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

1. 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]?

2. 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

3. 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.

4. 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.)

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