Math help

[ISportsFan]

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)

[Guest Agent of Oblivion]

RDRR

[ISportsFan]

RDRR

Actually, that's a single variable calculus Simpsons joke.

 

Jason

[Guest Agent of Oblivion]

Hey, I tried. My calculator's busted. I was using a slide rule.

[Slayer]

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

[Murmuring Beast]

The conclusive answer? Middle finger to the teacher; kick, wham, stunner....

 

 

OK, so that won't solve anything, but it would look great.

[ISportsFan]

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