f(x)=2sin^2x+cos^2x+sinxcosx,x∈R,f(π/12)f(x)=2sin^2x+cos^2x+sinxcosx,x∈R求:1) f(π/12)的值2) f(x)的最小值及相应x的值3) f(x)的递增区间
问题描述：
f(x)=2sin^2x+cos^2x+sinxcosx,x∈R,f(π/12)
f(x)=2sin^2x+cos^2x+sinxcosx,x∈R
求:
1) f(π/12)的值
2) f(x)的最小值及相应x的值
3) f(x)的递增区间
答
f(x) = 2sin²x+cos²x+sinxcosx = sin²x+1+sin2x/2 = 3/2+(sin2x-cos2x)/2
1、f(π/12)=(7-√3)/4
2、f‘(x)=cos2x+sin2x=√2(sin(2x+π/4))=0解得x=kπ/2-π/4，对应最小值x=kπ-π/4，f(x)=1
3、f'(x)>=0时x在[kπ-π/4,3kπ/2-π/4],递增
答
把函数化成单名函数：
f(x)=3/2+√2/2sin(2x+π/4)
下面应该很好做了。
答
f(x) = 2sin^2x+cos^2x+sinxcosx
= 1 + sin^2x + 1/2sin2x
= 1 + (1 - cos2x)/2 + 1/2sin2x
= 3/2 + 1/2(sin2x - cos2x)
= 3/2 + 1/sqrt(2) (sin(2x - π/4))
所以f(π/12) = 3/2 + 1/sqrt(2) * sin(-π/12) = (7 - sqrt(3))/4
min(f(x)) = 3/2 - 1/sqrt(2) = (3 - sqrt(2))/2 此时 2x - π/4 = -π/2, x = -π/8
f(x)在[ -π/8 + kπ, 3π/8 + kπ]上递增