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To find the critical points of any given function $f$, we should first set the first derivative $f'=0$ and solve to find the values of x.

Now to find at which x=a we have a maxima or a minima, we should find the $nth$ derivative such that $f^n(a)≠0$.

  • if n is odd, a is neither a local maximum nor a local minimum.
  • if n is even and $f^n(a)>0$, then f has a local minimum at a.
  • if n is even and $f(n)(a)<0$, then f has a local maximum at a.

What is the intuition behind these 3 points, in other words why these rules work out.

N.B. - I understand how it works till n=2. But when $f''$ turns out to be zero and we continue to find the next derivatives, that's when I start to feel I don't understand what is actually happening.


marked as duplicate by StackTD, Hans Lundmark, José Carlos Santos, user7530, GNUSupporter 8964民主女神 地下教會 Jan 17 '18 at 21:38

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Preliminary remark. I don't know what you mean by "extreme points". If you are interested in the minimal and maximal values of $f$ on a given interval $[a,b]$ you should create a candidate list containing $a$,$b$ and the zeros of $f'$. Then compare the values of $f$ in the points of this list. No higher derivatives needed.

You should compute and analyze the higher derivatives at a critical point $c$ only if you are really interested in the behavior of $f$ in the neighborhood of this point. You then need the first nonvanishing derivative at $c$ (apart from $f(c)$). If this is $f^{(n)}(c)$ then Taylor's theorem tells us that $$f(x)-f(c)=(x-c)^n\bigl(f^{(n)}(c)+o(1)\bigr)\qquad(x\to c)\ ,$$ since all low order terms of the Taylor development at $c$ vanish. The claims in your question then immediately follow from inspection of this formula: Look at the resulting sign of $f(x)-f(c)$ if $x$ is immediately to the left or to the right of $c$.

  • $\begingroup$ I meant critical points by extreme points. I will edit that anyways. But is there a graphical explanation that I can easily understand? $\endgroup$ – Sami Jan 17 '18 at 14:21

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