How do we use Rolle's theorem to prove that there is at most one solution of
$\displaystyle x^5 -5x +b=0$
in [-1,1] for any b $\displaystyle \in$ $\displaystyle R$
and for what values of b does this have a solution in [-1,1]?
How do we use Rolle's theorem to prove that there is at most one solution of
$\displaystyle x^5 -5x +b=0$
in [-1,1] for any b $\displaystyle \in$ $\displaystyle R$
and for what values of b does this have a solution in [-1,1]?
If $\displaystyle -1\leq \alpha < \beta\leq 1$ were roots of $\displaystyle f(x)=x^5-5x+b$ then, would exist $\displaystyle \xi\in (\alpha,\beta)$ such that $\displaystyle f'(\xi)=5\xi^4-5=0$ , but $\displaystyle \xi=\pm 1$ (contradiction). Try the second question.
Fernando Revilla
Let $\displaystyle f(x)=x^5-5x+b$
$\displaystyle f'(x)=5x^4-5=5(x^4-1)\leq 0$ in the interval $\displaystyle [-1, 1]$.
So f(x) is decreasing in the interval $\displaystyle [-1, 1]$.
Therefore, there is at most one solution of $\displaystyle x^5-5x+b=0$ in $\displaystyle [-1, 1]$ for any $\displaystyle b\in R$.
PS: This method does not use Rolle's theorem, though.
$\displaystyle f(-1)=4+b$
$\displaystyle f(1)=-4+b$
For the conditions of the intermediate value theorem to be satisfied, $\displaystyle f(-1)$ and $\displaystyle f(1)$ should have opposite signs.
$\displaystyle 4+b\geq0\geq-4+b$ (Note that f is a decreasing function.)
$\displaystyle -4\leq b\leq 4$