A Continuous Functional Equation

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[@mystychief]

Looks linear. Suppose f(x)=ax+b, then from equation 1: af(x+y)+b=ax+b+ay+b=a(x+y)+2b. So f(x+y)=(a(x+y)+b)/a=(x+y)+b/a or f(t)=t+c or f(x)=x+c. Since a is obviously 1 then b=c.

[@abshariadam]

5:05 😂😂😂

[@marat61]

f(f(x + y)) = f(x) + f(y)
let y = 0
f(f(x)) = f(x) + f(0)
let t = f(x)
f(t) = t + C

[@scottleung9587]

Nice!

[@SyberMath]

Thanks!

[@carly09et]

I get f(x) = x but I can't say why it is unique (it is in that f(0) = 0)

[@phill3986]

☮️👍✌️😃😃✌️👍☮️

[@KanuniSultanSuleyman243]

2:49 we find the solution why did you continue?

[@wernischluep7352]

f(0) should be a. you know why...

[@joysanghavi13]

Put y =0 and then x=f^(-1)z. Solve for z. Pretty easy

[@Sir_Isaac_Newton_]

what

[@TheOiseau]

My god, what a convoluted mess. You had the answer at 2:49, why is the video 10 minutes long?