How to calculate the estimation error at extreme points? By measuring $y$'s, we can estimate the associated parameter $x$ in a way that $y=f(x)$. If some errors are involved in the measurement of $y$, i.e., $y_{\rm meas}=\overline{y}\pm\Delta y$, then the error of estimated value of $x$ exists, and eventually limits the estimation precision, i.e., $x_{\rm est}=\overline{x}\pm\Delta x$. Assuming the error is small such that the function $f(x)$ can be modeled as a linear function, we can use the linear error propagation method to determine the estimation error of $x$, i.e., 
\begin{equation}
\Delta x=\left. \frac{\Delta y}{\vert \partial \overline{y}/\partial x\vert}\right\vert_{x=\overline{x}}.
\end{equation}
This implies that the error $\Delta x=\infty$ at the extreme points, i.e., when $\frac{\partial y}{\partial x}=0$. Does this mean that we cannot estimate the true value of $x$ at all? 
In fact, I want to know how to calculate the estimation error $\Delta x$ when $\frac{\partial y}{\partial x}=0$. Is there any particular method than the linear error propagation method? 
 A: One cannot use the lowest order linear error propagation formula when $\frac{df}{dx}=0$. Traditional one-parameter error propagation is based on the Taylor Series expansion (in your notation)
\begin{align}
\Delta y & = \sum_0^\infty \left.\frac{d^nf}{dx^n}\right|_{\bar{x}}\frac{(\Delta x)^n}{n!}\\
& = \left.\frac{df}{dx}\right|_{\bar{x}}\frac{(\Delta x)}{1!}+\left.\frac{d^2f}{dx^2}\right|_{\bar{x}}\frac{(\Delta x)^2}{2!}+\ldots
\end{align}
The usual error propagation formula works when the magnitude of the first term is larger than the sum of the rest of the series, which is impossible if $\frac{df}{dx}=0$. One must then look at higher order non-zero terms. For example, if $f(x)=x^2$ and the measurement is $y_{meas}=0 ± 0.1$, we find that $\left.\frac{df}{dx}\right|_{\bar{x}=0}$ is zero, but $\left.\frac{d^2f}{dx^2}\right|_{\bar{x}=0}=2$ is not. (It is, in fact, the only non-zero term in the series in this very simple example.) So
$$\Delta y = 2 \frac{(\Delta x)^2}{2!} = (\Delta x)^2$$
or
$$\Delta x = \sqrt{\Delta y}$$
i.e. $\Delta x =0.1$ if $\Delta y = 0.01$.
A more direct way to propagate the uncertainties from y to x is by solving  $f(x)=\bar{y}±\Delta y$ for $x$. If $x_{high}$ and $x_{low}$ are the highest and lowest solutions bracketing the solution to $f(\bar{x})=\bar{y}$, the inferred $x$ uncertainty could be stated as $+(x_{high}-\bar{x})$ and $-(\bar{x}-x_{low})$.  (These uncertainties will not in general be Gaussian, so be careful how you use them.)  This method also gives $\Delta x =0.1$ for the above $f(x)=x^2$ example.
Direct propagation works even if all the derivatives $f(x)$ are zero at the point of interest, i.e. it is a flat function (https://en.wikipedia.org/wiki/Flat_function). For example, consider the step function $y=floor(x)$, i.e. $y=n$ for $n\le x <n+1$, where $n$ is any integer.  If you measure $\bar{y}=3\pm0.01$, you know that $3\le \bar{x}<4$, so $\Delta x = 0.5$ is the full range uncertainty and $x_{est}=3.5\pm0.5$.  
