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Mathematics 18 Online
OpenStudy (anonymous):

How do you find the domain and range for a) sqrt ( (1/x)+1 ) b) 1/ sqrt (x+1) ?

OpenStudy (anonymous):

for (a) x cannot be 0 if not its undefined

OpenStudy (anonymous):

range is y cannot be 1

OpenStudy (anonymous):

(b) is x cannot be -1 if not denominator is 0 n 1/0 is undefined

OpenStudy (anonymous):

range is y cannot be 0

OpenStudy (anonymous):

Can you show me how did you get that?

OpenStudy (anonymous):

for a) I got (1/x) + 1 >= 0 ? Did I set it up correctly?

OpenStudy (anonymous):

Umm u dont even actually have to set up.. just apply this few rules (1) anything divided by 0 ( a fraction) is undefined so for (a) square root of (1/x + 1) u know that 1 over 0 is undefined when x = 0 so ur domain would be x cannot be equal to 0.. as simple as that

OpenStudy (anonymous):

the denominator cannot be zero and function is only 1 to 1

OpenStudy (anonymous):

Do I need to think about the negatives as well?

OpenStudy (anonymous):

yes.. for example 1/(x+1).. ur x cannot be negative one cause -1 +1 = 0

OpenStudy (anonymous):

For a) if the denominator was negative, it would make the equation inside the sqrt negative

OpenStudy (anonymous):

umm brother... its 1 OVER x...

OpenStudy (anonymous):

-2, -3, etc...

OpenStudy (anonymous):

even if x is -100000000000 its 1 OVER it.. thats like 0.00000000000 bla bla bla

OpenStudy (anonymous):

Oh nvm

OpenStudy (anonymous):

it doesnt matter a fraction is always smaller than a whole number when the numerator is smaller than the denominator

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