Given that the curve y = k sin(x) - (1/2) sin(2x) has a turning point at x =( n/3) , where k is a real number.
Find the value of k and determine whether the turning point is a maximum point or a minimum point.
p.s. x = n/3 ((n means pie..... 180 degree)
please help me solve the question......thanks a lot
amaths differentiation~~difficult ar!!
2007-10-21 1:26 am
回答 (1)
2007-10-21 1:48 am
✔ 最佳答案
y = k sin(x) - (1/2) sin(2x)dy/dx = kcosx -1/2 [cos(2x)] (2) = kcosx - cos 2x
Given that x = pi/3 is a turning point
dy/dx at x = pi/3 is kcos(pi/3) - cos(2pi/3) = 0
k/2 + 1/2 = 0
k = -1
d^2y/dx^2 = -ksinx + 2sin2x = sinx + 2sin2x
d^2y/dx^2 at x = pi/3 is sinpi/3 + 2sin 2pi/3 = 3rt3/2 > 0
Hence x = pi/3 is a minimum point
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