Let f(x) = x^2 - 3x + 2. Find the range of f(sin(x)).
Correct: A
The vertex of the parabola f(x) is at x = 3/2, and f(3/2) = (3/2)^2 - 3(3/2) + 2 = 9/4 - 9/2 + 2 = -1/4. Since the minimum value of sin(x) is -1 and the maximum value is 1, the range of sin(x) is [-1, 1]. We evaluate f(-1) = 1 + 3 + 2 = 6, and f(1) = 1 - 3 + 2 = 0. Since -1/4 is between f(1) and f(-1), the range of f(sin(x)) is [-1/4, 6].