Charlie Conner

2022-09-01

Let

$f(x)={x}^{2}\mathrm{sin}\left(\frac{1}{x}\right)$

for $x\ne 0$ and $f(0)=0$.

Show that $0$ is a critical point of $f$ that is not a local maximum nor a local minimum nor an inflection point.

$f(x)={x}^{2}\mathrm{sin}\left(\frac{1}{x}\right)$

for $x\ne 0$ and $f(0)=0$.

Show that $0$ is a critical point of $f$ that is not a local maximum nor a local minimum nor an inflection point.

Toby Barron

Beginner2022-09-02Added 7 answers

Prove it is a critical point (that the derivative exists and it is $0$).

Prove it is neither a maximum or a minimum by proving that for any $\epsilon >0$ there are $x$ and $y$, $0<x<\epsilon ,0<y<\epsilon $ such as $f(x)>f(0)$ and $f(y)<f(0)$ (and $f(-x)<f(0)$ and $f(-y)>f(0)$) so there is no neighborhood in which $0$ is maximum or minimum.

This also implies why it is not an inflection point.

Prove it is neither a maximum or a minimum by proving that for any $\epsilon >0$ there are $x$ and $y$, $0<x<\epsilon ,0<y<\epsilon $ such as $f(x)>f(0)$ and $f(y)<f(0)$ (and $f(-x)<f(0)$ and $f(-y)>f(0)$) so there is no neighborhood in which $0$ is maximum or minimum.

This also implies why it is not an inflection point.

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