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
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.