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# Hi! An explanation of this problem would be highly appreciated!

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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!

Nov 23, 2020

### 2+0 Answers

#1
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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!

Nov 24, 2020
#2
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Thanks for sharing your solution guest

Melody  Nov 24, 2020