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TuringTest
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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 \(f'(x)\)
 one year ago
 one year ago
TuringTest Group Title
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 \(f'(x)\)
 one year ago
 one year ago

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

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

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
use the definition of a derivative
 one year ago

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

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

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
limit f(x)f(0) something...
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
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 :)
 one year ago

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
can I use the derivative of an odd function is even?
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
don't need it
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
use the definition of a derivative again
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
that's what I'm doing, let me toy for a sec...
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
I found f(x) ;)
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
...if you would be so kind
 one year ago

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

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

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
\[f(y)f(x)=f(y)+f(x)\]
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
does that help?
 one year ago

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
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 ? :)
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
always a pleasure :)
 one year ago
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