y=x∧2-3x+4=(x-3/2)²+7/4;∵(x-3/2)²≥0;∴y≥0+7/4=7/4;所以值域为[7/4,+∞);f（x）=√x∧2-2x+4=√(x-1)²+3;∵(x-1)²≥0;∴√(x-1)²+3≥√3;所以值域为[√3,+∞)很高兴为您解答,skyhunt...