f(x)=2sin^2x+cos^2x+sinxcosx,x∈R,f(π/12)

题目

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

答案

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