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Suppose we have a strictly increasing and continuous function $f(x)$ such that for some $\bar{x}: \exists\ f'(\bar{x}) = 0$ while the derivative exists and is positive in a small neighborhood of $\bar{x}$. Function $y(x)$ is given implicitly by $$af(x)+bf(y)=(a+b)f(\bar{x})$$ where $a, b$ are some positive constants. I am interested in $\frac{dy(x)}{dx} \bigg|_{x = \bar{x}}$.

From the Implicit function theorem we know that $$\frac{dy(x)}{dx} = -\frac{af'(x)}{bf'(y)}$$ as long as $f'(y) \neq 0$ which is not true at the point $x = \bar{x}$, since $y(\bar{x}) = \bar{x}$ and $f'(\bar{x}) =0$.

My question is: Suppose that the derivative above actually exists, moreover there exist finite, non-zero left and right second derivatives $f''_{+}(\bar{x})$ and $f''_{-}(\bar{x})$ which have different signs.

Would it be true to write: $$\frac{dy(x)}{dx}\biggl|_{x=\bar{x}+0} = \lim_{x \to \bar{x}+0} -\frac{af'(x)}{bf'(y(x))}\ ?$$ And then to use L'Hospital rule to get $$\frac{dy(x)}{dx}\biggl|_{x=\bar{x}+0} = \lim_{x \to \bar{x}+0} -\frac{af''_{+}(x)}{bf''_{-}(y(x)) y'(x)}$$ Concluding that $$\frac{dy(x)}{dx}\biggl|_{x=\bar{x}+0} = -\sqrt{\frac{a}{b}\bigg|\frac{f''_{+}(\bar{x})}{f''_{-}(\bar{x})}\bigg|}$$ where the minus sign follows from the fact that clearly $y(x)$ is decreasing since $f(x)$ is increasing and $a, b > 0$

If the conclusion above is incorrect which additional assumptions I need for it to work?

Would be very grateful for any help. Thanks in advance!

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  • $\begingroup$ What is the context? Is this a (graded) homework? $\endgroup$
    – H. H. Rugh
    Jan 1, 2021 at 18:56
  • $\begingroup$ @H.H.Rugh No. I encountered smth similar to that in my research work, and then decided to explore the question in general, but didn't find anything alike. $\endgroup$
    – D F
    Jan 2, 2021 at 5:02

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This is not a case for using l'Hospital. Instead use Taylor expansion on the two sides. Set $\lambda_+=f''_+(\bar{x})>0$ and $\lambda_-= f''_-(\bar{x})<0$.

For $h>0$: $f(\bar{x}+h) -f(\bar{x})= \frac12 \lambda_+ h^2+o(h^2) = \frac12(\lambda_+ + \epsilon(h)) h^2$.

For $k<0$: $f(\bar{x}+k) - f(\bar{x})= \frac12 (\lambda_-+\epsilon(k))k^2$.

Suppose now that $x=\bar{x}+h$ with $h>0$ (small) which implies that $y=y(x)=\bar{x}+k$ with $k=k(h)<0$. Then $$ a (\lambda_+ +\epsilon(h)) h^2 + b (\lambda_- + \epsilon(k)) k^2 = 0,$$ which implies $y'_+(\bar{x}) = \lim_{h \rightarrow 0^+} k(h)/h = -\sqrt{-\frac{a\lambda_+}{b\lambda_-}}$ similarly $y'_-(\bar{x}) = \lim_{h \rightarrow 0^-} k(h)/h = -\sqrt{-\frac{a\lambda_-}{b\lambda_+}}$. Note that the derivatives from the left and from the right may be different.

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  • $\begingroup$ Thanks a lot! That makes sense. I am wondering, am I correct that for that approach to be valid, we need $f(x)$ to be 2 times differentiable in the neighborhood of $\bar{x}$ except maybe at $\bar{x}$? $\endgroup$
    – D F
    Jan 4, 2021 at 7:32
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    $\begingroup$ Indeed, in the above calculation, $f''_+>0$ and $f''_-<0$ should exist at $\bar{x}$ but as they have opposite signs, $f''$ does not exist at that point. If you impose other (more pathological) local assumptions on $f'$ (e.g. fractional power laws) you may arrive at other solutions to the problem. $\endgroup$
    – H. H. Rugh
    Jan 4, 2021 at 7:51

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