Question

16.函数 f(x)=2sin2xsinx 的值域是

Answer 1

f(x) = 2sin2xsinx = 4cosx(sinx)^2

= 4cosx[1-(cosx)^2] = 4[cosx - (cosx)^3]

记 u = cosx， 则 g(u) = 4(u-u^3),-1≤u≤1,

g' = 4(1-3u^2), 驻点 u = ±1/√3

g( 1/√3) = 8√3/9,g( -1/√3) =-8√3/9.

g(1)=0,g(-1)=0.

值域是 [-8√3/9, 8√3/9].

Answer 2

f(x) = 2sin2xsinx = 4cosx(sinx)^2
= 4cosx[1-(cosx)^2] = 4[cosx - (cosx)^3]
记 u = cosx， 则 f = 4(u-u^3),
df/du = 4(1-3u^2), 驻点 u = ±1/√3
u = cosx = 1/√3 时，f = 8√3/9
函数 f(x) = 2sin2xsinx 的值域是 [-8√3/9, 8√3/9]