1△k=1-√k+2k=4;
2(√k)^2-√k-3=0;
(√k+1)·(2√k-3)=0;
∵√k≥0,∴√k+1>0.
∴2√k-3=0.
则:√k=3/2;
k=9/4;
∴:函数f(x)=(9/4)△(4/9)·x
=(9/4)-√[(9/4)·(4/9)x]+2·(4/9)·x
=(9/4)-√x +8x/9
=(8/9)[(√x)^2 -(9/8)·√x+81/32]
=(8/9)[(√x - 9/16)^2 + 81×7/256]
≥(8/9)×(0+ 81×7/256)
=63/32.
值域是[63/32,+∞)
.