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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)?

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  1. amistre64
    • 5 years ago
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    the domain will be any combination of x and y values that keep the inside of the ln() greater than zero.

  2. amistre64
    • 5 years ago
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    sqrt() needs positive number only...

  3. amistre64
    • 5 years ago
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    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?

  4. amistre64
    • 5 years ago
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    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?

  5. anonymous
    • 5 years ago
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    yes

  6. amistre64
    • 5 years ago
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    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)

  7. amistre64
    • 5 years ago
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    since the range is controled by the ln(?) function, it will spit out everything from ln(1)and up....

  8. amistre64
    • 5 years ago
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    ln(1) = 0 right?

  9. anonymous
    • 5 years ago
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    yes

  10. amistre64
    • 5 years ago
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    then the range; the output of this thing; will be from zero to infinity [0,inf)

  11. amistre64
    • 5 years ago
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    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?

  12. anonymous
    • 5 years ago
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    yeah i think it does

  13. amistre64
    • 5 years ago
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    double check that and you should have your answers :)

  14. anonymous
    • 5 years ago
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    thank you so much!

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