A relation which intercepts with....

[Saracen Rue]

I am wondering if it's possible for a relation to intercept with both it's derivative and indefinite integral at the same location (not including e^x as it shares the same rule as both it's integral and derivative). This is also assuming the constant of the indefinite integral is equal to zero.

Basically;
Define f(x) where the solution to f(x) = f'(x) is the same as the solution to f(x) = ∫f(x)dx, f(x) ≠ e^x and c = 0

 

[fresh_42]

$f(x) = f '(x)$ is basically one way (among many) to define $e^x$ (up to a constant or the requirement$f(0)=1$).
What exactly are you looking for?

 

[Saracen Rue]

$f(x) = f '(x)$ is basically one way (among many) to define $e^x$ (up to a constant or the requirement$f(0)=1$).
What exactly are you looking for?

I'll give an example;
Let f(x) = sin^4(x). In this scenario, f(x), f'(x) and ∫f(x)dx all intersect the x-axis at (0, 0). Therefore, there is a three-way intersection between f(x), f'(x) and ∫f(x)dx at said point. I'm looking for other functions or relations which also have a three-way intersection between f(x), f'(x) and ∫f(x)dx. Preferably not at the origin, but I don't mind if they are.

 

[fresh_42]

How about $f(x)=(x-a)^n$ with $n > 1$?

 

[Saracen Rue]

How about $f(x)=(x-a)^n$ with $n > 1$?

Yes, that is an excellent example of what I meant. Thank you :)