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I have to prove using Rolle theorem that the equation $x^3-3x+4=0$ does not have more than one solution in $[-1,1]$.

By looking at similar problems (here for example) i supposed that the equation does have two solution $x_1$ e $x_2$ such as $f(x_1)=f(x_2)=0$

Then i computed $f'(x) = 3x^2-3$

How can i proceed now? Similar problems that i have seen proved the contradiction by showing that $f'(x)$ was stricly positive or negative

Michele Assirelli
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    So what does Rolle's theorem tell you about $f'$ in the interval $(0, 1)$? Is that possible with $f'(x) = 3 x^2 -3$? – Martin R Nov 20 '23 at 15:58
  • Hint : If $f(x_1)=f(x_2)=0$ with $-1 \leq x_1 < x_2 \leq 1$, then by Rolle's theorem, $f'$ must have a root in $(x_1, x_2)$, so it must have a root in $(-1,1)$. Is it the case ? – TheSilverDoe Nov 20 '23 at 15:58
  • Why are you using Rolle's theorem? You have a cubic polynomial with real coefficients. – John Douma Nov 20 '23 at 16:03
  • $f'(x)$ does have two roots for $x_1=1$ and $x_2=1$. So where is the absurd? That the interval should not include $-1$ and $+1$? I'm not familiar with the notation of interval with () parenthesis. – Michele Assirelli Nov 20 '23 at 16:04
  • Also i am using Rolle theorem because this is a problem from a text book and the request is to use Rolle theorem – Michele Assirelli Nov 20 '23 at 16:05
  • @MicheleAssirelli Yes, the conclusion in Rolle's theorem is that $f'$ has a root in the interval $]x_1, x_2[$ (excluding the bounds !). – TheSilverDoe Nov 20 '23 at 16:10

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You are certainly in the right direction, we have to prove that the equation $x^3 - 3x + 4$ does not have 2 roots $\in [-1,1]$. Rolle's theorem is stated as : if $\exists$ 2 values of x (let a,b) such that f(a) = f(b) then $\exists$ some c $\in (a,b)$ such that f'(c) = $0$.

Now let us assume there are 2 or more roots of (let f(x) = )$x^3-3x+4 \in [-1,1]$. Let us say 2 of them are $\alpha \ \& \ \beta$ with $\alpha < \beta$.

$\therefore f(\alpha) = f(\beta) = 0$.

As both $\alpha \ \& \ \beta \in [-1,1] \implies [\alpha, \beta] \subseteq [-1,1]$.

Now using Rolle's theorem in $[\alpha, \beta]$ (as $f(\alpha) = f(\beta)$) we get $\exists$ some c $\in (\alpha, \beta)$ such that $f'(c) = 0$

You correctly calculated f'(x) to be $3x^2-3$.

Taking 3 common we get $f'(x) = 3(x^2-1)$. Factorising this, we get $f'(x) = 3(x-1)(x+1)$

But we know $(x-1)(x+1) \leq 0 \in [-1,1]$

So we find that f'(x) is only $0$ at boundary points of interval [-1,1] and for x $\in (-1,1)$ it is < $0$.

This is a contradiction to the above statement that $\exists$ some c $\in (-1,1)$ such that $f'(c) = 0$.

Thus not more than 1 root of f(x) exists in [-1,1].

Note :- x $\in (a,b) \implies x \in [a,b] - \{a,b\}$ i.e. x can be anything between but not including a and b.

Ham Lemon
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