Question

已知函数f(x)=sinx+cosx,f'(x)是f(x)的导函数

(1)求函数F(x)=f(x)f'(x)+f的平方(x)的最大值和最小正周期.(2)若f(x)=2f'(x),求tan(x-4分之太)的值

Answer 1

解：（1）已知函数f（x）=sinx+cosx，则f′（x）=sinx-cosx．
代入F（x）=f（x）f′（x）+[f（x）]2
易得
F(x)=cos2x+sin2x+1=2sin(2x+π4)+1
当 2x+π4=2kπ+π2⇒x=kπ+π8(k∈Z)时， [F(x)]max=2+1=3
最小正周期为 T=2π2=π
（2）由f（x）=2f'（x），易得sinx+cosx=2cosx-2sinx．
解得 tanx=13
∴ tan(x-π/4)=[tanx-tan(π/4)]/[1+tanxtan(π/4)]=(13-1)/(1+13)=12/13

Answer 2

1. f(x) = sinx+cosx, f '(x) = cosx - sinx
F(x) = f(x)f'(x) + f²(x) = cos²x - sin²x + 1 + sin2x = 1 + cos2x + sin2x
= 1 + √2 sin(2x + π/4)
最大值 1 + √2 ， 最小正周期 π
2. f(x) =2 f '(x) => sinx + cosx = 2(cosx - sinx)
=> 3 sinx = cosx => tanx = 1/3
tan(x - π/4) = (tanx - 1) / (1 + tanx * 1) = (-2/3) / (4/3) = -1/2