Question

函数f(x)=cos(1/2x–π/4)则f‘(3π)=

Answer 1

f(x)=3cos(1/2x－π/6) x∈R
(1)f(x)=Acos(wx－π/6)x
T=2π/w=4π
(2)π<=1/2x－π/6<=2π
7π/6<=0.5x<=13π/6
7π/3<=x<=13π/3
单调递增区间为[7π/3+4kπ,13π/3+4kπ]
(3)1/2x－π/6=kπ
x=π/3+2kπ
对称轴为x=π/3+2kπ
1/2x－π/6=kπ+π/2
1/2x=kπ+2π/3
x=2kπ+4π/3
对称中心(2kπ+4π/3,0)
(4)取得最大值时
1/2x－π/6=2kπ
x=π/3+4kπ
解集是{x|x=π/3+4kπ}