Question

Describe the maximal domain of definition for each of the following functions.  

(a) f (x, y) = y/x   

(b) f (x, y) = x + y /x -  y

(c) f (x, y) = x+y/ x² + y² - 1  

(d) f (x,y ) =  √y -x² 

(e)  f (x, y) =√x in (y-sin x)  

(f) f (x,y ) = √y in(- x)

Answer 1

We will denote by D( f ) the maximal domain of definition.

(a) The function can be defined everywhere, where x ≠ 0. Therefore

D( f ) = f{(x, y) : x≠0}=R²\ {x=y}; 

the domain consists of the whole plane except for the y-axis. 

b) The function can be defined everywhere, where x - y ≠ 0. Therefore

D( f ) = f{(x, y) : x ≠ y}=R²\ {x=y};

the domain consists of the whole plane except for those points lying on the line x = y.

(c) Here the denominator is 0, when x² + y² = 1. Therefore

D(f) = {(x,y) :x²+y² ≠ 1}= R²\{x2+y²=1};

the domain conists of the whole plane except for the circle around the origin with radius 1.

(d) For the square root to be well-defined we need the inequality y-x²≥0 to be satisfied therefore  

D(f)={(x,y):y ≥x²};

the domain consists of the part of the plane, that lies on and above the parabola y = x².

(e) Here we need x ≥ 0 for the square root to be well-defined and y> sin x to evaluate the logarithm. Note that one inequality is strict, while the other is not. Thus the domain is

D(f ) = {(x, y) : x ≥ 0 and y > sin x} ; 

(f) To be able to evaluate f (x, y) we need −x > 0 as well as y − ln(−x)≥  0 to be satisfied. These inequalities can be rewritten as

D(f ) = {(x, y) : x < 0 and y≥ ln(−x)} .

Note again, that one inequality is strict and the other is not.