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
Best Answer
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.