# Using Rolle's theorem to prove at most one solution of quadratic

• Feb 11th 2011, 05:49 AM
maximus101
Using Rolle's theorem to prove at most one solution of quadratic
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]?
• Feb 11th 2011, 06:08 AM
FernandoRevilla
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
• Feb 11th 2011, 06:12 AM
alexmahone
Quote:

Originally Posted by maximus101
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 $\displaystyle \in$ $\displaystyle R$

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.
• Feb 11th 2011, 06:22 AM
alexmahone
Quote:

Originally Posted by maximus101
and for what values of b does this have a solution in [-1,1]?

$\displaystyle f(-1)=4+b$

$\displaystyle f(1)=-4+b$

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.)

$\displaystyle -4\leq b\leq 4$