Question

已知函数f(x)=sinx+cosx,f'(x)是f(x)的导函数。

已知函数f(x)=sinx+cosx,f'(x)是f(x)的导函数，求f'(x)及y-f'(x)的最小正周期，求当x属于【0，π、2】，F(x)=f（x)f'(x)+f^2(x)的值域
速度啊。。。。。。。。。快。。。。。。。。。。。十分钟之类解附加20分
是求当x属于【0，π/2】

Answer 1

f(x) = sinx+cosx
f'(x) = cosx -sinx
= √2((1/√2)cosx - (1/√2)sinx)
= √2(cos(x+π/4))
f'(x) 的最小正周期 = 2π
y-f'(x)
=sinx+cosx -(cosx-sinx)
=2sinx
y-f'(x) 的最小正周期 = 2π

F(x)= f(x)f'(x)+[f(x)]^2
=(sinx+cosx)(cosx-sinx)+ (cosx-sinx)^2
= (cosx)^2 - (sinx)^2 + 1-2sinxcosx
= cos2x - sin2x +1
= √2((1/√2)cos2x-(1/√2)sin2x) +1
=√2(cos(2x+π/4)+1
F(x)=f（x)f'(x)+f^2(x)的值域 = [1-√2,2]

Answer 2

f'(x)=cosx- sinx 最小正周期为2π
y-f'(x) =2sinx 最小正周期为2π

F(x)=f（x)f'(x)+f^2(x) =cos2x +sin2x +1
=根号2* sin(2x+π/4) +1 x∈[0, π/2], 2x+π/4 ∈[π/4, 5π/4]
F(x)的值域为[0, 1+根号2]