Application of derivatives

**pankaj** · November 14th 2008, 04:15 PM

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

**NonCommAlg** · November 14th 2008, 05:47 PM

Originally Posted by pankaj

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

it's just a simple result of Rolle's theorem ... and i think that 4 should be actually 3.

**pankaj** · November 15th 2008, 05:19 AM

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.

**NonCommAlg** · November 15th 2008, 09:04 AM

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

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

that $f'''(\gamma)=0.$ but obviously $\alpha < \gamma < \beta,$ which contradicts our assumption that $\alpha, \beta$ are two consecutive zeros of $f'''.$ so $f$ can have at most 3 zeros between $\alpha, \beta. \ \ \Box$

**pankaj** · November 15th 2008, 04:52 PM

Somehow I was missing point you made out inthe last sentence.
Nice proof.Graphical method on which I was working was not very convincing.

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