(x) = 4x^{3} + 4x - 6$ has exactly one real root? 

I think the best way is to show $f'(x) = 12x^2 + 4 > 0$ for all $x \in \mathbb{R}$. Thus, $f'(x)$ has zero real roots. Thus, $f(x)$ has at most one real root. 
I thought about trying to show that if $f$ is a polynomial and $f'$ has $n$ real roots, then $f$ has $n + 1$ roots by using Rolle's Theorem or Mean Value Theorem, but I don't think this fact, in general, is true. I would need to prove this statement.
Can someone please help me prove this fact?

$f(x)=4x^3+4x-6$
$f(0)=-6$ and $f(-1)=-14$
By IVT, there exists at least one real root, $x\in(-1,0)$ such that $f(x)=0$
Now try to prove by using contradiction.
If not, there exists al least $2$ real roots $x_1,x_2$, such that $f(x_1)=f(x_2)=0$
Since $f(x)$ is differentiable, by using Rolle's Theorem, there exists a number $k\in(x_1,x_2)$ such that $f^{\prime}(k)=0$. But $f^{\prime}(x)=12x^2+4>0\ \forall x\ne0$
