f(x) = |x + 2| - 3
0 = |x + 2| - 3
0 = x + 2 - 3
0 = x - 1
+ 1 + 1
1 = x
Domain: x > 1 and x < 1
Solution Set of the Domain: {x| 1 > x > 1} and {x|x 1 < x < 1} or (1, 1)
f(x) = |x + 2| - 3
f(x) = |0 + 2| - 3
f(x) = |2| - 3
f(x) = 2 - 3
f(x) = -1
Range: x ≤ 3
Solution Set of the Range: {x|x ≤ 3} or (-∞, 3]
