已知函数f(X)=(ax-x)lnx-1/2ax+x,当a=0时,求曲线y=f(X)在e
已知函数f(X)=(ax-x)lnx-1/2ax+x,当a=0时,求曲线y=f(X)在e
1,由f(x)=ax∧2-(a+2)x+lnx 得 f′(x)=2ax-(a+2)+1/x 当a=1,x=1时 f(1)=-2 f′(1)=2-3+1=0 曲线y=f(x)在点(1,f(1))处的切线方程 y=-2 2,f'(x)=2ax-(a+2)+1/x =[2ax^2-(a+2)x+1]/x,=2a(x-1/2)(x...