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Hello @Ilya, thanks for your reply. I think this should work out. TO get back g(x) from a function $f(x)=-\log g(e^{-x})=x+\eps$ I could replace $x$ by $-logx$, and I end up getting $g(x)=x.e^{-\eps}$. Similarly, $f(x)=1.1x+\eps$ on $(0,\,1.7)$ would give $g(x)=x^{1.1}.e^{-\eps}$. This function would not be differentiable at the kink. I think I missed including the conditions $g(0)=0$ and $g(1)=1$ in my initial post, and I apologize for that. But, given this extra information, $g(1)=1.e^{-\eps} \neq 1.$
@AthanagorWurlitzer, thank you for your comment. I have also thanked you on the MSE website. I missed out those two conditions, and have edited the post.
