Timeline for Concavity of the solution of a parametric implicit function

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Aug 23, 2014 at 18:47	comment	added	Joonas Ilmavirta		@Allen, I'm sorry for being slow. (Your first comment did reach me.) I'm a bit confused myself, frankly. Assuming the inverse function theorem is applicable throughout, the denominator can't vanish and thus has constant sign. I claimed that $\partial_xf+k\partial_xg$ having constant sign for $k\in\{0,1\}$ implies constant sign for $k\in[0,1]$. This seems to require more assumptions on $\partial_xf$ and $\partial_xg$, since they are indeed evaluated on different curves for different $k$. But if you work with the whole curve family $F=c$, $c\in\mathbb R$, at once, my argument works better.
Aug 22, 2014 at 21:55	comment	added	Allen		Thanks for your solution. I am still a bit confused on the increasing property. The implicit derivative holds on the curve $(x,y(x;k))$, depending on $k$. By knowing the sign on the curves $(x,y(x;0))$ and $(x,y(x;1))$, how do we induce the sign on the curve $(x,y(x;k))$ [as e.g., $\partial_x f(x,y(x;k))$ is now evaluated at $(x,y(x;k))$ instead]?
Aug 20, 2014 at 21:23	history	answered	Joonas Ilmavirta	CC BY-SA 3.0