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I want to prove that the equation $f(x)=x^3-x-1=0$ has only one real root, which is on the intervall $[1,2]$.

$$$$

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?

Andronicus
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Mary Star
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6 Answers6

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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.

Bernard
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    There is a $\LaTeX$ problem in your answer. – MathLover Mar 16 '18 at 21:38
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    @MathLover: Yes, II forgot a \backslash. Thanks for pointing it! – Bernard Mar 16 '18 at 22:35
  • We have that at $x=-\frac{\sqrt{3}}{3}$ the function gets its local maximum. Why do we get $f(x)\le-\dfrac{2\sqrt 9}3-1$ for $x\le \dfrac{\sqrt 3}3$ and not for $x\le -\dfrac{\sqrt 3}3$ ? I got stuck right now. – Mary Star Mar 16 '18 at 22:38
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    Because $f(x)$ decreases on $\biggl[-\dfrac{\sqrt 3}3,\dfrac{\sqrt 3}3\biggr]$. – Bernard Mar 16 '18 at 22:49
  • Ah, so we have the following: For $x\leq -\frac{\sqrt{3}}{3}$ we have that $f(x)\leq f\left (-\frac{\sqrt{3}}{3}\right )<0$, so in that interval there is no root. In the interval $\biggl[-\dfrac{\sqrt 3}3,\dfrac{\sqrt 3}3\biggr]$ the max and the min are negtive, so in that interval there is no root. For $x\geq \frac{\sqrt{3}}{3}$ we have that the function is increasing so the function has at most one root there. Therefore the root that we found in $[1,2]$ is unique. Is everything correct? – Mary Star Mar 16 '18 at 23:03
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    Quite correct.${}$ – Bernard Mar 17 '18 at 00:18
  • Great!! Thank you!! :-) – Mary Star Mar 17 '18 at 00:18
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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.

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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.

user
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  • We have that $f''\left (-\frac{\sqrt{3}}{3}\right )<0$ and $f''\left (\frac{\sqrt{3}}{3}\right )>0$ so we have that the function $f$ has a maximum at $x=-\frac{\sqrt{3}}{3}$ and a minimum at $x=\frac{\sqrt{3}}{3}$. How do we continue from here? – Mary Star Mar 16 '18 at 22:43
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    Note that to check for max and min it suffices the sign of $f'(x)$ which is negative between $x=-\frac{\sqrt{3}}{3}$ and $x=\frac{\sqrt{3}}{3}$ and positive otherwise. Now evaluate $f(x)$ at $x=-\frac{\sqrt{3}}{3}$ and $x=\frac{\sqrt{3}}{3}$ and observe that both are negative. – user Mar 16 '18 at 22:59
  • Ah ok! Thank you very much!! :-) – Mary Star Mar 16 '18 at 23:03
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    @MaryStar You are welcome! Can you finish now? That's a standard way also for not polynomial functions. – user Mar 16 '18 at 23:08
  • Yes!! Ah ok! So, also for functions with trigonometric terms like here: https://math.stackexchange.com/questions/2694265/the-equation-has-exactly-two-positive-roots-how-do-we-get-a-contradiction ? – Mary Star Mar 17 '18 at 00:20
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    @MaryStar Yes, indeed it is a general method! – user Mar 17 '18 at 00:30
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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.

Fimpellizzeri
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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 $x > 2$ $\frac{dx^3}{dx} = 3x^2 > \frac{d(x + 1)}{dx} = 1$.

Andronicus
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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 $[1,2]$.

SphericalTriangle
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