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已知函数f(x)=lnx+a/x,(a∈R),当a=1,且x≥1时,证明:f(x)≤1

谢明娟 1年前 悬赏15滴雨露 已收到2个回答 我来回答 举报

zhoujinling1984 幼苗

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函数f(x)应是如右形式:f(x)=(lnx+a)/x,否则函数的值域为无穷大;
f'(x)=(lnx+a)/x=[(1/x)*x-(lnx+a)]/x²=-(lnx)/x;{a=1};
当x≧1时,f'(x)≦0,f(x)是单调递减函数,其最大值是在区间左端x=1处f(x)=(ln1+1)/1=1;
所以 f(x)≤1;

1年前

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崇Ginny 幼苗

共回答了33个问题 向TA提问 举报

f(x)=lnx+a/x
f(x)=lnx+1/x
x=1时 f(x)=lnx+1/x=1
f'(x)=1/x-/x^2=(1/x)(1-1/x)>=0
f(x)>=1

1年前

1
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