+1 vote
in Class 11 by kratos

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)

1 Answer

+5 votes
by kratos
 
Best answer

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 **** 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.

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