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If $f(x)$ is a twice differentiable function, continuous in it's domain such that $f(a)=0$, $f(b)=2$, $f(c)=-1$, $f(d)=2$ and $f(e)=0$ where $a<b<c<d<e$;

Then find the minimum number of zeroes of $g(x)=(f'(x))^{2}+f''(x).f(x)$ $\forall$ $x \in [a,e]$.

I cannot decide how I should proceed. I realize that we will have $f'(z)=0$ for three values of $z$ in $[a,e]$ (using Rolle's theorem, I believe). This will give me:

$f''(z).f(z)=0$

How do I proceed from here? I don't think I can differentiate $g(x)=0$ to obtain another equation, since $f'''(x)$ is undefined.

Any help is appreciated, thanks!

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darkv0id
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1 Answers1

7

Hint:

Consider the function $h(x)=f(x)f'(x)$.

P..
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  • a good hint, +1 – Rustyn Jan 16 '13 at 16:17
  • @Pambos Clearly, $h'(x)=g(x)$. What next? Can I use $\int^x_0 g(x)\ dx= \int^x_0 h'(x)\ dx=0$ to obtain $f'(x)f(x))=0$ ? – darkv0id Jan 16 '13 at 16:36
  • Apply Rolle's theorem on $h$. – P.. Jan 16 '13 at 16:41
  • @Pambos Applying Rolle's Theorem, I will obtain three zeroes of $g(x)$, corresponding to $z_1$ and $a$; $z_2$ and $z_1$; and $e$ and $z_2$,where $z_1$, $z_2$ and $z_3$ are the zeroes of $f(x)$, is that correct? – darkv0id Jan 16 '13 at 16:54
  • @Pambos Is my approach (integrating $g(x)$) mathematically sound? – darkv0id Jan 16 '13 at 16:56
  • @darkv0id: We want to find the minimum number of zeros of $h(x)=f'(x)f(x)$ and apply Rolle's theorem on $h$ to find the minimum number of zeros of $g$, right? But the minimum number of zeros of $h(x)$ is the sum of the minimum number of zeros of $f(x)$ and $f'(x)$. For $f'(x)$ you have it right, is three (using Rolle's theorem, yes); but for $f(x)$ is not $3$, is $4$ (here use the intermediate value theorem). So the total for $h$ is ... – P.. Jan 16 '13 at 17:16
  • @Pambos Oh I get it now, we will get a total of seven points including extrema and zeroes of $f(x)$, giving us a total of six points where $h'(x)=g(x)=0$. – darkv0id Jan 16 '13 at 17:56
  • @Thank you for your help (and patience for putting up with me) ! – darkv0id Jan 16 '13 at 17:56
  • That's right six points. You are welcome @darkv0id. – P.. Jan 16 '13 at 19:14