已知f(x)等于根号下的X的平方减一,试判断f(x)在[1,+∞)上的单调性,并证明 求详细步骤
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f(x)在[1,+∞)上单调递增。
证:
设1≤x1<x2
f(x2)-f(x1)
=√(x2²-1)-√(x1²-1)
=[√(x2²-1)-√(x1²-1)][√(x2²-1)+√(x1²-1)]/[√(x2²-1)+√(x1²-1)]
=(x2²-x1²)/[√(x2²-1)+√(x1²-1)]
=(x2+x1)(x2-x1)/[√(x2²-1)+√(x1²-1)]
x1≥1 √(x1²-1)]≥0
x2>x1 x2>1 x2²-1>0 √(x2²-1)]>0
√(x2²-1)+√(x1²-1)>0
x2>0 x1>0 x2+x1>0
x2>x1 x2-x1>0
(x2+x1)(x2-x1)/[√(x2²-1)+√(x1²-1)]>0
f(x2)-f(x1)>0
f(x2)>f(x1)
f(x)在[1,+∞)上单调递增。
证:
设1≤x1<x2
f(x2)-f(x1)
=√(x2²-1)-√(x1²-1)
=[√(x2²-1)-√(x1²-1)][√(x2²-1)+√(x1²-1)]/[√(x2²-1)+√(x1²-1)]
=(x2²-x1²)/[√(x2²-1)+√(x1²-1)]
=(x2+x1)(x2-x1)/[√(x2²-1)+√(x1²-1)]
x1≥1 √(x1²-1)]≥0
x2>x1 x2>1 x2²-1>0 √(x2²-1)]>0
√(x2²-1)+√(x1²-1)>0
x2>0 x1>0 x2+x1>0
x2>x1 x2-x1>0
(x2+x1)(x2-x1)/[√(x2²-1)+√(x1²-1)]>0
f(x2)-f(x1)>0
f(x2)>f(x1)
f(x)在[1,+∞)上单调递增。
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