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y=sinx+cosx的极值点用导数求

题目详情
y=sinx+cosx的极值点 用导数求
▼优质解答
答案和解析
y = sinx + cosx = √2sin(x + π/4)
y' = √2cos(x + π/4),y'' = -√2sin(x + π/4)
y' = 0 => cos(x + π/4) = 0
x + π/4 = 2kπ + π/2 或 x + π/4 = 2kπ - π/2
x = 2kπ + π/4 或 x = 2kπ - 3π/4
y''|(2kπ + π/4) = -√2 < 0,取得极大值
y''|(2kπ - 3π/4) = √2 > 0,取得极小值
极小点为(2kπ - 3π/4,-√2)
极大点为(2kπ + π/4,√2)