關於三角微分應用的問題.

設f(x)=-2sin2x-4cosx.
(a)求y=f(x)的極大點和極小點.
設0 < x <2π
f(x)=-2sin2x-4cosx.
f'(x)=-4cos2x+4sinx
f''(x)=8sin2x+4cosx
f'(x)=0
sinx=cos2x
sinx=1-2(sinx)^2
sinx=-1 or sinx=1/2
x=3π/2 or (x=π/6 or x= 5π/6)
f''(3π/2) =0 ,therefore x=3π/2 is rejected as it's neither maxima nor minimum ..
f"(π/6)>0
f"(5π/6) <0
當x=π/6,y=-3√3
當x=5π/6,y=3√3
所以f(x)最小值為(π/6,-3√3),最大值為(5π/6,3√3)


(b)描繪y=f(x)的圖像.
設x=0,y=-2sin0-4cos0=-4
當y=0,c0sx(1+sinx)=0 x=π/2,3π/2
x截距為π/2,3π/2 ,y截距為-4
如圖所示.

圖片參考：http://hk.geocities.com/stevieg_1023/fx.gif

[其中x軸係以孤度為單位]