A function which infinitely changes concavity but is everywhere non-decreasing

Asked Mar 03 '21 at 06:50

Active Mar 03 '21 at 13:51

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For the infinitely changing concavity part, I have come up with this specific example $y = ^4\sin\frac{1}{x}$.

Derivative of $\sin\frac{1}{x}$ is $-\frac{\cos\frac{1}{x}}{x^2}$, and $x^2$ will always be positive, however $\cos\frac{1}{x}$ is only positive in the first quadrant.

And for the everywhere non-decreasing part, I have not idea where to start.

How should I define the domain to fit the requirement? Please leave some hints!

Much appreciated!

edited Mar 03 '21 at 13:51

DMcMor

9,407

asked Mar 03 '21 at 06:50

Hailey Chan

Precisely what do you mean by infinitely changes concavity? – Andrew Chin Mar 03 '21 at 06:53
I'm sure he means the second derivative changes its sign infinitely many times. – hugh_maths Mar 03 '21 at 09:04
You can find some examples in Equation for a smooth staircase function, such as $x+ \sin(x)$. – Vepir Mar 03 '21 at 09:08
Assuming you mean a function with "infinitely many flex points". – Vepir Mar 03 '21 at 09:17
I see. What about the everywhere non-decreasing requirement? I am not sure what does that means. – Hailey Chan Mar 03 '21 at 11:00
Everywhere non-decreasing means there are no values $a,b$ such that $a \lt b$ and $f(a) \gt f(b)$. It means it goes flat or upwards at all points. – Ross Millikan Mar 03 '21 at 14:51
You can show that $y=f(x)$ is increasing (non-decreasing) on $(a,b)$ by showing that the derivative is nonnegative, i.e. $f’\left( x \right) \ge 0;\forall;x \in \left( {a,b} \right).$ – Vepir Mar 03 '21 at 20:12

A function which infinitely changes concavity but is everywhere non-decreasing

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