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anonymous
 5 years ago
I have two functions: f(x,y)=((x^2)*y) / ((x^4)+y^2) and g(x,y)=((x^2)*y) / ((x^2)+y^2).
Can I define any of these functions to be continuous? How would that be done?
anonymous
 5 years ago
I have two functions: f(x,y)=((x^2)*y) / ((x^4)+y^2) and g(x,y)=((x^2)*y) / ((x^2)+y^2). Can I define any of these functions to be continuous? How would that be done?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Continuous at the origin. :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, you just have to look for where you would have "problems". You don't want the denominator in a fraction to be zero, you don't want insides of radicals to be negative, and you don't want zeroes or negatives inside of logarithms. There are no radicals or logarithms in either of your functions, so we'll look at the denominator being zero. In f(x, y), how could your denominator be zero? Answer: If x^4 + y^2 = 0. This is possible only when (x, y) = (0, 0). So, the function is continuous everywhere except at (x, y) = (0, 0). In g(x, y), how could your denominator be zero? Answer: If x^2 + y^2 = 0. This is possible only when (x, y) = (0, 0). So, the function is continuous everywhere except at (x, y) = (0, 0).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks, I knew that part allready though, but how could I define the functions in question so that they are continuous at the point (0,0)/origin?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Say: \[f(x, y) = x^2y/(x^4+y^2), x^4 + y^2\neq0\] \[g(x, y) = x^2y/(x^2+y^2), x^2 + y^2 \neq0\] This kind of discontinuity is nonremovable, so you really can't.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What would the motivation for that be? I dont quite follow the reasoning.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Why would the discontinuity be nonremovable? The discontinuity is infinite to be specific. If you choose (x, y) points that get closer and closer to (0, 0), the value of z gets bigger and bigger. You can't really "fix" an infinite discontinuity. Think of like f(x) = 1/x^2. You can't fix the discontinuity at x = 0. Also, because there's nothing to cancel out in the numerator, you can't remove the x^4 + y^2 from the bottom in f(x, y). Same rationale applies for g(x, y).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[defn the fctn \to be =0 at (0,0)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ha, whoops. I guess the discontinuity is not infinite. Jumped the gun there. The better way to put it would be that the limit does not exist at that point for f(x, y). If you tried to approach from y = 0, you would get the limit as being 0. If you tried to approach from y = x, you would get the limit as being infinite. Since these paths don't agree, you can conclude that the limit doesn't exist at (0, 0) for f(x, y). I'm not sure about g(x, y), though...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I guess the paths agree (or at least the ones I've tried) for g(x, y). I suppose you could set up a piecewise function where g(x, y) = 0 when (x, y) = (0, 0) and g(x, y) = (x^2y)/(x^2 + y^2) when (x, y) =\= (0, 0).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hey, here in f(x,y) & g(x,y) the functions are in 0/0 form. so we can't say it is infinite. we have to find the limit, though i am unable to find it.
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