# Proof using Rolle's Theorem to show there is c such that f$^4$(c) = 0, for a < c < b

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 ^_^

• Thanks for thoroughly explaining your work. May 5, 2013 at 23:29

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