I think the domain of the first function is all R amd its range (0,+∞), because if we simplify we get the square root of the 1 divided by the absolut value of x, which is always positive, so its square root has a solution which is always positive too. The second one has the domain (-∞,o) and the range (o,+∞), becuase the x is multiplied by -1, so if we take a positive number as x it will result in the square root of a negative number which is not included in R. And of course, if we take a negative value for the x, yhe y will always result positive.
For the first function, Domf(x)can not be the whole R set. Can you find f(0)? For the second one, f(x) is possible only for x>0, so the function domain can not include negative numbers.
I think the answer is: a) |x|^-1/2 = 1/sqrt of |x| So, it's 1 divided by any positive, and that means that the range goes from 0 to infinite (0,+∞) and the domain is R - {0}
b) (-x)-1/2 = 1/sqrt of -x That means that x must be negative because you can only get a real result if you do a sqrt of a positive number: domain (-∞,0) and the range is R - {0} I'm not completely sure about the 2nd one.
You are right in domains and the first range, but for the second function, it's not possible to get any posive value (f(x)>0 in all cases); so the range of f(x)is (0,+∞).
I think the domain of the first function is all R amd its range (0,+∞), because if we simplify we get the square root of the 1 divided by the absolut value of x, which is always positive, so its square root has a solution which is always positive too.
ReplyDeleteThe second one has the domain (-∞,o) and the range (o,+∞), becuase the x is multiplied by -1, so if we take a positive number as x it will result in the square root of a negative number which is not included in R. And of course, if we take a negative value for the x, yhe y will always result positive.
For the first function, Domf(x)can not be the whole R set.
ReplyDeleteCan you find f(0)?
For the second one, f(x) is possible only for x>0, so the function domain can not include negative numbers.
I think the answer is:
ReplyDeletea) |x|^-1/2 = 1/sqrt of |x|
So, it's 1 divided by any positive, and that means that the range goes from 0 to infinite (0,+∞) and the domain is R - {0}
b) (-x)-1/2 = 1/sqrt of -x
That means that x must be negative because you can only get a real result if you do a sqrt of a positive number: domain (-∞,0) and the range is R - {0}
I'm not completely sure about the 2nd one.
You are right in domains and the first range, but for the second function, it's not possible to get any posive value (f(x)>0 in all cases); so the range of f(x)is (0,+∞).
ReplyDelete