f(x)=ax+1+(lnx)/xa属于R若f(x)在区间内单调增求a的范围

Question

f(x)=ax+1+(lnx)/x a属于R若f(x)在区间内单调增 求a的范围

Answer 1

答：
f(x)=ax+1+(lnx)/x在区间内单调递增
求导：f'(x)=a+1/x^2-(lnx)/x^2>=0
设g(x)=(lnx-1)/x^2<=a恒成立
求导：g'(x)=(1/x)*(1/x^2)-2(lnx-1)/x^3
=(1-2lnx+2)/x^3
=(3-2lnx)/x^3
解g'(x)=0得：3-2lnx=0,x=e^(3/2)
00,g(x)是增函数
x>e^(3/2),g'(x)<0,g(x)是减函数
所以：x=e^(3/2)时g(x)取得最大值
a>=g[e^(3/2)=[lne^(3/2)-1]/[e^(3/2)]^2=(3/2-1)/e^3=1/(2e^3)
所以：a>=1/(2e^3)