This is a question of convention - specifically the convention used in Stewart's Calculus.

In Stewart's calculus (the latest version), in chapter 4.1, definition 6 defines a critical number of a function $f$ to be a number $c$ in the domain of $f$ such that either $f'(c) = 0$ or $f'(c)$ does not exist.

Suppose $f$ is obtained from a nice function by restricting to a closed interval $[a,b]$. Say, $g(x) = x$ and $f = g|_{[0,1]}$. Does Stewart consider 0 and 1 to be critical points of $f$? In other words, does $f'(0)$ and $f'(1)$ exist?

The book seems to be extremely elusive above this, going so far as to not including any exercises that might elucidate which convention he uses.

I'm asking this because I'm an instructor trying to decide what convention to take. I'm hoping that by choosing one convention over the other, I do not inadvertently subtly contradict Stewart somewhere down the road.

In Stewart's description of "The closed interval method" (Section 4.1), he first asks you to find the values of $f$ at critical numbers in $(a,b)$, and then to compute the values of $f$ at the endpoints $a$ and $b$ (and doesn't simply say compute values of $f$ at all critical numbers). The fact that he does this would suggest that he does not consider the endpoints to be critical points. On the other hand, taking this convention would seem to imply that $f'(a)$ does exist, but he defines $f'(a) := \lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}$, and he only ever defines the limit of a function at points on the interior of its domain, so technically according to Stewart this limit is "undefined", so "does not exist", but then it is very puzzling why he phrases the closed interval method in the way that he does.

  • $\begingroup$ Stewart is not claiming that $f'(a)$ or $f'(b)$ exist or don't exist. He is not calling the endpoints critical numbers. Just because something is not a critical point does not mean its derivative has to exist - that is not logically equivalent to its converse $\endgroup$ Commented Mar 31, 2020 at 6:42
  • $\begingroup$ @NinadMunshi His definition of a critical point is a point where either the derivative vanishes or it does not exist. Thus, this means that not being a critical point means the derivative exists and is nonvanishing. A definition is an "if and only if". $\endgroup$ Commented Mar 31, 2020 at 6:45
  • $\begingroup$ A definition is not always an "if and only if" $\endgroup$ Commented Mar 31, 2020 at 6:46
  • $\begingroup$ @NinadMunshi Give me one example of a peer-reviewed math paper where a definition is not an if and only if. $\endgroup$ Commented Mar 31, 2020 at 6:47
  • $\begingroup$ You're giving me a textbook written for first year calculus students, it seems a little unfair to expect me to provide you a paper no? Focusing on the purpose of a definition is more important than the pathologies of an imprecise definition. When Stewart is mentioning limits not existing, we both know intuitively he is referring to cusps and vertical tangents (as he is assuming continuity). End points don't fit in that picture. Logically, it would not be a good idea to focus on that pathology as it is not the point of the concept - because Stewart did not aim for a real analysis audience $\endgroup$ Commented Mar 31, 2020 at 6:56

1 Answer 1


You are correct. The derivative does not exist at the endpoints based on Stewart's definition, and therefore by Stewarts' definition endpoints are critical points. However, as Ninad Munshi, suggests, do not focus on this. If a student asks, you can have the same discussion we are having here. Whether you test endpoints to find a global max or min or you include them as "critical points" is only a matter of naming.


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