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)?

Question

amistre64 · Answer

the domain will be any combination of x and y values that keep the inside of the ln() greater than zero.

amistre64 · Answer

sqrt() needs positive number only...

amistre64 · Answer

separate 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 · Answer

regardless, 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?

anonymous · Answer

yes

amistre64 · Answer

Since 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 · Answer

since the range is controled by the ln(?) function, it will spit out everything from ln(1)and up....

amistre64 · Answer

ln(1) = 0 right?

anonymous · Answer

yes

amistre64 · Answer

then the range; the output of this thing; will be from zero to infinity [0,inf)

amistre64 · Answer

open 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?

anonymous · Answer

yeah i think it does

amistre64 · Answer

double check that and you should have your answers :)

anonymous · Answer

thank you so much!