已知函数f(x)=a的x次方+x-2/x+1(a>1),证明:函数f(x)在(-1,正无穷)上为单调递增
已知函数f(x)=a的x次方+x-2/x+1(a>1),证明:函数f(x)在(-1,正无穷)上为单调递增
数学人气:148 ℃时间:2019-10-29 12:52:19
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a的取值范围?已知函数f(x)=a的x次方+x-2/x+1(a>1),证明:函数f(x)在(-1,正无穷)上为单调递增直接求导,f′(x)=(a^x)×㏑a+2/x²,由a>1,有f′(x)>0,只能证明在(-1,0),(0,+∞)单调递增可是它要在(-1,正无穷)上单调递增我看错题目了,不好意思哈 应该是对x求导,得:f′(x)=(a^x)×㏑a+3/(x+1)²,由a>1,有f′(x)>0对于x>-1恒成立,即为f(x)在(-1,+∞)单调递增
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