Mean Value Theorem to Second Derivative
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.
$\endgroup$ 21 Answer
$\begingroup$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 $c$
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