y=√(x+1)
=>x>=-1
设x2>x1>=-1
=>√(x2+1)>0,(√(x2+1)+√(x1+1))>0
x2-x1>0
√(x2+1)-√(x1+1)
=(x2-x1)/(√x2+1)+√(x1+1))>0
=>√(x2+1)>√(x1+1)
因此函数单调递增.
y=√(x+1)
=>x>=-1
设x2>x1>=-1
=>√(x2+1)>0,(√(x2+1)+√(x1+1))>0
x2-x1>0
√(x2+1)-√(x1+1)
=(x2-x1)/(√x2+1)+√(x1+1))>0
=>√(x2+1)>√(x1+1)
因此函数单调递增.