Question

曲线f(x)=[f'(1)/e]*e^x-f(0)x+1/2(x^2)在点(1,f(1))处的切线方程为

Answer 1

f(x)=[f'(1)/e]*e^x-f(0)x+1/2(x^2)
令x=0代入，得
f(0)=f'(1)/e
f'(1)=ef(0)
令x=1,得
f(1)=f'(1)-f(0)+1/2
=ef(0)-f(0)+1/2

原方程两边求导，得
f'(x)=[f'(1)/e]*e^x-f(0)+x
令x=1，得
f'(1)=f'(1)-f(0)+1
f(0)=1
所以
f'(1)=e
f(1)=e-1+1/2=e-1/2

从而
切线方程为
y-(e-1/2)=e(x-1)
y-e+1/2=ex-e
即
y=ex -1/2

Answer 2

这类题目总是用这几部的，（1）令x=0;有f(0)=f'(1)/e-f(0);
(2)令x=1,有f(1)=f'(1)-f(0)+1/2;
(3)求导（有时是积分）；f'(x)=f'(1)/e*e^x-f(0)+x,所以f'(1)=f'(1)-f(0)+1;
三元一次方程组，解得f(0)=1,f'(1)=e,f(1)=e-1/2;
所以切线方程为y=f'(xo)(x-xo)+f（xo）=ex-1/2;