At what points in the interval [0, 2π], does the function sin 2x attain its maximum value?
The function f(x) = sin 2x attains its maximum value at x = π/4 and x = 5π/4 in the interval [0, 2π].
Explanation
This question tests the application of derivatives to find maxima. Following the methodology shown in the context for similar trigonometric functions, we find critical points by setting f'(x) = 0, then determine the nature of these points using the sign of the derivative in intervals. The maximum value of sin 2x is 1, which occurs when 2x = π/2 or 5π/2 within the given range.
Solution Steps
Step 1: Let f(x) = sin 2x. Then f'(x) = 2 cos 2x.
Step 2: For critical points, set f'(x) = 0, which gives cos 2x = 0.
Step 3: In the interval [0, 2π], we have 2x ∈ [0, 4π]. So 2x = π/2, 3π/2, 5π/2, 7π/2, giving x = π/4, 3π/4, 5π/4, 7π/4.
Step 4: Using the first derivative test: f'(x) > 0 in (0, π/4), f'(x) < 0 in (π/4, 3π/4), f'(x) > 0 in (3π/4, 5π/4), f'(x) < 0 in (5π/4, 7π/4), and f'(x) > 0 in (7π/4, 2π).
Step 5: Since f'(x) changes from positive to negative at x = π/4 and x = 5π/4, these are points of local maxima. The maximum value is f(π/4) = f(5π/4) = 1.