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?
 A: 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 $f'(x)>0$ 
A: $f(x)=4x^3+4x-6$
$f(0)=-6$ 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 $f^{\prime}(x)=12x^2+4>0\ \forall x\ne0$
A: Your $f'(x)$ is strictly positive  which means your function is strictly increasing. A strictly increasing function does not have more than one real root.
Because otherwise it is not going to be one-to-one. Simply put, with more than one real root you have to have a turning point somewhere between those roots which makes your function both increasing and decreasing between the roots.
A: Pedestrian:
$f'(x)>0$ implies $f$ is strictly increasing, i.e.for
$x_1 < x_2$  we have  $f(x_1) < f(x_2)$.
Assume a strictly increasing function has more than one zero.
Let $x_1 < x_2$ be $2$ zeroes of the function:
$f(x_1)=f(x_2)= 0$.
A contradiction.
P.S. Can there be a double zero if $f'(x) >0$?
