A Surprisingly Easy Functional Equation 🤯🤯🤯 

great video!

Mad math brain teaser! Love it!❤ Thank you!

cool problem... keep it up!

Never seen this type of problem before. thanks

Wow

The solution to the problem at the end : *f(x) = 1-x^2*

That’s Awesome! 🤗❤️✨

forgor to substitute the x in the denominator and got x instead.💀

Crazy, need some time to digest it, so I got the main point

x->√(1-x²) => f(√(1-x²))+2f(√(1-[√(1-x²)]²))=1+[√(1-x²)]² => f(√(1-x²))+2f(√(1-(1-x²)))=1+(1-x²) Con xϵ[0,1] porque 1-x² debe ser ≥0 y x debe ser ≥0 para que √(1-(1-x²))=√(x²) sea x y no -x : f(√(1-x²))+2f(x)=2-x² Como f(x)+2f(√(1-x²))=1+x², se tiene que: f(√(1-x²))=(1/2)[1+x²-f(x)], por lo tanto: (1/2)[1+x²-f(x)]+2f(x)=2-x² (3/2)f(x)+(1/2)[1+x²]=2-x² (3/2)f(x)=(1/2)[4-2x²-1-x²] ∴ f(x)=(1/3)[3-3x²]=1-x². Funciona para xϵ[-1,1] porque 1-x²=1-(-x)²=f(-x), extendiendo el dominio.

Hi Mathemadix, is it in anyway possible to get in touch with you over the email?