Question

已知函数f(x)=e的x次方-1-x求y=f(x)在点(1,f(1))处的切线方程

Answer 1

f(1)=e^1-1-1=e-2

f'(x)=e^x-1

K=f'(1)=e^1-1=e-1

所以切线方程是

y-(e-2)=(e-1)(x-1)=(e-1)x-e+1

y=(e-1)x-e+1+e-2

=(e-1)x-1