Are the endpoints of the domain of a function counted as critical points?

Question

Do the end points of a domain come under critical points? I know we say critical point is a point where the derivative is zero or the derivative doesn't exist.

For example: $$ f:[0,\pi] \to [-1,1], f(x) = \sin(x).$$ Does this have 1 critical point or 3 critical points (0 and $\pi$ included) ?

NOTE: This question is limited to only Single Variable Functions. Although I really would love an insight to this for Multivariable as well.

Answer 1

Edited

$$f'(x) = \cos(x) = 0 \iff x = \frac{\pi}{2}$$ The function $f$ has three critical points.

1. A local maximum: $x = \pi/2$ (at which $f(\pi/2) = 1$.)
2. The endpoints of the domain of $f$ (that is, $[0,\pi]$): $x = 0$ and $x = \pi$ .

Since the other answer has elaborated on OP's understanding on the definitions using the differentiability of $f$, there's no point repeating its arguments. Instead, I'll cite from a university course web page to show why we need to include the endpoints of the domain of $f$ if $f$ is defined at those points. By doing so, we learn the definitions by heart instead of by memory.

The goal of the procedure of finding critical points is to identify points in the domain at which a (global and/or local) extremum could possibly occur.

1. vanishing derivatives:
2. endpoints of interval: (image source: http://tutorial.math.lamar.edu/Classes/CalcI/MinMaxValues_Files/image002.png)
3. derivative undefined: , including points of discontinuity

Source: © CalculusQuest™

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As @mathcounterexamples.net points out in Can critical points occur at endpoints? E.g. $f(x) = \frac{1}{x}$ at the interval $[1,4]$, the definition of critical points can vary. Although OP's definition comes from Wikipedia's page on critical points, it actually originates from p.84 of Demidovǐc and Baranenkov's Problems in mathematical analysis.

The converse is not true: points at which $f'(x) = 0$, or $f'(x)$, does not exist (critical points) are not necessarily extremal points of the function $f(x)$.

Example 5 in p.86 seems contradictory to what we've known.

$y:[-1\frac12, 2\frac12] \to \Bbb R$ defined as $y = x^3-3x+3$. In the solution, an explicit expression for $y'$ is first given, then it says "the critical points of $y$ are $x = \pm 1$".

Edited again: As @MichaelRybkin points out, the author actually means the greatest and least values on $[-1\frac12, 2\frac12]$ of $y: \Bbb{R} \to \Bbb{R}$ defined by $y = x^3 - 3x + 3$.

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Final remark: Personally, I prefer © CalculusQuest™'s definition, which includes the endpoints of the domain since that makes much more sense with our goal.

Answer 2

Yes, the function has 3 critical numbers. One is where the derivative of the function $f(x)=\sin{x}, x\in[0,\pi]$ is zero and the other two happen to be the endpoints $x=0$ and $x=\pi$ because the function $f(x)=\sin{x}, x\in[0,\pi]$ is non-differentiable at those points.

Do you remember what it means for a function to be differentiable at a point? The function has to have a derivative at that point. What is the derivative of the function $f(x)=\sin{x}, x\in[0,\pi]$ at $x=0$? Well, it should be:

$$ \lim_{x\to0}\frac{\sin{x}-\sin{0}}{x-0}=\lim_{x\to0}\frac{\sin{x}}{x} $$

Which is nothing more than two one-sided limits (if those two limits exist and are equal to each other, the limit itself exists):

$$ \lim_{x\to0^-}\frac{\sin{x}}{x},\ \lim_{x\to0^+}\frac{\sin{x}}{x} $$ But the first of those two limits for all intents and purposes is nonexistent because all $x$ values that lie to the left of $0$ are not in the domain of the function $f(x)=\sin{x}, x\in[0,\pi]$. For a limit to exist, you need two one-sided limits. But you've got only one! Thus, the derivative at $x=0$ does not exist which makes it a critical number. The exact same idea applies to the other endpoint.