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
 A: You're on the right track. If there were two roots, there would have to be a zero of $f'$ between them. So if the zeros of $f'$ are at $p$ and $q$ (with $p < q$), perhaps you can show that for $x < p$, , $f(x) < 0$ for some obvious reason. And maybe for some other reason you can show the same when $x$ is between $p$ and $q$. In that case, the only possible roots satisfy $x > q$. 
What kinds of "obvious" reason would show that $f(x) < 0 $ when $x < p$? Well, suppose that $f(x) = -100x^6 - x$, and you'd like to show that $f(x) < 0$ for $x < -1$. Well, the first term is a number no greater than $-100$, and the second is a number that's much smaller, so overall it's negative. Maybe more compelling is the argument that $f(x) = -100 x^6 -x = -x(100x^5 - 1)$. The first factor of this is positive (for $x < -1$), while the second factor is less that $100x^5 \le 100(-1)^5  = -100$. 
Something like that might work to help you establish that you're function's got only one root. 
A: 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.
A: The fact that the function has points where the derivative is zero, does not mean that the function has a root, see for instance $y=x^2+1$.
The use of Rolle's theorem only tells you that there are two points for which the derivative is zero, hence there must be $3$ or less roots, not that there is only $1$ root.
A: Well, note that you can't have $f(x)=(x-\alpha)(x-\beta)^2$ for any real $\alpha,\beta,$ because if that were the case, then by product rule, we would have $$f'(x)=2(x-\alpha)(x-\beta)+(x-\beta)^2=(2x-2\alpha)(x-\beta)+(x-\beta)^2=(3x-2\alpha-\beta)(x-\beta),$$ and so $f'(\beta)=0.$ But the zeroes of $f'$ are $\pm\frac1{\sqrt{3}},$ neither of which is a zero for $f.$ Hence, either $f$ has exactly one real root, or has three distinct real roots.
You supposed by way of contradiction that $f$ had at least two real roots. Hence, it has three distinct real roots by the work above--say $\alpha_1,\alpha_2,\alpha_3$ where $\alpha_1<\alpha_2<\alpha_3.$ But then $f'$ has a root in the interval $(\alpha_1,\alpha_2)$ and a root in the interval $(\alpha_2,\alpha_3)$ by Rolle's Theorem, so in particular, we can conclude that $-\frac1{\sqrt{3}}<\alpha_2<\frac1{\sqrt{3}}.$ Now, since $f'(x)\le0$ on $\left[-\frac1{\sqrt3},\frac1{\sqrt3}\right],$ then can conclude that $f\left(-\frac1{\sqrt3}\right)\ge f(\alpha_2)=0.$ but $$f\left(-\frac1{\sqrt3}\right)\approx-3.23,$$ yielding the desired contradiction.
