高中数学】求值域g(x)=(6t+5)(t-1)/(2-4t^2),且t∈[0,1]

2个回答

  • 采用分离常数法:g(t)=(6t^2-t-5)/(2-4t^2)=【6(t^2-1/2)-t-2]/[-4(t^2-1/2)]

    =-3/2+(t+2)/[4(t^2-1/2)]=-3/2+1/4*1/[(t^2-1/2)/(t+2)]=-3/2+1/4/(-4+t+2+7/2/

    (t+2)] 令x=t+2,因为 t∈[0,1],但 2-4t^20===>t根号2/2===>x∈[2,2+根号2/2) U(2+根号2/2,3].

    g(t)=-3/2+1/4/(x+7/2/x-4) 钩形函数Y=x+7/2/x,x∈[2,2+根号2/2) U(2+根号2/2,3].单调递增.===>

    x+7/2/x∈[15/4,4) U (4,25/6]===>x+7/2/x-4 ∈[-1/4,0)U (0,1/6]===>1/(x+7/2/x-4) ∈(-无穷,-4】U 【6,+无穷)===>原函数值域为:(-无穷,-5/2] U [0,+无穷).