问题 填空题
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______.
答案

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

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