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Please help me with this problem on demonstrations.

By using Rolle's Theorem, show that $$f(x)=x^{10}+ax-b\quad,\quad {\rm where}\;\; a,b\in \mathbb{R}$$ has at most two real roots.

Thanks in advance.


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If there are at least three real roots, then you can apply Rolle's theorem twice and get two real roots of $f'(x) = 0$. Why is this a problem? It's important here that $9$ is odd. – Dylan Moreland Aug 12 '11 at 2:22
Thanks. But How I can find $x_1$, $x_2$ such that $f(x_1)=f(x_2)$ and such that $ - \sqrt[9]{9} \in \left\langle {{x_1},{x_2}} \right\rangle $? – mathsalomon Aug 12 '11 at 2:56
I'm not sure where you got $\sqrt[9]{9}$. Could you explain? We could head to chat, I suppose. – Dylan Moreland Aug 12 '11 at 3:08
sorry, I wanted to say : $$-\sqrt[9]{a} \in \left\langle {{x_1},{x_2}} \right\rangle$$ which is a root of $f'(x)$ – mathsalomon Aug 12 '11 at 3:17
That's a root, yes. I would avoid saying anything involving $x_1, x_2$ at this point. Can there be another root? – Dylan Moreland Aug 12 '11 at 3:51

Hint: How many real roots does $f'(x)$ have?

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$f'(x)$ has a root, $x=-\sqrt[9]{a}$, but how do I show formally that this implies that $f(x)$ has two real roots? – mathsalomon Aug 12 '11 at 2:49
@math: The question asks to prove at most two real roots, which is different from proving there are two real roots. Proceed by contradiction: Assume there are at least three. Now using Rolle's theorem, can you say something about $f'(x)$? – Aryabhata Aug 12 '11 at 2:52
Oh, I see. I will try using the absurd method. thanks :) – mathsalomon Aug 12 '11 at 3:04
I've never heard anyone refer to reductio ad absurdum as "the absurd method" before... – Rahul Aug 12 '11 at 4:51
That's what I meant "reductio ad absurdum," excuse me, not fluent in English. – mathsalomon Aug 12 '11 at 5:25

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