Given that an extreme value can only occur at a critical point and in the following case $$f(x) = \frac{1}{x} \quad [1,4]$$ we definitely have two absolute extreme values (maximum and minimum), are $x = 1$ and $x = 4$ critical points?

If so, what is the reasoning for that: the fact that they are in the domain of $f$ and they are not differentiable at those points?

Is my reasoning correct?

  • 1
    $\begingroup$ It's just a matter of definition. $\endgroup$ – Kenny Lau Aug 12 '18 at 12:56
  • $\begingroup$ @KennyLau not sure what you mean here. $\endgroup$ – bru1987 Aug 12 '18 at 12:58
  • $\begingroup$ There is not much mathematical value in the question "can critical points occur at endpoints" because it is merely a matter of definition. $\endgroup$ – Kenny Lau Aug 12 '18 at 12:58
  • $\begingroup$ Critical points are usually defined as points where the first derivative vanishes, so no end points can be critical points (as there is no derivative). But you can always talk about extrema end points. $\endgroup$ – DonAntonio Aug 12 '18 at 13:04
  • 1
    $\begingroup$ Definition the same in every book? That may be a good idea, but not likely to be implemented in our lifetime. $\endgroup$ – GEdgar Aug 12 '18 at 13:11

A critical point is a point at which the derivative vanishes. So definitely, $1$ and $4$ are not critical points.

Now those points are at the boundary of the domain of $f$ and are extremas.

However, consider a point $x$ which is a minimum or a maximum of a differentiable function $f$ and which belongs to the interior of the the domain of $f$. Then $f^\prime(x)=0$.

In summary

An extrema belonging to the interior of the domain of a differentiable map is a critical point.

An extrema may not be a critical point, if it belongs to the frontier of the domain. Example: the function of the question.

And obviously the derivative of a function can vanish at a point belonging to the frontier. In that case the point is a critical point.

Lastly a critical point may not be an extrema. Example $f: x \mapsto x^3$ at $x=0$.

  • $\begingroup$ That was very helpful my friend, I appreciate your answer. Have a great day! $\endgroup$ – bru1987 Aug 12 '18 at 13:08
  • $\begingroup$ One question though: so the sentence "extreme values can only occur at critical points" is incorrect. Agreed? $\endgroup$ – bru1987 Aug 12 '18 at 13:12
  • $\begingroup$ You’re right. Enjoy too day, night or whatever it is where you’re. $\endgroup$ – mathcounterexamples.net Aug 12 '18 at 13:14
  • $\begingroup$ Ok got it. Thank you. $\endgroup$ – bru1987 Aug 12 '18 at 13:15
  • $\begingroup$ This is quite late, but would you mind taking a look at oregonstate.edu/instruct/mth251/cq/Stage7/Lesson/critical.html, in which the instructor suggests the endpoint(s) of an interval inside the domain is a boundary point. That would contradict the 2nd paragraph of your summary. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 20 at 20:03

I want to point out that it is not sure that you have an extreme point in the critical points. One example is $f(x)=x^3$ which has a $x=0$ as critical point but obviously it's not an extreme.

When we are trying to find a critical point in a certain domain we set $f'(x)=0$. Then (for the fact that I mentioned that not every critical point is extreme) we plug those critical points and the endpoints of the given domain(in case this domain has endpoints)in the function and see which of them is the smaller value(minimum) and which the bigger(maximum). You realise that altough the endpoints may not be critical points, they can behave as extreme points. That happens to the function you mention.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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