While finding the critical points of a function, the points at which the function $f$ is undefined are not considered critical points. However, the point at which $f’$ is undefined can be considered an inflection point. I don’t understand how this can be the case since I thought that finding the inflection point was analogous to applying the first derivative test to the derivative of a function.
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$\begingroup$ I do consider the points where the derivative doesn't exist as critical points: they are very important in order to describe the behavior of the function. There is no “compelling” mathematical definition of critical point. $\endgroup$– egregCommented Nov 30, 2021 at 23:41
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The correct definition of inflection point is a point where concavity of the function changes. This may well happen at a point where the function is not differentiable. Consider, for example, $$f(x) = \begin{cases} x^2, & x<0 \\ \sqrt x, & x\ge 0. \end{cases}$$ This function is concave up for $x<0$ and concave down for $x>0$, and so $0$ is an inflection point, even though $f'(0)$ does not exist.
Many calculus books lead readers to think that we must have second derivatives changing from positive to negative (or vice versa) in order to test for inflection points; even worse, some readers think that any point with zero second derivative must be an inflection point. (Consider $f(x)=x^4$ at the origin.)
answered Nov 30, 2021 at 23:56
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$\begingroup$ Geometrically, I do understand the reasoning behind why such a point would be an inflection point. However, the point would not be included in the domain of the second derivative. How is this different than also considering a point where the function f itself is not defined as an inflection point? $\endgroup$ Commented Dec 1, 2021 at 0:07
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$\begingroup$ No. Everything we talk about is in the domain of the function. Domain of $f’$ is another matter. $\endgroup$ Commented Dec 1, 2021 at 0:58