Well, /sci/? What's a function whose value at x is equal

f(x) = 0
no problem.

>>7958930
The exponential function. Easy

>>7958968
¯\_(ツ)_/¯

By pluging x=0 into the equation we get f(0) = 0. If we further assume that the function is C^1 then by fundamental theorem of calculus we can differentiate the integral equation to get f'(x) = f(x). The theorem of Picard-Lindelöf implies that the IVP
f'(x) = f(x)
f(0) = 0
has a unique solution which is f(x) = 0.

File: tricked.jpg (17KB, 200x200px) Image search: [Google]

17KB, 200x200px

>>7958968

>>7958968
I suppose you mean k*e^x.

>>7958930
The constant infinity function.

>>7959010
[math]f:\mathbb{R}\to \mathbb{R}[/math]

not

[math]f:\mathbb{R}\to \infty[/math]

Infinity is not in the real numbers.

>>7959003
Where k=0 maybe.

>>7959031
It's possible to make a perfectly consistent theory of integration for functions to the nonnegative extended reals, and in that setting his answer works.

>>7959073
Oh? And when x=0? Or when x < 0? You need to finangle your definition of definite integral a good bit before it works.

>>7959073
Learn measure theory m8

>>7958930
take the derivative of both sides

>>7959082
>tfw forgot 0*inf=0

Also, if you want f''(x)=f (x) instead of just f'(x)=f (x), sinh(x) and cosh(x) work as well.

f(x)=0 is the only answer.

>>7958930

f(x)= 0.5x^2

>>7959242
indef integral of 0.5x^2=x^3/6

which is only true at x=0,x=3