# A relation which intercepts with...

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
Mentor
2021 Award
##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?

##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
Mentor
2021 Award
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 :)