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anonymous
 5 years ago
How do I find domain and range of f(x,y)=ln(sqrt(1+x^2+y^2)? and what is the boundary of the domain and what type of set is it (open, closed, neither)?
anonymous
 5 years ago
How do I find domain and range of f(x,y)=ln(sqrt(1+x^2+y^2)? and what is the boundary of the domain and what type of set is it (open, closed, neither)?

This Question is Closed

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the domain will be any combination of x and y values that keep the inside of the ln() greater than zero.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sqrt() needs positive number only...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0separate it into 2 functions and define the domains of each function: f(u) = ln(u) and g(x) = sqrt(1+x^2+y^2) are we to assume that y is an implicit funtion of x? or an independant variable?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0regardless, sqrt() has to have a domain that is greater than zero for the function to operate inside of normal parameters. Any of this makeing sense?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Since x and y are both squared, anything we use for them will either be a zero or a positive number; which makes sqrt(?) always equal to or greater than 1 which is fine. any values can be used so the domian is (inf,inf)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0since the range is controled by the ln(?) function, it will spit out everything from ln(1)and up....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0then the range; the output of this thing; will be from zero to infinity [0,inf)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0open or closed just depends on their definitions.... which i am not too certain about :) I mean, open means without bounds, but does a onesided infinity count?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0double check that and you should have your answers :)
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