I want to prove that the equation $f(x)=x^3-x-1=0$ has only one real root, which is on the intervall $[1,2]$.
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I have done the following:
$f(1)=-1<0$ and $f(2)=5>0$ so $f(1)\cdot f(2)<0$ and so from Bolzano's Theorem we have that the function has at least one root on $[1,2]$.
We suppose that there are two roots, $a$ and $b$. Then we have that $f(a)=f(b)=0$.
The function $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$. So, from Rolle's Theorem there is a $c\in (a,b)$ such that $f'(c)=0 \Rightarrow 3c^2-1=0$.
How can we get a contradiction?
Best Answer
Without derivative:
Let $r$ be the root. Then the polynomial factorizes as
$$(x-r)(x^2+rx+r^2-1)$$ and there are other real roots iff
$$\Delta=r^2-4(r^2-1)\ge0,$$
$$|r|\le\frac2{\sqrt3}.$$
But $f\left(\dfrac2{\sqrt3}\right)<0$ implies $r>\dfrac2{\sqrt3}$, a contradiction.