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

**maximus101** · February 11th 2011, 05:49 AM

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]?

**FernandoRevilla** · February 11th 2011, 06:08 AM

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

**alexmahone** · February 11th 2011, 06:12 AM

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.

**alexmahone** · February 11th 2011, 06:22 AM

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.)

$-4\leq b\leq 4$

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