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

• Feb 11th 2011, 06:49 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

$x^5 -5x +b=0$

in [-1,1] for any b $\in$ $R$

and for what values of b does this have a solution in [-1,1]?
• Feb 11th 2011, 07:08 AM
FernandoRevilla
If $-1\leq \alpha < \beta\leq 1$ were roots of $f(x)=x^5-5x+b$ then, would exist $\xi\in (\alpha,\beta)$ such that $f'(\xi)=5\xi^4-5=0$ , but $\xi=\pm 1$ (contradiction). Try the second question.

Fernando Revilla
• Feb 11th 2011, 07:12 AM
alexmahone
Quote:

Originally Posted by maximus101
How do we use Rolle's theorem to prove that there is at most one solution of

$x^5 -5x +b=0$

in [-1,1] for any b $\in$ $R$

Let $f(x)=x^5-5x+b$

$f'(x)=5x^4-5=5(x^4-1)\leq 0$ in the interval $[-1, 1]$.

So f(x) is decreasing in the interval $[-1, 1]$.

Therefore, there is at most one solution of $x^5-5x+b=0$ in $[-1, 1]$ for any $b\in R$.

PS: This method does not use Rolle's theorem, though.
• Feb 11th 2011, 07:22 AM
alexmahone
Quote:

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

$f(-1)=4+b$

$f(1)=-4+b$

For the conditions of the intermediate value theorem to be satisfied, $f(-1)$ and $f(1)$ should have opposite signs.

$4+b\geq0\geq-4+b$ (Note that f is a decreasing function.)

$-4\leq b\leq 4$