Does a stationary point (derivative equals zero) have to be local max/min/inflection

Question

I was wondering whether a function can have a point in which its derivative is zero but the point is neither local min, local max or inflection point. I was considering the function $$f(x)=\begin{cases}x^2\sin\frac1x,\ x\neq 0;\\
0,\ x=0.\end{cases}$$
The derivative at $x=0$ is zero (can be proven using limit definition of derivative). But for every $h>0$ there are positive and negative numbers for $f$ in $(-h,h)$, and therefore this isn't a local maximum or minimum. Is the point $x=0$ an inflection point or not? How can this be proven?
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Thank you

Answer 1

Good question!

• The given function $f'(x)$ is zero but discontinuous^† at $0.$

  So, for the same reason you cited as why it isn't a turning point, the stationary point $0$ isn't an inflection point either. (Also, neither the sign test nor second-derivative test are applicable.)

• Another example of a non-inflection (curvature doesn't change sign), non-turning stationary point is the point $0$ of the function $g(x)=3.$

• On the other hand, if a function has a continuous first derivative at, and is neither constant left of nor right of, its stationary point, then the stationary point is surely either a turning or inflection point.

• Incidentally, the function $h(x)=|x|$ has a non-stationary turning point at $0.$

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^{† $$f'(x)=\begin{cases}2x\sin\frac1x-\cos\frac1x,\ x\neq 0;\\ 0,\ x=0.\end{cases}$$ Let $\left\langle x_n:n\in\mathbb N\right\rangle=\left\langle \frac1{2n\pi}:n\in\mathbb N\right\rangle$ so that $x_n\rightarrow0$ and $n\neq0.$ But $f'(x_n)=-1\nrightarrow0=f'(0).$}