已知函数f(x)=x2-(2a+1)x+alnx. (Ⅰ)当a=2时,求曲线y=f(x)在点(1,f(1))处的切线方程; (Ⅱ)求函数f(x)的单调区间.
已知函数f(x)=x2-(2a+1)x+alnx.
(Ⅰ)当a=2时,求曲线y=f(x)在点(1,f(1))处的切线方程;
(Ⅱ)求函数f(x)的单调区间.
(Ⅰ)当a=2时,求曲线y=f(x)在点(1,f(1))处的切线方程;
(Ⅱ)求函数f(x)的单调区间.
数学人气:542 ℃时间:2019-08-18 15:48:02
优质解答
(I)当a=2时,f(x)=x2-(2a+1)x+alnx=x2-5x+2lnx,∴f′(x)=2x−5+2x,∴f′(1)=2-5+2=-1,∵f(1)=1-5=-4,∴曲线y=f(x)在点(1,f(1))处的切线方程为:x+y+3=0.(II)f′(x)=2x−(2a+1)+ax=2x2−(2...
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