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?

Your intuition for the first one  is correct!
$f(0)<0$ and $f(1)>0$, so by IVT $f$ has a root say $x_0$
Suppose $f$ has another root $x_1 \neq x_0$ with $x_0<x_1$ .Then $f(x_0)=f(x_1)=0$ and by Rolles theorem $\exists$ $c \in (x_0,x_1)$ such that $f'(c)=0$, contradicting to the fact $f'(x)>0$ 
