设根号下13-4x=t,则t≥0
则t^2=13-4x,x=(13-t^2)/4
则y=(13-t^2)/2-3-t=-(t^2)/2-t+7/2=-1/2[t^2+2t]+7/2
=-1/2[(t+1)^2-1]+7/2=-1/2(t+1)^2+4
因为t≥0,y=f(t)图象为开口向下,顶点为(-1,4)的抛物线
所以f(t)max=f(0)=7/2
则原函数的值域为(-无穷大,7/2]
求函数y=2x-3-根号下13-4x的值域
设根号下13-4x=t,则t≥0
则t^2=13-4x,x=(13-t^2)/4
则y=(13-t^2)/2-3-t=-(t^2)/2-t+7/2=-1/2[t^2+2t]+7/2
=-1/2[(t+1)^2-1]+7/2=-1/2(t+1)^2+4
因为t≥0,y=f(t)图象为开口向下,顶点为(-1,4)的抛物线
所以f(t)max=f(0)=7/2
则原函数的值域为(-无穷大,7/2]