To Prove : $ \frac{f'(c)}{f(c)} = \frac{1}{a-c} + \frac{1}{b-c} $
I think I should proceed in following way :
Define $g(x) = ln|f(x)| + ln|x-a| + ln|b-x|$
such that
$g'(x) = \frac{f'(x)}{f(x)} - \frac{1}{a-x} - \frac{1}{b-x} $
Then, using Rolle's Theorem , I have to prove that :
$\exists c \in (a,b)$ such that $g'(c) = 0$
But I am not able to solve it any further .
Let $f$ be continuous on $ [a,b]$, differentiable on $ (a,b) $ and $ f(x) \neq 0 \forall x \in (a,b) $ Prove $ \exists c \in (a,b) $ such that
3
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calculus
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0Use $g(x) = \ln|f(x)| \pm (\ln|x-a| + \ln|b-x|)$ – 2017-01-29
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1The function you defined is not continuous on $[a, b]$. $g(a)$ and $g(b)$, are not defined. – 2017-01-29
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0Ohh, so how can I can I modify g(x) so as to make it continuous and give the same derivative? – 2017-01-29
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0In case $\lim_{x\rightarrow a} f'\ne +\infty$ and $\lim_{x\rightarrow b} f'\ne -\infty$, $h(x) = \frac{f'(x)}{f(x)} - \frac{1}{a-x} - \frac{1}{b-x}$ would be positive near $a^+$ and negative near $b^-$, and by intermediate value theorem would have a root. – 2017-01-29
1 Answers
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Let $$g(x) = f(x) (x-a) (b - x) $$ then $g(a) =g(b) = 0$ and hence there is a $c\in (a, b) $ for which $g'(c) =0$. Clearly this means that $$ f'(c) (c-a) (b - c) + f(c) (b-c) - f(c) (c-a) =0$$ Dividing by $f(c)(c-a) (b-c) $ we are done.