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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 $[1,2]$.

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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?

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    $\begingroup$ Show that $$f'(x)>0$$ for all $x$ with $$1<x<2$$ $\endgroup$ – Dr. Sonnhard Graubner Mar 16 '18 at 21:17
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    $\begingroup$ The discriminant of $x^3-x-1$ is $-23<0$ and $f(1)f(2)<0$, end of story. $\endgroup$ – Jack D'Aurizio Mar 16 '18 at 21:21
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    $\begingroup$ @MaryStar Please remember that you can choose an answer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – user Mar 17 '18 at 23:28
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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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  • $\begingroup$ How do you get $x^2+rx+r^2-1$? $\endgroup$ – MathLover Mar 17 '18 at 5:17
  • $\begingroup$ @MathLover: by long division. $\endgroup$ – Yves Daoust Mar 17 '18 at 9:55
  • $\begingroup$ Could you please,can you write it paper or can you upload it to any site and give me a link? ... or can you write it in $ \LaTeX$ I dont know your method..But I want to know..Thank you.! $\endgroup$ – MathLover Mar 17 '18 at 12:13
  • $\begingroup$ @MathLover: is it too much effort to lookup two words ? $\endgroup$ – Yves Daoust Mar 17 '18 at 13:06
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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.

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    $\begingroup$ There is a $\LaTeX$ problem in your answer. $\endgroup$ – MathLover Mar 16 '18 at 21:38
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    $\begingroup$ @MathLover: Yes, II forgot a \backslash. Thanks for pointing it! $\endgroup$ – Bernard Mar 16 '18 at 22:35
  • $\begingroup$ 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. $\endgroup$ – Mary Star Mar 16 '18 at 22:38
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    $\begingroup$ Because $f(x)$ decreases on $\biggl[-\dfrac{\sqrt 3}3,\dfrac{\sqrt 3}3\biggr]$. $\endgroup$ – Bernard Mar 16 '18 at 22:49
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    $\begingroup$ Quite correct.${}$ $\endgroup$ – Bernard Mar 17 '18 at 0:18
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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.

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  • $\begingroup$ 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? $\endgroup$ – Mary Star Mar 16 '18 at 22:43
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    $\begingroup$ 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. $\endgroup$ – user Mar 16 '18 at 22:59
  • $\begingroup$ Ah ok! Thank you very much!! :-) $\endgroup$ – Mary Star Mar 16 '18 at 23:03
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    $\begingroup$ @MaryStar You are welcome! Can you finish now? That's a standard way also for not polynomial functions. $\endgroup$ – user Mar 16 '18 at 23:08
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    $\begingroup$ @MaryStar Yes, indeed it is a general method! $\endgroup$ – user Mar 17 '18 at 0:30
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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.

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

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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 $[1,2]$.

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  • $\begingroup$ I'm confused. Why the downvotes? This answer is correct! $\endgroup$ – José Carlos Santos Mar 16 '18 at 21:45

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