The function \(f : \mathbb{R} \to \mathbb{R}\) satisfies \(f(x) f(y) = f(x + y) + xy\)

for all real numbers \(x\) and \(y\) Find all possible functions \(f\).

I'm not really sure how to solve the problem; an explanation would be highly appreciated! Thank you in advance!

Guest Nov 23, 2020

#1**+1 **

Never mind, I got it :).

For anyone who stumbles across this in the future and is curious how to approach it, here was my solution:

Setting \(x = y = 0\), we get \(f(0)f(0) = f(0) + 0\), so \(f(0)^2= f(0)\). Thus, this is only true when \(f(0) = 1\).

To substitute this value into our equation, we can set \(y = -x\). It becomes

\(\begin{align*} f(x) f(-x) &= f(0) -x^2 \\ &= 1 -x^2 \\ &= (1+x) (1-x). \\ \end{align*}\)

Therefore, \(f(x) = \boxed{1+x} \) or \(\boxed{1-x}\).

If any part is wrong or could've been done more efficiently, please let me know. Hope this helps someone!

Guest Nov 24, 2020