g (x) = (√5-x) + I 3 - x I / (x² - 1)
*the 5-x is all under the sq root
*3-x is an absolute value
*the 5-x is all under the sq root
*3-x is an absolute value
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domain is just what is permitted as an input.
you can't divide by zero and (for purposes here) you can't take the square root of a negative number.
lok at the denominator. it factors to (x+1)(x-1)
so x cannot be -1 or 1
in the numerator, I assume the first term is sqrt(5-x). If so, then 5-x can't be negative, so x can't be greater than 5.
In formal notation that translates to: (-inf,-1) U (-1,1) U (1,5]
you can't divide by zero and (for purposes here) you can't take the square root of a negative number.
lok at the denominator. it factors to (x+1)(x-1)
so x cannot be -1 or 1
in the numerator, I assume the first term is sqrt(5-x). If so, then 5-x can't be negative, so x can't be greater than 5.
In formal notation that translates to: (-inf,-1) U (-1,1) U (1,5]
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Denominator cannot be -1 or 1, or else it will be undefined.
√(5-x) has to be positive or else it will be undefined.
So exclude -1, 1, and any value higher than 5 (5 works)
So, (-∞,-1)u(-1,1)u(1,5]
√(5-x) has to be positive or else it will be undefined.
So exclude -1, 1, and any value higher than 5 (5 works)
So, (-∞,-1)u(-1,1)u(1,5]