I am having trouble understanding a question.

A function given as :

Let $f (x+y)=f(x)f(y)$ for all $x$ and $y.$ If $f (5)=2$ and $f'(0)=3$ then $f'(5)$ is equal to what?

So the trouble which I am having is.. Here it is given $f(5) =2$ which would mean $x+y$ is $5.$ Which can be $2+3, 1+4$ & $0+5.$ But none of these on being multiplied as $f(x) f(y)$ give $2.$

Or if I think this way: $f(x)f(y)$ is $2.$ And I get the quadratic equation $x (5-x)^2=2.$ I get an irrational number.

I am sorry if the basic way I thought about it is absolutely wrong. I think I am thinking about it in a wrong way. So I need help understanding the given function first then.

  • $\begingroup$ Do you know any basic function that satisfy the property $f(x+y)=f(x)f(y)$? $\endgroup$ – Marcel Dec 14 '15 at 15:37
  • $\begingroup$ No. That is what I have to figure out in the question. That is why I am having problem figuring it out. There are four options given with this question and the answer too. $\endgroup$ – Freya Dec 14 '15 at 15:38
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    $\begingroup$ The fact that you wrote "a function given as" makes me suspect that you may not be understanding the question correctly. The question doesn't actually tell you what the function $f$ is. It only gives you partial information about it. $\endgroup$ – David Dec 14 '15 at 15:39
  • $\begingroup$ @Freya Notice that $e^{x+y}=e^xe^y$. $\endgroup$ – Marcel Dec 14 '15 at 15:41
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    $\begingroup$ Possibly duplicate math.stackexchange.com/questions/1495336/… $\endgroup$ – Nizar Dec 14 '15 at 15:45

Hint. Differentiate the function $g(x) = f(5 + x)$ at $x = 0$ in two different ways.

Edit. As Abstraction points out in a comment, no function $f$ satisfying the hypotheses can actually exist.

  • $\begingroup$ Or, in general: ${d \over dy}f(x+y) = f(x)f'(y) = f'(x+y)$ and for $y=0$ we have $3f(x)=f'(x) \implies f(x)=e^{3x}$ and $f(5)=e^{15}=2$... something's not right here, is it? $\endgroup$ – Abstraction Dec 14 '15 at 15:59
  • $\begingroup$ Hi David. Thank you for your hint. I did it in one way , another way (?) I haven't done yet. But then in a comment a question as this has been referred to , so I got the help I needed. $\endgroup$ – Freya Dec 14 '15 at 16:13
  • $\begingroup$ @Abstraction You are correct. You should add that as an answer. $\endgroup$ – David Dec 14 '15 at 16:50

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