"If $x^*$ is a local maximum, then $f''(x^*)$ is less than zero." Why is this false?

Question

Since $\,f''(x^*)<0$, implies local max. Can you give me examples where $x^*$ is a local max but $\,f''(x^*)=0$.

Answer 1

Note that $x^{*}\mbox{ a local maximum}\implies f''(x^{*})<0$ being false is not the same as the existence of a local maximum $x^{*}$ for which $f''(x^{*}) = 0$; you only need a local maximum $x^{*}$ such that $f''(x^{*})<0$ is false.

Consider $f(x) = -|x|$. Then for any $a>0$, on the domain $(-a,a)$, the local maximum is $f(x^{*} = 0) = 0$, but $f$ is not twice differentiable at $x = 0$, that is, $f''(x^{*}=0)$ doesn't even exist, and $f''(x^{*} = 0) < 0$ is false automatically.

If you really want an example of a local maximum where the second derivative is zero, let $a>0$ and consider $f(x) = x^{3}$ on $[-a,0]$. The local maximum is again at $x^{*} = 0$, but $f''(x) = 6x$, and $f''(x^{*} = 0) = 0$.

Answer 2

A function $f(x)$ has a local maximum at the point $x^∗$ if there exists some $\epsilon>0$ such that $f(x^∗) \geq f(x)$ for all $x$ within a distance $\epsilon$ of $x^*$.

Consider the constant function $f(x)=c$. It is easy to see that every point in $\mathbb{R}$ is a local maximum of $f(x)$. We also have $f^{\prime\prime}(x)=0$ for all $x\in\mathbb{R}$.