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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.

Tejus
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Vulcan
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    Please read a LaTeX manual. You should use the dollar signs for the math only, and not for all the text. – TMM Dec 09 '12 at 14:23
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    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.) – user642796 Dec 09 '12 at 14:30
  • That's ok,Thank you^^ – Vulcan Dec 09 '12 at 14:30
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    $g(x)=\frac{f(x)-f''(x)}{f'(x)}$, 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? – vesszabo Dec 09 '12 at 15:07
  • @vesszabo: the supremum will be $\infty$ – Simon Markett Dec 09 '12 at 15:11
  • I am a engineering student(also not a course in real analysis),so i don't have the concept of $\sup$. – Vulcan Dec 09 '12 at 15:15
  • Uhh, yes, I indeed misinterpreted, sorry. – vesszabo Dec 09 '12 at 15:15
  • If $f(x)-f''(x)=0$ for every $x\in [a,b]$, then $f(x)=\exp(x)$ or $f(x)=\exp(-x)$, but this is impossible. Since $g(x)=\frac{f(x)-f''(x)}{f'(x)}$ if $f'(x)\neq0$, we have to show that the range of $\frac{f(x)-f''(x)}{f'(x)}$ is $(-\infty,\infty)$, because in the opposite case we can choose $g$ as a constant function such that there is no solution. – vesszabo Dec 09 '12 at 15:46
  • That means there is no solution for $f''(x)+g(x)f'(x)-f(x)=0$?? – Vulcan Dec 09 '12 at 15:54
  • I don't understand your second and third sentence,why we have to show the range of $g(x)$ is $(-\infty,\infty)$.Could you elaborate more? – Vulcan Dec 09 '12 at 16:01
  • Not necessarily, I only tried to rephrase the statement. Not the range of $g$, but of right-hand-side, the fraction. – vesszabo Dec 09 '12 at 16:05

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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.

Simon Markett
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