Question

曲线f(x)=

f′(1)

e

ex-f(0)x+

1

2

x2在点（1，f（1））处的切线方程为______．

Answer 1

由题意，f′(x)=

f′(1)

e

ex-f(0)+x，f(0)=

f′(1)

e

∴f′(1)=

f′(1)

e

e-

f′(1)

e

+1=e
∴f(x)=ex-1+

1

2

x2
∴f(1)=e-

1

2

∴所求切线方程为y-e+

1

2

=e（x-1），即y=ex-

1

2

故答案为：y=ex-

1

2