Vtyjkyuk

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 $F:YrightarrowmathbbR$ 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 $a$.

The proofs of derivative tests for optimality are based on Taylor's theorem. The Taylor series of bump functions do not provide useful information, so the tests are not helpful. You can study the function values of a sequence that converges to $x$ to rule out maximality or minimality. If $f(x_i)<f(x)$ (or vice versa) for all $i$ and $x_irightarrow x$ as $irightarrowinfty$, then $x$ cannot be a minimum (or maximum).