
#1
Oct1807, 06:04 PM

P: 440

I'm having a bit of trouble grasping the domain and range of functions of 2 variables. Does anyone know of any helpful tutorials that will help me get the hang of this concept? Any help appreciated.




#2
Oct1807, 06:10 PM

P: 1,705

tutorial starts here and now
[tex]f(x,y)=\frac{1}{\sqrt{x^2+y^2}}[/tex] what is the domain and range of this function? 



#3
Oct1807, 08:23 PM

P: 440

Its domain is x^2 + y^2 must be greater than or equal to zero, and its range is from minus to plus infinity?...




#4
Oct1807, 11:02 PM

P: 1,705

Domain and range, functions of 2 variablesthe range is actually discontinuous but again you're close edit the domain is actually discontinuous can the function ever be negative? did you just guess? 



#5
Oct1807, 11:10 PM

P: 1,076





#6
Oct1807, 11:23 PM

P: 1,705





#7
Oct1807, 11:42 PM

P: 1,705

you've probably confused the hell out of him




#8
Oct1907, 11:53 AM

P: 440

ice's post kind of inspired me to take the bull by the horns, so I opened up a calc book of mine, different from my class text, and I think I've got a handle on it. The domain would be D = {(x,y)x^2 + y^2 >0}, and the range would be R = { z  z = f(x,y), x^2 + y^2 >0}. Is that right?




#9
Oct1907, 03:29 PM

Sci Advisor
HW Helper
P: 3,680

So for which (x, y) in R^2 is it true that x^2 + y^2 > 0? The strict quadrants, right... [itex]x\neq0\neq y[/itex]. What is the range simplified the same way? 



#10
Oct1907, 03:43 PM

P: 440

Edit: Also, since the function takes an pair (x, y) from R^2, and turns it into R^1, is that what they call a mapping? 



#11
Oct1907, 11:11 PM

P: 1,705

the domain for this function is all values for x and all values of except the point (0,0), i think you know why. and the range is [itex](\infty,0)[/itex] and [itex] (0,\infty)[/itex] notice unbounded below and open notation, the parenthesis instead of brackets, and the open and unbounded above, again the parenthesis instead of brackets. do you know why the range doesn't include zero? try this now [tex] f(x,y)=lnxy[/tex] what is the domain and range of this function, describe it in words, set builder notation, and what it would look like on a graph 



#12
Oct2007, 01:08 AM

P: 440

A few questions, are you saying that the range of the function does not include zero, or that it does? And shouldn't it be for all z since z = f(x,y) ? Ok, for [tex] f(x,y)=lnxy[/tex] The domain of this function is the set of all x and y pairs such that x  y is greater than zero. The range of the function is the set z such that z = f(x,y), and z is greater than zero, and also x and y must fulfill the constraint for the domain. So D = {(x, y)  x > y}, and R = {z  z = f(x, y), x > y}. And graphically, I think it would exist in the first octant, and have a positive slope. 



#13
Oct2007, 11:55 AM

P: 1,705

look how i wrote the range of the previous function. it is pointless to write the range implicitly as you have. give me in interval notation. you're essentially saying that the range of the function f(x,y) is all values of f(x,y). and about the domain graphically you're wrong 


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