At x = (5pi)/6, f(x) = 2 sin3x + 3 cos3x is ______.

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At x = (5pi)/6, f(x) = 2 sin3x + 3 cos3x is maximum. Explanation: We have f(x) = 2 sin 3x + 3 cos 3x f'(x) = 2 cos 3x · 3 – 3 sin 3x·3 = 6 cos 3x – 9 sin 3x f'(x) = – 6 sin 3x · 3 – 9 cos 3x · 3 = – 18 sin 3x – 27 cos 3x `"f''"((5pi)/6) = - 18 sin 3((5pi)/6) - 27 cos 3((5pi)/6) = - 18 sin ((5pi)/2) - 27 cos((5pi)/2) = -18 sin(2pi + pi/2) - 27 cos(2pi + pi/2) = -18sin pi/2 - 27 cos pi/2` = – 18 · 1 – 27 · 0 = – 18 < 0 maxima Maximum value of f(x) at x = (5pi)/6 `"f"((5pi)/6) = 2 sin 3((5pi)/6) + 3 cos 3((5pi)/6) = 2 sin (5pi)/2 + 3 cos (5pi)/2 = 2 sin (2x + pi/2) + 3cos(2pi + pi/2) = 2 sin pi/2 + 3 cos pi/2` = 2

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