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Let $f'(x)$ and $g'(x)$ satisfy the hypothesis of mean value theorem, then prove that

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

where $g''(c)\neq 0$. I have tried various things like Cauchy MVT with $f'(x)$ and $g'(x)$, substituting $h(x)=\dfrac{1}{g'(x)}$ ,$h(x)=f'(x)+\dfrac{a}{g'(x)}$ where $a\in\mathbb{R}$, but it didn't lead anywhere. Any help will be appreciated. Please note that I am allowed to use only Cauchy's MVT, LMVT and Rolle's theorem.

Nosrati
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Archis Welankar
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  • See also https://math.stackexchange.com/questions/993286/mean-value-theorem-to-second-derivative – Nosrati Sep 01 '17 at 17:13

1 Answers1

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Let $$ k=\frac{f(b)-f(a)-(b-a)f'(a)}{g(b)-g(a)-(b-a)g'(a)}. $$ Define $$ F(x)=f(x)-f(a)-(x-a)f'(a)-k[g(x)-g(a)-(x-a)g'(a)]. $$ Then $$ F(a)=0, F(b)=0. $$ By Rolle's Theorem, there is $\xi\in(a,b)$ such that $F'(\xi)=0$, namely $$ f'(\xi)-f'(a)-k[g'(\xi)-g'(a)]=0$$ from which one has $$ k=\frac{f'(\xi)-f'(a)}{g'(\xi)-g'(a)}.$$ By Cauchy's Theorem, there is $c\in(a,\xi)$ such that $$ k=\frac{f'(\xi)-f'(a)}{g'(\xi)-g'(a)}=\frac{f''(c)}{g''(c)}.$$ Done.

xpaul
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