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for (a) x cannot be 0 if not its undefined
range is y cannot be 1
(b) is x cannot be -1 if not denominator is 0 n 1/0 is undefined
range is y cannot be 0
Can you show me how did you get that?
for a) I got (1/x) + 1 >= 0 ? Did I set it up correctly?
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
the denominator cannot be zero and function is only 1 to 1
Do I need to think about the negatives as well?
yes.. for example 1/(x+1).. ur x cannot be negative one cause -1 +1 = 0
For a) if the denominator was negative, it would make the equation inside the sqrt negative
umm brother... its 1 OVER x...
-2, -3, etc...
even if x is -100000000000 its 1 OVER it.. thats like 0.00000000000 bla bla bla
it doesnt matter a fraction is always smaller than a whole number when the numerator is smaller than the denominator