Question

f(x)=2cos²x+3sinx+3,x∈[π/6,2π/3]的值域

Answer 1

f(x)=2-2sin²x+3sinx+3
=-2(sinx-3/4)²+48/8

x∈[π/6,2π/3]

1/2<=sinx<=1
所以
最大是49/8
最小是6
值域 [6,49/8]

Answer 2

f(x)=2cos²x+3sinx+3
=2(1-(sinx)^2) + 3sinx+3
= -2(sinx)^2+3sinx+5
= -2( sinx - 3/4)^2+ 49/8
x∈[π/6,2π/3]

sinx∈[1/2, 1]
max f(x) = 49/8
f(π/6) = -1/8+49/8 = 6
f(2π/3) = 2(3/4) + 3√3/2 +3 = (1/2)(3√3 + 9)
值域= [(1/2)(3√3 + 9), 49/8]

Answer 3

f(x)=2(1-sin²x) + 3sinx+3
=-2sin²x+3sinx+5
=-2(sinx-3/4)²+49/8
∵x∈[π/6,2π/3]
∴f(x)∈【49/8,6】