已知函数f(x)=x^2+2xtanθ-1,x∈(-1,根号3],其中θ∈(-π/2,π/2).

(1)当θ=-π/6时,tanθ=-√3/3f(x)=x^2-2√3x/3-1=(x-√3/3)^2-4/3f(x)在[-1,√3/3]单调减,在[√3/3,√3]单调增f(-1)=1+2√3/3-1=2√3/3f(√3)=0所以f(x)在x=-1处取得最大2√3/3在x=√3/3处取得最小-4/3(2)f(x)=x^2+2xtanθ-1=(x+tanθ)^2-1-(tanθ)^2当-tanθ≤-1即tanθ≥1,即π/4≤θ<π/2时f(x)在[-1,√3]单调增当-tanθ≥√3,即tanθ≤-√3,即-π/2<θ≤-π/3时f(x)在[-1,√3]单调减综上所述:θ的取值范围(-π/2,-π/3]∪[π/4,π/2) 收起