Prove that the equation $x^3-x-1=0$ has only one real root

Question

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?

Answer 1

The solutions of $3c^2-1=0$ are $\pm1/\sqrt{3}$ outside $[1,2]$.

That gives you a contradiction with the assumption that there are two roots inside $[1,2]$.

Answer 2

HINT

We can observe that

• $f(x)$ is continuous and limits for $x\to \pm \infty$ are $\pm \infty$ thus we have at least one real root

then consider

• $f'(x)=3x^2-1=0\implies x=\pm \frac{\sqrt 3}{3}$ and study max/min

then by IVT you'll be able to prove that $f(x)$ has exactly one root.

Answer 3

We have that $f'(x)=3x^2-1$ so $f$ has critical points at $x_0=-1/\sqrt3$ and $x_1=1/\sqrt3$. It's easy to check that these are not roots of $f$, and hence $f$ has no double root.

It follows that $f$ has either $1$ real root, or $3$ real roots: one in $(-\infty,x_0)$, one in $(x_0,x_1)$ and one in $(x_1,+\infty)$.

On the other hand, $f'\geq 0$ in $(-\infty, x_0)$ so it is increasing and

$$f(x_0) = -\frac1{3\sqrt3}-\frac1{\sqrt3}-1<0$$

so $f<0$ in $(-\infty, x_0)$. It follows that $f$ must have exactly $1$ real root.

Answer 4

The derivative $3x^2-1$ has roots $\pm\dfrac{\sqrt 3}3$, and $- \dfrac{\sqrt 3}3$ corresponds to a local maximum, which happens to be $- \dfrac{2\sqrt 9}3-1<0$. Hence $f(x)\le-\dfrac{2\sqrt 9}3-1$ for $x\le \dfrac{\sqrt 3}3$, then it is monotonically increasing to $+\infty$. Therefore, it has only one (real) root.

Answer 5

One can notice, that $f(1)f(2) = -5 < 0$, so there must be root of the equation. Also $f'(x) = 3x^2 - 1$, so it is monotonically increasing in range [1, 2] - there is only one root.

Next step is to prove that $x^3 < x + 1 $ for $x<1$. For $0 < x < 1$ we have $x + 1 > x > x^3$. For $-1 < x < 0$ following is met: $x + 1 > 0 > x ^ 3$. Finally for $x < -1$ we have $1 + x > x > x^3$.

Last step is to prove, that $x^3 > x + 1$ for x > 2, but this is obvious, since $f(2) = 5 > 0$ and for every $x > 2$ $\frac{dx^3}{dx} = 3x^2 > \frac{d(x + 1)}{dx} = 1$.

Answer 6

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.