If solution of $f'(x) /g'(x)= \lambda $ is an extremum point of $y = f(x) - \lambda \,g(x) $ then can it be shown that $f(x)$ and $g(x)$ are monotone functions?
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I think this question is unclear. What do you mean by monotonously changing function? Just monotone? Locally? What are the hypotheses on $f$, $g$? – Fimpellizzeri Jan 29 '17 at 06:43
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$f(x)-λg(x)=0$ implies $\frac{f(x)}{g(x)}=λ$ and so $\frac{g'(x)}{g(x)}=\frac{f'(x)}{f(x)}$ Hint: domain of $\ln{x}$: $x>0$ – M.Diggerson Jan 29 '17 at 06:48
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What even is an extremum point of $f(x)-\lambda g(x)=0$? Is that the constant, zero function? – Fimpellizzeri Jan 29 '17 at 07:08
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Sorry, edited it. – Narasimham Jan 29 '17 at 07:20
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Not even locally, no. A 'simple' counter-example can be obtained by taking $\lambda=0$ and $f$ given by
$$f(x)=x^2\cdot \exp\left({\sin\left(\frac1x\right)}\right),$$
where of course $f(0)=0$. Then $f(x)>0$ when $x\neq 0$, so that $x=0$ is a (global) minimum, but $f$ is never monotone near $x=0$.
You can check that $f$ is everywhere differentiable, although its derivative is not continuous. That said, if you take $f(x)=x^4\cdot \exp\left({\sin\left(\frac1x\right)}\right)$, then all of these hold and additionally $f'$ is everywhere continuous (differentiable in fact).
Fimpellizzeri
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I was implying non-zero $ \lambda$. For $ \lambda = 1,f = x, g = -1/x, f- \lambda g $ has minimum at $ x = 1$ – Narasimham Jan 29 '17 at 11:53
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You have probably not yet asked this clearly yet. Let $h$ be any function, say $h(x)=x^2\cdot \exp\left({\sin\left(\frac1x\right)}\right)$. Let $\alpha,\beta\neq0$ and let $f=\alpha\cdot h$ and $g=\beta\cdot h$. Then with $\lambda =\alpha/\beta$, every $x$ is a solution to $f'/g'=\lambda$, so the answer still applies. – Fimpellizzeri Jan 29 '17 at 17:12
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What is true only in the immediate neighborhood,need not be so in entire domain, right? – Narasimham Jan 29 '17 at 20:11
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That is also true, but even if you were only concerned about them being locally monotone (as opposed to globally monotone, which is stronger), that is still not true (with those hypotheses). – Fimpellizzeri Jan 29 '17 at 23:35