Question

已知函数f(x)=根号3sin2x+2cos平方x+1.求f(x)的单调递增区间.？

Answer 1

f(x)=根号3sin2x+2cos平方x+1
=√3sin2x+cos2x+2
=2sin(2x+π/6)+2
所以
当2kπ-π/2,1,f(x)=√3sin2x+2(cosx)^2+1
= √3sin2x +cos2x + 2
= 2(sin(2x+π/6) + 2
调递增区间
2nπ- π/2 <= 2x+π/6 <= 2nπ+ π/2
nπ- π/3 <= x <= nπ+ π/6,2,f'(x)=√3cos2x*2-4cosx*sinx=0
2√3cos2x-4cosx*sinx=0
√3cos2x-2cosx*sinx=0
√3cos2x-sin2x=0
tan2x=√3
2x=60°
x=30°
f''=-4√3sin2x-4cos2x
x>30,f''<0，是增函数,2,f(x)=根号3sin2x+2cos^2x+1=根号3sin2x+cos2x+2=2(根号3/2sin2x+1/2cos2x)+2=2sin(2x+Pai/6)+2
单调递增区间是：-Pai/2+2kPai<=2x+Pai/6<=Pai/2+2kPai
即为：［Kpai,-Pai/3,kPai+Pai/6],2,根号3sin2x+2cos平方x+1=2sin(2x+π/6)-2
递增区间为[kπ-π/3，kπ+π/6],1,