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I was solving some curve sketching questions and wanted to ask something that I am a little confused about. Why, when we find the first derivative of an equation and set it equal to zero to find the critical points, we refer to the points as "possible" max and min. Are there times when the critical point is not a max or min? Thank you in advance!

amWhy
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Lulu
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    Consider $ y = x ^ 3 $ – player3236 Feb 24 '21 at 18:06
  • Look at https://math.stackexchange.com/q/1721620/879543 – pawel Feb 24 '21 at 18:32
  • Consider $y=0$ - its first derivative is zero, but does it suffice to determine that the point is a minimum or maximum...? – CiaPan Feb 24 '21 at 21:08
  • @CiaPan: Unfortunately, depending on how you define 'minimum', or 'maximum', either every point is a minimum or maximum, or every point is neither. – Joe Feb 24 '21 at 22:44
  • @Joe You're right, but that was not what I asked about. Does $y' = 0$ suffice to tell minimum from maximum in this case? – CiaPan Feb 25 '21 at 16:48
  • @CiaPan Sorry, I was just making a remark about the confusing terminology. Of course you are correct in saying that you need use other information to determine whether it is a minimum or a maximum. In this case, we can consider how for every neighbourhood of every point $a$, we have $f(x)\geq f(a)$, and $f(x) \leq f(a)$. – Joe Feb 25 '21 at 17:20

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  • The graph of $f(x)=x^2$ has a minimum at $x=0$.
  • The graph of $f(x)=-x^2$ has a maximum at $x=0$.
  • The graph of $f(x)=x\sin x$ does not have a minimum or maximum at $x=0$. Rather, it has a local minimum.
  • The graph of $f(x)=x^3$ does not have a minimum, maximum, local minimum, or local maximum at $x=0$. It has a point of inflection.

Just knowing that $f'(x)=0$ at some point is not enough. We need more information. Moreover, not all critical points occur when $f'(x)=0$. The graph of $f(x)=|x|$ is not even differentiable at $x=0$, but it certainly has a minimum there.

Joe
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