3
$\begingroup$

Is there a function $f(x)$ which is defined near $x = c$ and infinitely differentiable near $x = c$ and satisfy the following properties:

For any positive real number $\delta$,

there exist real numbers $x, x^{'}$ such that $c - \delta < x, x^{'} < c$ and $f(x) > f(c)$ and $f(x^{'}) < f(c)$ .

For any positive real number $\delta$,

there exist real numbers $x, x^{'}$ such that $c < x, x^{'} < c + \delta$ and $f(x) > f(c)$ and $f(x^{'}) < f(c)$ .

$\endgroup$
2
$\begingroup$

Yes. Define the function $f $ for positive $x $ by $$ f (x) = e^{-1/x}\sin (1/x), $$ by $f (0) = 0$ and by $f (-x) = f (x) $. Take $c=0$.

$\endgroup$
0
1
$\begingroup$

Let $f(x) = 0$, for $ x \leq 0$ and $f(x) = e^{-\frac{1}{x}} \sin \frac{1}{x}$ for $x > 0.$

For any positive integer $n$ and $x > 0$ $f(x) = \dfrac{e^{-\frac{1}{x}}}{x^{n+1}}\times x^{n+1} \sin \frac{1}{x}$ which shows $f(x)$ is differentiable $n$ times at $0$ and $f^{n}(0)=0.$

$f(x)$ changes sign on $(0,\delta)$ for any $\delta > 0$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.