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Consider function of one independent variable $x$, where $y=f(x)$.Assume function to continuous and all its $n$ derivatives exist and are continuous on given domain.

Now if first non-zero derivative, at $x=a$, is even and is positive we say $x=a$, is point of minima, and if it is negative we say it is maxima.

Note:If you can't have first derivative zero for $x=a$, then also there is possibility to get a maxima or minima. That means being first derivative to be zero is not necessary condition for a point $x=a$ to be maxima or minima.

Now consider function of two independent variables $z=f(x,y)$. Here we talk in terms of so called "partial derivatives".

According to some theorem, here if a point $(x,y)=(a,b)$ is maxima or minima then it is "guarantee" that both its first order partial derivatives is zero. Therefore, having if both partial derivatives is zero at point $(x,y)$ is necessary condition for having a maxima or minima.

Why can't here third order or higher order partial derivatives are not deciding maxima or minima as in function of one variable? Why only first and second order derivatives are used to decide the maxima or minima? Why we cant also have higher order derivative test as , we have in function of one independent variable?

Mårten W
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Jay Patel
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  • For a smooth function like the ones you're considering, the sentence "Note:If you can't have first derivative zero for x=ax=a, then also there is possibility to get a maxima or minima. That means being first derivative to be zero is not necessary condition for a point x=ax=a to be maxima or minima." seem wrong: for if $x=a$ is a maximum for a smooth function $f:\mathbb R \to \mathbb R$, then $f(a) = 0$. In short: for a smooth function $f$ on the reals to have a max or min at $a$, we must have $f'(a) = 0$. Of course, it may not be a strict maximum (as any constant function shows). – John Hughes Oct 10 '16 at 11:11
  • @JohnHughes Sir, What is mean by smooth Function. – Jay Patel Oct 10 '16 at 11:43
  • Infinitely differentiable, with all derivatives being continuous. (In other words: what you said in your first sentence.) – John Hughes Oct 10 '16 at 12:30
  • See my answer here: https://math.stackexchange.com/a/2870187/71829 – Keshav Srinivasan Aug 03 '18 at 05:42

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