laluxiang
幼苗
共回答了11个问题采纳率:81.8% 举报
a = (sinx,1),b = (1,cosx)
ƒ(x) = a • b
= (sinx)(1) + (1)(cosx)
= √2sin(x + π/4)
ƒ'(x) = √2cos(x + π/4)
F(x) = ƒ(x)ƒ'(x) + ƒ²(x)
= [√2sin(x + π/4)][√2cos(x + π/4)] + [√2sin(x + π/4)]²
= sin[2(x + π/4)] + 1 - cos[2(x + π/4)]
= sin(2x + π/4) - cos(2x + π/2) + 1
= cos(2x) - [- sin(2x)] + 1
= sin(2x) + cos(2x) + 1
= √2sin(2x + π/4) + 1
最小正周期T = (2π)/2 = π
最大值 = 1 + √2
最小值 = 1 - √2
1年前
5