(1)当θ=-π/6时,tanθ=-√3/3
f(x)=x^2-2√3x/3-1=(x-√3/3)^2-4/3
f(x)在[-1,√3/3]单调减,在[√3/3,√3]单调增
f(-1)=1+2√3/3-1=2√3/3
f(√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
(1)当θ=-π/6时,tanθ=-√3/3
f(x)=x^2-2√3x/3-1=(x-√3/3)^2-4/3
f(x)在[-1,√3/3]单调减,在[√3/3,√3]单调增
f(-1)=1+2√3/3-1=2√3/3
f(√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