Question

求带根号函数的值域

y=√（x-1) + √(2x+1) 要有过程，谢谢

Answer 1

x-1>=0
2x+1>=0
=>
x>=1
and
x>=-1/2
故x>=1
又因为该函数单调增，则x=1的最小值，值域为[√3,无穷)