已知f(x)=2cos2x(即2倍cosx的平方)+2√3sinxcosx+1,求f(x)的单调区间和f(x)在(0,π/4)值域

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已知f(x)=2cos2x(即2倍cosx的平方)+2√3sinxcosx+1,求f(x)的单调区间和f(x)在(0,π/4)值域

已知f(x)=2cos2x(即2倍cosx的平方)+2√3sinxcosx+1,求f(x)的单调区间和f(x)在(0,π/4)值域
已知f(x)=2cos2x(即2倍cosx的平方)+2√3sinxcosx+1,求f(x)的单调区间和f(x)在(0,π/4)值域

已知f(x)=2cos2x(即2倍cosx的平方)+2√3sinxcosx+1,求f(x)的单调区间和f(x)在(0,π/4)值域
f(x)=2(cosx)^2 + 2√3sinxcosx + 1
=2(cosx)^2 -1 + 2√3sinxcosx + 2
=cos2x + √3sin2x + 2
=2[(1/2)cos2x + (√3/2)sin2x] + 2
=2sin(2x + π/6) + 2
2kπ - π/2≤ 2x + π/6 ≤2kπ + π/2
2kπ - 2π/3≤ 2x ≤2kπ + π/3
当kπ - π/3≤ x ≤kπ + π/6时 ,f(x)单调递增
2kπ + π/2≤ 2x + π/6 ≤2kπ + 3π/2
2kπ + π/3≤ 2x ≤2kπ + 4π/3
当kπ + π/6≤ x ≤kπ + 2π/3时,f(x)单调递减
∵x∈(0,π/4)
∴2x + π/6 ∈(π/6 ,2π/3)
当x在(0,π/4)时,f(x)的值域是(3,4]

f(x)=cos(2x)+√3sin(2x) +2
= 2*(1/2cos(2x)+√3/2sin(2x))+2
=2*sin(2x+π/6)+2
(2x+π/6)属于(π/6+2π/3)
f(x)的值域是(-√3/+2,4)
加分哦

f(x)=cos(2x)+√3sin(2x) +2
= 2*[1/2cos(2x)+√3/2sin(2x)]+2
=2*sin(2x+π/6)+2
因为-1<=sin(2x+π/6)<=1
-2<=2*sin(2x+π/6)<=2
0<=2*sin(2x+π/6)+2<=4
所以0<=f(x)<=4
当x属于(0,π/4...

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f(x)=cos(2x)+√3sin(2x) +2
= 2*[1/2cos(2x)+√3/2sin(2x)]+2
=2*sin(2x+π/6)+2
因为-1<=sin(2x+π/6)<=1
-2<=2*sin(2x+π/6)<=2
0<=2*sin(2x+π/6)+2<=4
所以0<=f(x)<=4
当x属于(0,π/4)时
sin(0+π/6)即1/21<2*sin(2x+π/6)<√3
3<2*sin(2x+π/6)+2<2+√3
所以f(x)在(0,π/4)值域为(3,2+√3)

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