Partial Derivatives and continuity

Jun 2006
12
0
NSW Australia
I do not understand the second part of the following question. Can someone please explain it to me. Thanks

Find all points in the plane where f is not defined. Is it possible to extend the domain of f to the entire plane (by defining values of f(x,y) at points where it it currently undefined) in such a way that the 'new' function is continuous on the entire plane?

See the attached gif (click on function and Answer.gif) for the Function and 1st part answer of the question. The 2nd part answer of the question is these points can be added to the domain of f(x,y) by defining f(x,y)=2y.

Function: f(x,y) = (x^2 - y^2)/(x - y)
Answer: 1st part {x, y E R^2 | x=y}
 

Attachments

Last edited:

matheagle

MHF Hall of Honor
Feb 2009
2,763
1,146
It's just like \(\displaystyle f(x)={x^2-4\over x-2}\)

which is ALMOST the same as the line y=x+2, but has a hole in it at x=2, since you cannot have a zero in your denominator.

In order to make f(x) continuous you plug in x=2 into x+2, giving you 4, filling in that hole.

\(\displaystyle f(x,y)={x^2-y^2\over x-y}\) is not defined along the line y=x.

Everywhere else it is equal to \(\displaystyle g(x,y)=x+y\)

IF you let y=x, then x+y becomes 2x, or 2y along that line.
 
Last edited: