Question

已知函数f(x)=2根号3sinXcosX+2cos^2X-1(X属于R) 若x∈[-5π/12，π/3],求f（x）的取值范围

求这样的题的步骤是什么

Answer 1

f(x) = sqrt(3) sin2x + cos2x = 2 sin(2x + pi/6)
2x+pi/6 = 0, x = -pi/12 时，f(x) = 0
2x+pi/6 = pi/2, x = pi/6 时，f(x) = 2
x = -5pi/12, 2x+pi/6 = -2pi/3, f(x) = -sqrt(3)
x=pi/3, 2x+pi/6 = 5pi/6, f(x) = 1
范围是-sqrt(3)->1

Answer 2

解：f(x)=2√3sinXcosX+2cos^2X-1
=√3sin2X+cos2X
=2sin(2X+π/6)
∵x∈[-5π/12，π/3],
∴2X+π/6 ∈[-2π/3，5π/6]
∴2sin(2X+π/6) ∈[-2，2]
∴f（x）的取值范围是 [-2，2]