## Denebel 3 years ago How do you find the domain and range for a) sqrt ( (1/x)+1 ) b) 1/ sqrt (x+1) ?

1. SNSDYoona

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

2. SNSDYoona

range is y cannot be 1

3. SNSDYoona

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

4. SNSDYoona

range is y cannot be 0

5. Denebel

Can you show me how did you get that?

6. Denebel

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

7. TheViper
8. SNSDYoona

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

9. panlac01

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

10. Denebel

Do I need to think about the negatives as well?

11. SNSDYoona

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

12. Denebel

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

13. SNSDYoona

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

14. Denebel

-2, -3, etc...

15. SNSDYoona

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

16. Denebel

Oh nvm

17. SNSDYoona

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