y=sinx+cosx的极值点 用导数求
问题描述：
y=sinx+cosx的极值点 用导数求
答
y'=cosx-sinx=-√2sin(x-π/4)，
令 y'=0，得 x-π/4=kπ，
x=kπ+π/4，k∈Z
k=2n时，
f(2nπ+π/4)=sin(π/4)+cos(π/4)=√2
所以极大值点为(2nπ+π/4，√2)
同理，k=2n-1时，极大值点为(2nπ-3π/4，-√2)
答
y = sinx + cosx = √2sin(x + π/4)y' = √2cos(x + π/4),y'' = -√2sin(x + π/4)y' = 0 => cos(x + π/4) = 0x + π/4 = 2kπ + π/2 或 x + π/4 = 2kπ - π/2x = 2kπ + π/4 或 x = 2kπ - 3π/4y''|(2kπ +...