IDENTITYING DOMAIN & RANGE -
FUNCTION DOMAIN (x) RANGE (y)
y = x² (- ∞, ∞) [0, ∞)
y = 1/x (- ∞, 0) ∪ (0, ∞) (- ∞, 0) ∪ (0, ∞)
y = √x [0, ∞) [0, ∞)
y = √(4 – x) (- ∞, 4] [0, ∞)
y = √(1 - x²) [- 1, 1] [0, 1]
Solution – The formula y = x² gives a real y-value for any real number x, so the domain is (- ∞, ∞). The range of y = x² is [0, ∞) because the square of any real number is non-negative and every non-negative number y is the square of its own square root, y = (√y)² for y ≥ 0
1
The formulae y = ------ gives a real y-value for every x except x = 0.
x
We cannot divide any number zero.
The range of y = 1/x, the set of reciprocals of all non-zero real numbers, is the set of all non-zero real
1
number, since y = -------
(1/y)
The formulae y = √x gives a real y-value only if x ≥ 0. The range of y = √x is [0, ∞) because every non-negative number is some number’s square root (namely, it is the square root of its own square).
In y = √(4 – x). the quantity (4 – x) cannot be negative. That is (4 – x) ≥ 0, or x ≤ 4. The formula gives real y-values for all x ≤ 4. The range of √(4 – x) is [0, ∞), the set of all non-negative numbers.
The formula y = √(1 - x²) gives a real y-value for every x in the closed interval form – 1 to 1. Outside this domain, (1 - x²) is negative and its square root is not a real number. The values of (1 - x²) vary from 0 to 1 on the given domain, and the square roots of these values do the same. The range of √(1 - x²) is [0, 1]