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The function \(f : \mathbb{R} \to \mathbb{R}\) satisfies 

\(f(x) f(y) = f(x + y) + xy\)

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

 

I've found that \(f(0)=1\) and if \(f(-1)\) is not \(0,\) then \(f(1)=0.\) I'm just stuck and any answer/help would be appreciated! (btw i got the values for \(f(0)\) and \(f(1)\) by plugging in values for x and y like 0 and 0, or x=1, y=1.)

 

------------Thanks! laugh

 Jul 12, 2020
edited by madyl  Jul 12, 2020
 #1
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I checked if there were constant solutions, and there are none, so I don't think there are any other solutions.

 Jul 12, 2020
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What do you mean checked constant solutions, im pretty sure there is a solution so can you show me your work?

madyl  Jul 12, 2020
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How about  f(x) = x + 1

 

f(x).f(y) = x + y + x.y + 1

 

f(x+y) + x.y = x + y + x.y + 1

 Jul 13, 2020

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