Find the domain and the range of f(x) = 3 sin⁻¹(2x) + π/2.

1. 1 · Write what is given and what is to be found. Given: f(x) = 3 sin⁻¹(2x) + π/2. To find: its domain and range.
2. 2 · Write the rule. sin⁻¹ u needs −1 ≤ u ≤ 1, and its value lies in [−π/2, π/2].
3. 3 · Find the domain. −1 ≤ 2x ≤ 1, so −1/2 ≤ x ≤ 1/2. Domain = [−1/2, 1/2].
4. 4 · Multiply the range by 3. −π/2 ≤ sin⁻¹(2x) ≤ π/2, so −3π/2 ≤ 3 sin⁻¹(2x) ≤ 3π/2.
5. 5 · Add π/2. −3π/2 + π/2 ≤ f(x) ≤ 3π/2 + π/2, so −π ≤ f(x) ≤ 2π.
6. 6 · Check and box the answer. At x = −1/2: 3(−π/2) + π/2 = −π. At x = 1/2: 3(π/2) + π/2 = 2π. Domain [−1/2, 1/2], range [−π, 2π].

Scoring tip: Start from the principal range of sin⁻¹ and change it one operation at a time: multiply first, then add.