second and first derivatives are $0$

Question

what happens if both the second and first derivatives at a certain point are $0$? Is it an infliction point, an extremum point, or neither? Can we say anything at all about a point in such a case?

Answer 1

You can't say anything. Example : $$x\longmapsto x^3\quad x\longmapsto x^4\quad \text{and}\quad x\longmapsto\begin{cases} x^3\sin\left(\frac{1}{x}\right)&x\neq 0\\ 0&x=0\end{cases}.$$

Answer 2

• if $\frac{df}{dx}(p) = 0$, then $x = p$ is called a critical point of $f(x)$, and we do not know anything new about the behavior of $f( x) $ at $x = p$.

• if $\frac{d^2f}{dx^2} (p) = 0$ at $x = p$, then we do not know anything new about the behavior of $f(x)$ at $x = p$.

As @arugula mentioned $f(x)=\begin{cases} x^3\sin(1/x)&x\neq 0\\ 0&x=0\end{cases}$

Answer 3

You can tell what it does (assuming this function is well-behaved at the point in question; without that this question becomes basically pointless) based on the order of the lowest non-zero derivative beyond the first.

An inflection point has a non-zero odd derivative: $f(x)=x^3$ has an inflection point at $0$ because its third derivative is non-zero.

An extremum has a non-zero even derivative: $f(x) = x^4$ has an extremum at $0$ because its fourth derivative is non-zero.