anonymous
  • anonymous
How do you find the domain and range for a) sqrt ( (1/x)+1 ) b) 1/ sqrt (x+1) ?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
for (a) x cannot be 0 if not its undefined
anonymous
  • anonymous
range is y cannot be 1
anonymous
  • anonymous
(b) is x cannot be -1 if not denominator is 0 n 1/0 is undefined

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anonymous
  • anonymous
range is y cannot be 0
anonymous
  • anonymous
Can you show me how did you get that?
anonymous
  • anonymous
for a) I got (1/x) + 1 >= 0 ? Did I set it up correctly?
TheViper
  • TheViper
http://cs.gmu.edu/cne/modules/dau/calculus/domains_ranges/dom2_frm.html
anonymous
  • 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
  • anonymous
the denominator cannot be zero and function is only 1 to 1
anonymous
  • anonymous
Do I need to think about the negatives as well?
anonymous
  • anonymous
yes.. for example 1/(x+1).. ur x cannot be negative one cause -1 +1 = 0
anonymous
  • anonymous
For a) if the denominator was negative, it would make the equation inside the sqrt negative
anonymous
  • anonymous
umm brother... its 1 OVER x...
anonymous
  • anonymous
-2, -3, etc...
anonymous
  • anonymous
even if x is -100000000000 its 1 OVER it.. thats like 0.00000000000 bla bla bla
anonymous
  • anonymous
Oh nvm
anonymous
  • 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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