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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 \(f'(x)\)

Mathematics
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I found \(f(0)=0\) but now I am not seeing how to proceed. Just a hint would be nice.
I further think I proved that the function is odd, though I think that doesn't really help me :/
use the definition of a derivative

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Other answers:

the fact that it is odd does seem to help with the last part
I was thinking of using the definition of the derivative somehow but couldn't figure out how to get it into the right form....
\[f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}\]
limit f(x)-f(0) something...
ah 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 :)
okay so that is f'(0)=1, on to the last part you say the oddness could help?
yes
can I use the derivative of an odd function is even?
don't need it
use the definition of a derivative again
that's what I'm doing, let me toy for a sec...
I found f(x) ;)
haha, 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
...if you would be so kind
if you know f'(x) and that f(0)=0 then finding f(x) is really easy :)
\[f'(x)=\lim_{y\to x}\frac{f(y)-f(x)}{y-x}\]
\[f(y)-f(x)=f(y)+f(-x)\]
does that help?
ok
If it helps, it's easier for me to see it when I write the derivative as \[f'(x)=\lim_{y\to0} \frac{f(x+y)-f(x)}{y}\]
that would be \[\lim_{y\to0}{f(y)+x^2y+xy^2\over y}=1+\lim...\]ah... it's \(f'(x)=x^2+1\), right ? :)
yes
sweet, 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
gotta go...fun problem...I'll post it later if you like
always a pleasure :)

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