Differentiation .～～＊

f(x)=√3 sin2x +cos 2x
f ' (x)=2√3cos2x-2sin2x
f' ' ' (x)=-4√3sin2x-4cos2x

To find turning point, set f ' (x) = 0
2√3cos2x-2sin2x = 0
√3cos2x-sin2x=0
sin2x/cos2x=√3
tan2x = √3
- π/2 ＝＜ｘ＝＜π/2, therefore, - π ＝＜2ｘ＝＜π
Therefore, 2x=π/3 or 2x=π/3 - π = -2π/3
x = π/6 or x = -π/3
When x = π/6,
f ' ' (x) = -4√3 (√3/2) -4(1/2) = -8 <0
f(x) = √3 (√3/2) + (1/2) = 2
Therefore, (π/6,2) is max. point
When x = -π/3,
f ' ' (x) = -4√3 (-√3/2) -4(-1/2) = 8 >0
f(x) = √3 (-√3/2) + (-1/2) = -2
Therefore, (-π/3,-2) is min. point

f(x)=√3 sin2x +cos 2x
f &#39; (x)=2√3cos2x-2sin2x
f&#39; &#39; &#39; (x)=-4√3sin2x-4cos2x

f &#39; (x) = 0
2√3cos2x-2sin2x = 0
√3cos2x-sin2x=0
2(√3/2cos2x-1/2sin2x)=0
cos(π/6)cos2x-sin(π/6)sin2x=0
simply....
x=π/6

f &#39; &#39; (x) x=π/6 &gt; 0
Therefore
(π/6,2) is min. point

If there are any mistake
please feel free to contact me

2006-11-04 21:35:43 補充：
f *(x) cannot be displayed here,therefore I use dy/dx and d^2y/dx*2 instead of f*(x) and f**(x) respectively dy/dx=2√3cos2x-2sin2xd^2y/dx*2=-4√3sin2x-4cos2xdy/dx= 0...simply....x=π/6d^2y/dx*2 [x=π/6] = 0and so on