Let $\displaystyle f(x)$ be a thrice continuously differentiable function.Prove or disprove that between two consecutive roots of $\displaystyle f'''(x)=0$ there exist atmost $\displaystyle 4$ roots of $\displaystyle f(x)=0$

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- Nov 14th 2008, 04:15 PMpankajApplication of derivatives
Let $\displaystyle f(x)$ be a thrice continuously differentiable function.Prove or disprove that between two consecutive roots of $\displaystyle f'''(x)=0$ there exist atmost $\displaystyle 4$ roots of $\displaystyle f(x)=0$

- Nov 14th 2008, 05:47 PMNonCommAlg
- Nov 15th 2008, 05:19 AMpankaj
I had drawn the graph and I found that there are indeed 3 points but I am not able to apply Rolle's theorem.I doubt if converse of Rolle's theorem is true.

- Nov 15th 2008, 09:04 AMNonCommAlg
suppose $\displaystyle \alpha < \beta$ are two consecutive zeros of $\displaystyle f'''.$ suppose $\displaystyle f$ has 4 zeros $\displaystyle a_i, \ 1 \leq i \leq 4,$ such that $\displaystyle \alpha \leq a_1 < a_2 < a_3 < a_4 \leq \beta.$ then by Rolle's theorem there exist $\displaystyle a_i < b_i < a_{i+1}, \ 1 \leq i \leq 3,$

such that $\displaystyle f'(b_i)=0, \ i=1,2,3.$ again by Rolle's theorem there exist $\displaystyle b_i < c_i < b_{i+1}, \ i=1,2,$ such that $\displaystyle f''(c_i)=0, \ i=1,2.$ finally, again, by Rolle's theorem there exists $\displaystyle c_1 < \gamma < c_2$ such

that $\displaystyle f'''(\gamma)=0.$ but obviously $\displaystyle \alpha < \gamma < \beta,$ which contradicts our assumption that $\displaystyle \alpha, \beta$ are two consecutive zeros of $\displaystyle f'''.$ so $\displaystyle f$ can have at most 3 zeros between $\displaystyle \alpha, \beta. \ \ \Box$ - Nov 15th 2008, 04:52 PMpankaj
Somehow I was missing point you made out inthe last sentence.

Nice proof.Graphical method on which I was working was not very convincing.