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f '(x) = - 6 sin 2x
f(x) + f'(x) = 10 + 3 cos 2x - 6 sin 2x
k = √(3² + (-6)²) = √(9+36) = √45 = 3√5
Jadi
f(x) + f '(x) = 10 + 3√5 cos (2x - β)
nilai minimum cosinus = - 1
Jadi nilai minimum
f(x) + f '(x) = 10 - 3√5