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The second-derivative test states that if $x$ is a real number such that $f'(x)=0$, then:

  1. If $f''(x)>0$, then $f$ has a local minimum at $x$.
  2. If $f''(x)<0$, then $f$ has a local maximum at $x$.
  3. If $f''(x)=0$, then the text is inconclusive.

But there's no need to despair if the second-derivative test is inconclusive, because there is the higher-order derivative test. It states that if $x$ is a real number such that $f'(x)=0$, and $n$ is the smallest natural number such that $f^{(n)}(x)\neq 0$, then:

  1. If $n$ is even and $f^{(n)}>0$, then $f$ has a local minimum at $x$.
  2. If $n$ is even and $f^{(n)}<0$, then $f$ has a local manimum at $x$.
  3. If $n$ is odd, then $f$ has an inflection point at $x$.

But the higher-order derivative test can also be inconclusive, if $f^{(n)}(x)=0$ for all $n$. My question, what can you do if the higher-order derivative test is inconclusive?

Is the first-derivative test the only option at that point, or are there other options?

EDIT: I’m interested in finding a method that depends only on the germ of $f$.

EDIT 2: Let me explain more precisely what I’m saying regarding the germ. Let $X$ be the set of all functions $f$ infinitely differentiable at $a$ where $f^{(n)}(a)=0$ for all $n$. Two functions $f$ and $g$ in $X$ belong to the same germ if there is an open interval $I$ containing $a$ such that $f(x) = g(x)$ for all $x$ in $I$. Now let $Y$ be the set of germs of $X$. I want to know if there exists a nontrivial function $F:Y\rightarrow\mathbb{R}$ such that if $F$ evaluated at a particular germ yields a positive number, then all the functions in the germ have a local minimum at $a$, and if it yields a negative number then all the functions in the germ have a local maximum at $a$.

Keshav Srinivasan
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  • The case you are talking about is for the polynomials of degree $m<n$. However, if you see that $f^n$$(x)$ $=$ $0$ $\forall n\in N$, then then the function must be a constant one, right ?? @Keshav Srinivasan – Anik Bhowmick Aug 04 '18 at 04:53
  • @AnikBhowmick I'm not talking about polynomials, I'm talking about smooth functions in general. If you have an infinitely differentiable function $f$ whose derivatives of all orders evaluated at a given point $x$ is zero, then the higher-order derivative test fails. But that doesn't mean that $f$ is a constant function. For instance consider the example function given here: https://en.wikipedia.org/wiki/Non-analytic_smooth_function – Keshav Srinivasan Aug 04 '18 at 05:01
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    I assume you meant to ask what can you do if the higher-order derivative test is inconclusive? – N. F. Taussig Aug 04 '18 at 08:14
  • @N.F.Taussig Yes – Keshav Srinivasan Aug 04 '18 at 12:34

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