ISportsFan 0 Report post Posted October 7, 2003 Just seeing if anyone can help me with this problem (piecewise function): f(x,y) =... xy x^2 + xy + y^2 when (x,y) not equal to (0,0) 0 when (x,y) = (0,0) I need to know the largest set on which the function is continuous. Thanks for any help. Jason Edit: First piece hard to read because of formatting issues (I tried though). It's (xy)/(x^2 + xy + y^2) Share this post Link to post Share on other sites
Guest Agent of Oblivion Report post Posted October 7, 2003 RDRR Share this post Link to post Share on other sites
ISportsFan 0 Report post Posted October 7, 2003 RDRR Actually, that's a single variable calculus Simpsons joke. Jason Share this post Link to post Share on other sites
Guest Agent of Oblivion Report post Posted October 7, 2003 Hey, I tried. My calculator's busted. I was using a slide rule. Share this post Link to post Share on other sites
Slayer 0 Report post Posted October 7, 2003 Not the solution (I'm not doing the whole thing for you, heh), but a little push to help you on your way. You can see the upper function is defined for all possible values of x and y except at (0,0), but with the piecewise addition defining a value of 0 at (0,0), you need to show that lim f(x,y) = 0 (x,y)->0 If it is, it's continuous on the entire graph, if not, then it's continuous everywhere except (0,0) Angelslayer: Resident Above-Average Math Major Share this post Link to post Share on other sites
Murmuring Beast 0 Report post Posted October 7, 2003 The conclusive answer? Middle finger to the teacher; kick, wham, stunner.... OK, so that won't solve anything, but it would look great. Share this post Link to post Share on other sites
ISportsFan 0 Report post Posted October 7, 2003 Not the solution (I'm not doing the whole thing for you, heh), but a little push to help you on your way. You can see the upper function is defined for all possible values of x and y except at (0,0), but with the piecewise addition defining a value of 0 at (0,0), you need to show that lim f(x,y) = 0 (x,y)->0 If it is, it's continuous on the entire graph, if not, then it's continuous everywhere except (0,0) Angelslayer: Resident Above-Average Math Major Thanks a lot. I was thinking really too hard on it, because of the bottom xy (as if x was a big negative and y was a positive, then could the bottom equal zero at some other point than (0,0)). But, then as you just pointed out the obviousness of it, I realized the dominant functions in the denominator are x^2 and y^2, which means that the function cannot equal zero in the denominator (and thus be undefined) anywhere except (0,0). Thanks for the kick in the pants (in a good way, obviously). Jason Share this post Link to post Share on other sites
