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The question is as follows:

Give 3 information:

(1) f is a polynomial (thus I claim f is continuous at every point)

(2) $f(a) = f'(a) = f''(a) = f'''(a) = 0$

(3) $f(b) = 0$

Goal: use Rolle's Theorem to show that there is c satisfying $a < c < b$ such that $f^4(c) = 0$

Here is my attempt:

1/ Recall Rolle's Theorem:

If f is continuous on $[a,b]$ and f is differentiable on $(a,b) $

[ i.e: f'(x) exists in a < x < b ], and $f(a) = f(b)$

Then there is c such that $a < c < b$ and $f'(c) = 0$

2/ By condition (2) and (3), $f(a) = f(b) = 0.$
So there is k satisfying $a < k < b$ and $f'(k) = 0$ by Rolle's

Now $f'(k) = f'(a) = 0$, then use Rolle's again, there is m satisfying $a < m < k < b$ and $f''(m) = 0$

Continue up to the 3rd derivative, where I should get $f'''(n) = f'''(a) = 0$ where $a < n < m < k < b$. Then use Rolle's again, I say there is c satisfying $a < c < n < m < k < b$ such that $f^4(c) = 0$. c definitely satisfies a < c < b, since c < something < b, that "something" namely is n, m, k.

**Would someone please check my proof for any mistakes? Somehow I feel my proof is a bit too obvious to be true >_< But since the problem asks me to specifically use Rolle's Theorem, this approach is the first way that I can think of.

Thank you in advance ^_^

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  • $\begingroup$ Thanks for thoroughly explaining your work. $\endgroup$ – Stefan Smith May 5 '13 at 23:29
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The proof that you sketched is exactly what I was thinking after I read the 3 points together with the request to use Rolle's theorem. So in principle, it looks OK to me :-).

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It's a perfect proof but what is lacking is to learn to use $\LaTeX$.

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