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 2 years ago
Suppose that \(f\) satisfies the equation \[f(x+y)=f(x)+f(y)+x^2y+xy^2\]for all real numbers \(x\) and \(y\). Suppose further that\[~~~~~~~~~~~~~\lim_{x\to0}\frac{f(x)}x=1\]a) Find \(f(0)\)
b) Find \(f'(0)\)
c) Find \(f'(x)\)
 2 years ago
Suppose that \(f\) satisfies the equation \[f(x+y)=f(x)+f(y)+x^2y+xy^2\]for all real numbers \(x\) and \(y\). Suppose further that\[~~~~~~~~~~~~~\lim_{x\to0}\frac{f(x)}x=1\]a) Find \(f(0)\) b) Find \(f'(0)\) c) Find \(f'(x)\)

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TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0I found \(f(0)=0\) but now I am not seeing how to proceed. Just a hint would be nice.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0I further think I proved that the function is odd, though I think that doesn't really help me :/

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.2use the definition of a derivative

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.2the fact that it is odd does seem to help with the last part

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0I was thinking of using the definition of the derivative somehow but couldn't figure out how to get it into the right form....

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.2\[f'(0)=\lim_{x\to0}\frac{f(x)f(0)}{x0}\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0limit f(x)f(0) something...

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0ah dang but... oh, okay I guess I was trying to get things algebraically manipulated there, but I see now that since x=f(x) at x=0 the limit \(is\) the definition got it :)

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0okay so that is f'(0)=1, on to the last part you say the oddness could help?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0can I use the derivative of an odd function is even?

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.2use the definition of a derivative again

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0that's what I'm doing, let me toy for a sec...

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0haha, dang okay then go ahead and give me a little more help on finding f'(x) they don't even ask for f(x), jeez

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0...if you would be so kind

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.2if you know f'(x) and that f(0)=0 then finding f(x) is really easy :)

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.2\[f'(x)=\lim_{y\to x}\frac{f(y)f(x)}{yx}\]

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.2\[f(y)f(x)=f(y)+f(x)\]

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.1If it helps, it's easier for me to see it when I write the derivative as \[f'(x)=\lim_{y\to0} \frac{f(x+y)f(x)}{y}\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0that would be \[\lim_{y\to0}{f(y)+x^2y+xy^2\over y}=1+\lim...\]ah... it's \(f'(x)=x^2+1\), right ? :)

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0sweet, thanks y'all I still want to see how your way was gonna go @Zarkon , but I'll work on that myself for a while

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.2gotta go...fun problem...I'll post it later if you like

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0always a pleasure :)
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