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已知F(x)=sin(2x+6/π)+2/3,x属于R

最小正周期2π/2=π 单调减区间 2kπ+π/2≤2x+π/6≤2kπ+3π/2 2kπ+π/3≤2x≤2kπ+4π/3 kπ+π/6≤x≤kπ+2π/3 y=sin(2x+π/6)+3/2 =sin[2(x+π/12)]+3/2 因此将sin2x向左移π/12得到sin[2(x+π/12)] 再向上移3/2个单位即可

f(x)=sin(2x+π/6)+3/2 -1

f(x) =sin(2x+π/6)+3/2 2kπ+π/2≤2x+π/6≤2kπ+3π/2 k∈Z 2kπ+π/6≤x≤2kπ+2π/3 因此,f(x)的单减区间为:[2kπ+π/6,2kπ+2π/3] (k∈Z)

(1)由函数 f(x)=sin(2x+ π 6 )+ 3 2 ,x∈R ,可得周期等于 T= 2π 2 =π.由 2kπ- π 2 ≤2x+ π 6 ≤2kπ+ π 2 (k∈Z) 求得 kπ- π 3 ≤x≤kπ+ π 6 (k∈Z) ,故函数的递增区间是 [ .(2)由条件可得 f(x)=sin(2x+ π 6 )+ 3 2 =sin[2(x+ π 12 )]+ 3 2 ....

(1)T=2π2=π.(2)由2kπ-π2≤2x+π6≤2kπ+π2,得kπ-π3≤x≤kπ+π6,k∈Z,∴函数的单调增区间为[kπ-π3,kπ+π6](k∈Z).(3)∵x∈[0,π2],∴2x+π6∈[π6,7π6],∴-12≤sin(2x+π6)≤1,∴当2x+π6=π2,即x=π6时函数有最大值1,当2x+π6=7π6时,即x=π2,函数有...

1. f(x)=sin(2x+π/6)+1/2 则最小正周期为T=2π/2=π。 2. y=(1/2)(cosx)^2+(√3/2)sinxcosx+1 =(√3/4)sin2x+(1/4)cos2x+5/4 =(1/2)[(√3/2)sin2x+(1/2)cos2x]+5/4 =(1/2)(sin2xcosπ/6+cos2xsinπ/6)+5/4 =(1/2)sin(2x+π/6)+5/4 f(x)最小正周期为T=2π/...

21.(3)x∈[0,π/3], u=2x+π/6的值域是[π/6,5π/6], v=f(x)=sinu-√3的值域是[1/2-√3,1-√3], 对任意x∈[0,π/3],f(x)^2-(2+m)f(x)+2+m

(1)∵f(x)=2sin(2x+ π 6 ),∴其最小正周期T= 2π 2 =π;∴由2kπ- π 2 ≤2x+ π 6 ≤2kπ+ π 2 得kπ- π 3 ≤x≤kπ+ π 6 (k∈Z),∴函数的增区间为[kπ- π 3 ,kπ+ π 6 ](k∈Z),(2)∵x∈( π 4 , 3π 4 ],∴2x+ π 6 ∈( 2π 3 , 5π 3 ],∴-1≤sin(2...

(1)f(x)=sin(2x+π/6)+3/2,最小正周期为2π/2=π,单增区间为2Kπ-π/2

解:f(x)=sin(2x+π/6)+1/2. x∈[0,π/2} ∵| sin(2x+π/6)|≤1,∴sin(2x+π/6)=1时,f(x)具有最大值,且f(x)max=1+1/2=3/2. 此时,2x+π/6=π/2. 2x=π/2-π/6. =π/3. ∴x=π/6. 即,当x=π/6,【符合x∈[0,π/2]的要求】函数f(x)=sin(2x+π/6)具有最大值f(x)max=...

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