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- anonymous

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

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- anonymous

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

- chestercat

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- anonymous

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

- anonymous

range is y cannot be 1

- anonymous

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

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- anonymous

range is y cannot be 0

- anonymous

Can you show me how did you get that?

- anonymous

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

- TheViper

http://cs.gmu.edu/cne/modules/dau/calculus/domains_ranges/dom2_frm.html

- 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

- anonymous

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

- anonymous

Do I need to think about the negatives as well?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

-2, -3, etc...

- anonymous

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

- anonymous

Oh nvm

- 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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