Question

已知函数f(x)=lnx-x+1.(1)书曲线y=f(x)在x=2处的切线方程

Answer 1

f(x)=lnx-x+1

f'(x)=1/x-1

k=f'(2)=1/2-1=-1/2

f(2)=ln2-2+1=ln2-1

∴切线方程是

y-(ln2-1)=-1/2(x-2)=-x/2+1

y=-x/2+1+ln2-1

=-x/2+ln2