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Let $f$ and $g$ be twice differentiable and continuous in $[a,b]$. Show that there exists a $c$ in $[a,b]$ such that:

$$\frac{f(b) - [f(a) + f'(a)(b-a)]}{g(b)-[g(a)+g'(a)(b-a)]} = \frac{f''(c)}{g''(c)}$$

I was thinking by MVT applied to both $f$ and $g$ I know $f(b) = [f(a) + f'(c)(b-a)]$ and $g(b)=[g(a)+g'(c)(b-a)]$. I was not sure why it necessarily be the same $c$ for $f$ and $g$, also I am not seeing the connection to the second derivative ratios. Any insight in the right direction would be appreciated.

UserX
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Aaron
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  • Regarding the equality of both $c$s, maybe the following us useful: http://en.m.wikipedia.org/wiki/Mean_value_theorem#Cauchy.27s_mean_value_theorem. – PhoemueX Oct 27 '14 at 12:38
  • I think it is too, I've been playing around with it for hours to no avail. Do you think Taylor's theorem applies here? – Aaron Oct 27 '14 at 17:19

1 Answers1

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The usual approach is to rewrite the statement so that it becomes a consequence of Rolle's theorem. That is, I want a function $h$ such that the desired inequality is equivalent to $h''(c)=0$

Also, looking at the fraction on the left suggests I should introduce the functions $F(x) = f(x) - [f(a) + f'(a)(x-a)]$ and $G(x) = g(x) - [g(a) + g'(a)(x-a)]$. Then

  • $F(a)=0$, $F'(a)=0$, $G(a)=0$, $G'(a)=0$,
  • $F''=f''$ and $G''=g''$
  • the desired identity is $F(b)/G(b) = F''(c)/G''(c)$.

The denominators should be cleared, to make the expression linear in derivatives.
$$ F(b)G''(c) = G(b)F''(c) $$ after which it's not hard to come up with $h$: $$ h(x) = F(b)G(x) - G(b)f(x) $$

It remains to use Rolle's theorem twice, using the observations that

  • $h(a)=0$
  • $h(b)=0$
  • $h'(a)=0$

The conclusion is $h''(c)=0$ for some $c$

  • Wow. The answer was staring me in the face. Thanks a lot. – Aaron Oct 27 '14 at 17:24
  • Hmm seems like we also need $h'(b) = 0$, but trying to reverse engineer it seems to lead back to what we wanted to show in the first place as well. – Aaron Oct 28 '14 at 03:52
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    No, you don't need that. First application of Rolle gives $h'(c)$ for some $c\in (a,b)$. Then you apply it to $h'$ on $[a,c]$. –  Oct 28 '14 at 03:53
  • Right, thanks, just figured it out as I wrote it out. That's very clever... – Aaron Oct 28 '14 at 03:57
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    Um, this is wrong as typed. You need $F(x) = f(x)-[f(a)+f'(a)(x-a)]$, etc. This is the (signed) vertical distance from the tangent line to the graph of $f$. ... You shouldn't accept answers before you work them out carefully :P You might also point out that you're applying the Cauchy Mean Value Theorem, rather than reproving it :) – Ted Shifrin Aug 08 '17 at 01:24
  • @TedShifrin I think there is another mistake in this answer. Don't we need $h'(b)=0$ also, for applying Rolle's theorem to get $f''(c)=0$ ? –  Aug 14 '17 at 05:21
  • @Blue You don't need it, as you get $h'(\xi)=0$ for some $\xi\in (a,b)$. Also, there's another typo in the definition of $h$. – Ted Shifrin Aug 14 '17 at 05:27
  • @TedShifrin Oh thanks, I get it. :) –  Aug 14 '17 at 05:34