Rolle's theorem prove polynomial has only 1 root Prove that $x^3-x-4=0$ has exactly one real root:
This is my working so far:
suppose $f(x) = x^3-x-4$
 has $2$ roots : $a,b$
 $f(a) = f(b) = 0$
$f'(x)=3x^2-1$
$f'(x)$ exists on $(a,b)$ so $f$ is differentiable on $(a,b)$
By Rolle's therorem there exists $c \in [a,b]$
such that $f'(c) = 3c^2-1=0$
However here I can solve to get $c = \frac{1}{\sqrt3}$
but I am supposed to get a contradiction where there is no such $c$ to make the derivative zero!!
please help
Thanks

Show $f(x)$ has a local maximum at $(-{1 \over \sqrt{3}}, {2 \over 3 \sqrt{3}} - 4)$ and a 
local minimum at $({1 \over \sqrt{3}}, -{2 \over 3 \sqrt{3}} - 4)$. These two points on the graph divide the graph into 3 portions for which $f(x)$ is either increasing or decreasing. Use this to show the graph intersects the $x$ axis exactly once. The idea is that all the "gyrations" in the graph are below the $x$-axis, and there is only one root, on the right where the function is increasing.
