3
$\begingroup$

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.

$\endgroup$

marked as duplicate by Xander Henderson, Brahadeesh, Lord Shark the Unknown, Lee David Chung Lin, Dbchatto67 Apr 21 at 4:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

6
$\begingroup$

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: local extrema
  2. endpoints of interval: endpoint extremum (image source: http://tutorial.math.lamar.edu/Classes/CalcI/MinMaxValues_Files/image002.png)
  3. derivative undefined: corner extremum enter image description here, including points of discontinuity enter image description here

Source: © CalculusQuest™


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.

oriignal book's example

$y:[-1\frac12, 2\frac12] \to \Bbb R$ 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 $y = x^3 - 3x + 3$.


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

$\endgroup$
  • 1
    $\begingroup$ This is much much better. Thanks $\endgroup$ – rajdeep dhingra Apr 20 at 20:56
  • 1
    $\begingroup$ The function $y=x^3-3x+3$ is defined on the entire real line. The interval $[-1.5,2,5]$ is the interval they're looking for mins and maxes on. It's not the domain of the function $y$. If $[-1.5,2,5]$ was the domain of the function $y$ and we were looking for mins and maxes on the interval that was equal to that domain, I believe we would have to consider them critical points because they would fit the definition of critical points. I think, practically, it would make no difference because endpoints still have to be checked. There is a possibility that mins and maxes can occur at endpoints. $\endgroup$ – Michael Rybkin Apr 21 at 2:05
  • 1
    $\begingroup$ I think "critical numbers" is a concept that's only applicable within the context of the analysis of graphs. $\endgroup$ – Michael Rybkin Apr 21 at 2:12
  • 1
    $\begingroup$ @MichaelRybkin Thanks for clarifying. That makes sense now. I still prefer the American way instead of the Soviet way. As OP says, the former, though being intuitive and unprecise ("possibly") at the first place, is much clearer than the later. Since comments on the SE network are ephemeral, I've incorporated your comments into my answer, so that interested readers can understand Wiki's definition better. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 21 at 9:23
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I get your logic. But don't we have a definition for derivative at end points. We only take one sided derivatives at the end points. Do you have any book/source to back up your answer ? Thank you for your help $\endgroup$ – rajdeep dhingra Apr 20 at 16:01
  • $\begingroup$ Well, you can look it up on Wikipedia: en.wikipedia.org/wiki/Differentiable_function Wikipedia is more or less a reliable and authoritative source of information when it comes to mathematics. $\endgroup$ – Michael Rybkin Apr 20 at 16:02
  • $\begingroup$ They have mentioned on that Wikipedia page that , the they are taking in an open interval U. Whereas what I am asking here is a closed interval case. $\endgroup$ – rajdeep dhingra Apr 20 at 16:04
  • $\begingroup$ It says quite clearly at the beginning of the second paragraph: More generally, if $x_0$ is a point in the domain of a function $f$, then $f$ is said to be differentiable at $x_0$ if the derivative $f '(x_0)$ exists. It follows that if the point $x_0$ is in the domain of the function and the derivative $f '(x_0)$ does not exist, then the function is non-differentiable at that point. $\endgroup$ – Michael Rybkin Apr 20 at 16:06
  • $\begingroup$ To put it simply, a derivative is basically a formula for the slope of the tangent line to a curve. If there is no curve to speak of to the left of a point, there cannot be a tangle line at that point. $\endgroup$ – Michael Rybkin Apr 20 at 16:13

Not the answer you're looking for? Browse other questions tagged or ask your own question.