On the closed interval [0,2? ] , find the maximum value of the function f(x) = 4

Question

On the closed interval [0,2? ] , find the maximum value of the function f(x) = 4sin x - 3 cos x. Please show work.

Explanation / Answer

f(x) = 4sinx - 3cosx f'(x) = 4cosx + 3sinx f'(x) = 0 => 4cosx + 3sinx = 0 => 3sinx = -4cosx => sinx/cosx = -4/3 => tanx = -4/3 => x = arctan(-4/3) => x = -53.13 degrees but this is not in [0, 360] note that, the period of tan function is 180, that is after every 180 the value gets repeated.. thus, -53.13 + 180 = 126.87 126.87 + 180 = 306.87 306.87 + 180 = 486.87 (out of bound) Thus, one of or both or no of 126.87 and 306.87 (which are in [0,2pi]) is the maximum value. to get that, we must find the 2nd derivative. f"(x) = -4sinx + 3cosx note that, f"(126.87) < 0 and f"(306.87) > 0 hence at x = 126.87 f(x) has maximum value

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