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Let $f$ be continuous on $[a,b]$, with continuous first and second derivatives on $[a,b]$. Suppose $f$ has at least 3 distinct zeroes in $[a,b]$. Show that $f''(x)+g(x)f'(x)-f(x)=0$ has at least one root in $[a,b]$, where $g$ is any continuous function over $[a,b]$.

Say $f(\alpha)=f(\beta)=f(\gamma)=0;\;\alpha,\beta,\gamma\in(a,b)$ and $\alpha<\beta<\gamma$ since they are distinct.
Hence, by Mean Value Theorem, I know $f'(\zeta_1)=0$ where $\zeta_1\in(\alpha,\beta)$; $f'(\zeta_2)=0$ where $\zeta_2\in(\beta,\gamma)$. And also for $f''(\zeta_3)=0$; where $\zeta_3\in(\zeta_1,\zeta_2)$.

I know I have to use intermediate value theorem, but I have no idea on how to use my result to do that. Thank you.

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    $\begingroup$ Please read a LaTeX manual. You should use the dollar signs for the math only, and not for all the text. $\endgroup$ – TMM Dec 9 '12 at 14:23
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    $\begingroup$ Please only use $\LaTeX$ to typeset the mathematical parts. (If I somehow changed something meaningful, I apologise, but I do not think anything was lost in the move from $\LaTeX$ to text.) $\endgroup$ – user642796 Dec 9 '12 at 14:30
  • $\begingroup$ That's ok,Thank you^^ $\endgroup$ – Vulcan Dec 9 '12 at 14:30
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    $\begingroup$ $g(x)=\frac{f(x)-f''(x)}{f'(x)}$, if $f'(x)\neq 0$. If you choose $g(x):=\sup\{\frac{f(x)-f''(x)}{f'(x)}\}+1$, then $g$ is a constant function, continuous, but there is no solution of your equation. Maybe I misinterpret something? $\endgroup$ – vesszabo Dec 9 '12 at 15:07
  • $\begingroup$ @vesszabo: the supremum will be $\infty$ $\endgroup$ – Simon Markett Dec 9 '12 at 15:11
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The crucial idea is that $g$ can be any continuous function. For example it can be something like $Cf'$ for $C$ really big. Then the function $f''+gf'-f=f''+C(f')^2-f$ will be dominated by the non-negative summand $C(f')^2$. So it is morally clear that we have to have a careful look a the locus where $f'=0$.

So let $x<y$ be two zeros of $f'$ such that there is precisely one zero of $f$ but no other zeros of $f'$ inbetween; without loss of generality we may assume that $f'>0$ on the interval $(x,y)$. Then $f(x)<0$, $f(y)>0$, $f''(x)>0$ and $f''(y)<0$. Therefore $f''(x)+gf'(x)-f(x)>0$ and $f''(y)+gf'(y)-f(y)<0$. By mean value we are done.

This is all still a bit unprecise but I have to go... the idea is there though. So we have be to able to make the choice in the last paragraph. If this is not possible we should get an easy special case.

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  • $\begingroup$ After drawing a picture,i got what you mean finally.Thanks $\endgroup$ – Vulcan Dec 9 '12 at 16:46

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