Answer:
There is an absolute maximum at x = π/2, absolute minimums at x = 7π/6 and 11π/6, and a relative maximum at x = 3π/2.
Step-by-step explanation:
f(x) = 2 sin x − cos(2x)
Relative extrema occur where f'(x) = 0.
f'(x) = 2 cos x − (-sin(2x) · 2)
f'(x) = 2 cos x + 2 sin(2x)
0 = 2 cos x + 2 sin(2x)
0 = cos x + sin(2x)
0 = cos x + 2 sin x cos x
0 = cos x (1 + 2 sin x)
cos x = 0 or sin x = -½
x = π/2, 7π/6, 3π/2, 11π/6
Evaluate the function at the endpoints and at each relative extrema.
f(0) = -1
f(π/2) = 3
f(7π/6) = -3/2
f(3π/2) = -1
f(11π/6) = -3/2
f(2π) = -1
There is an absolute maximum at x = π/2, absolute minimums at x = 7π/6 and 11π/6, and a relative maximum at x = 3π/2.
