求当x∈[π 6,5π 6]时,函数y=-2cos^2x 2sinx 3

令3x-π/6=kπ,得对称轴方程是x=kπ/3-π/18x∈[0,π/3](3x-π/6)∈[-π/6,5π/6]cos(3x-π/6)x∈[-√3/2,1]2cos(3x-π/6)x∈[-√3,2

1、最小正周期=2π/w=2π/2=π.2、根据题意：0

sin^2x+cos^2x=1所以f(x)=3-sinx-2(1-sin^2x)=2sin^2x-sinx+1=2(sinx-1/4)^2+7/8π/6

3(x的平方-7x+13)-2【(x+6)(x-5)+(3x-2)(x-3)】=-5x的平方-x+77=-5*49/4=39/4

(1)因为,x∈［0,π/2］,2x＋π/6∈［π/6,7π/6］,sin(2x＋π/6)∈［-1/2,1］,又a＞0所以,-2a＋2a＋b=-5a＋2a＋b=1解得：a=2,b=-5(2)由(1)知

sin(2x+π/6)的单调区间是在x∈[0,π/6]为增,在x∈[π/6,π/2]为减,因为系数-2a小于零,所以f(x)在x∈[0,π/6]为减,在x∈[π/6,π/2]为增

f(x)是R上的奇函数f(-π/6)=-f(π/6)=sinπ/6+cosπ/6=1/2+√3/3=(1+√3)/2

当x=√5-1x^2+5x-6=(x+6)(x-1)=(√5+5)(√5-2)再问：请问下x²+5x-6是怎样转化成(x+6)(x-1)，请说具体点再答：十字相乘法：x²+5x-6

f(x)=2cosxcos(x-π/6)-√3sin^2x+sinxcosx=2cosxcos(x-π/6)-√3sin^2x+sinxcosx=2cosx(√3/2cosx+1/2sinx)-√3i

f(x)=sin(2x+π/3)+sin2x=1/2sin2x+1/2*根号3cos2x+sin2x=根号3sin(2x+π/6)最小正周期=2π/2=π最大值=根号3最小值=-根号3

(1)因为,x∈［0,π/2］,2x＋π/6∈［π/6,7π/6］,sin(2x＋π/6)∈［-1/2,1］,又a＞0所以,-2a＋2a＋b=-5a＋2a＋b=1解得：a=2,b=-5(2)由(1)知

x∈[π/12,π/2]2x∈[π/6,π]2x+π/6∈[π/3,7π/6]sin(2x+π/6)∈[-1/2,1]2sin(2x+π/6)∈[-1,2]值域为：[-1,2]

f(X)≥1即sin(2X+6/π)+3/2≥1,令u=2X+6/π则sinu≥-1/2画个正玄图很容易看出2k∏-∏/6≤u≤2k∏+7∏/6,即2k∏-∏/6≤2X+6/π≤2k∏+7∏/6,得k

1/2x∈【-π/6,5/6π）1/2x-π/6∈【-π/3,2/3π）最大值=2sinπ/2=2最小值=2×sin(-π/3)=-2×√3/2=-√3

首先：x=π/2时,sinx=1＞x³/6=π³/48x=0时,sinx=x³/6=0分别对两个函数求二阶导数（没学过就直接“数形结合可知”）在（0,π/2）上,sinx

f(x)=(1/2)cosx-[(√3)/2]sinx-cosx=-(1/2)cosx-[(√3)/2]sinx=-sin[x+(π/6)]-1≤f(x)≤1

y=g(X)与y=f(x)的图像关于x=1对称可以得到g(x)=f(2-x)=√3sin(π/4(2-x)+π/6)=)=√3sin(π/2-π/4x+π/6)=√3cos(π/4x-π/6)0再问：

令　2kπ+π/2≤2x+π/6≤2kπ+3π/2则　kπ+π/6≤x≤kπ+2π/3令k=-1,0,得　f(x)在∈[-π,π]上的减区间为[-5π/6,-π/3]和[π/6,2π/3].

（1）最小正周期T=2π÷2=π;（2）x∈[0,π/2];2x+π/6∈[π/6,7π/6]∴sin(2x+π/6)∈[-1/2,1]∴最大值=3；最小值=-3/2;您好,很高兴为您解答,skyhu

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