已知函数f(x)=f'(1)e的(x-1)次方-f(0)x+1/2x平方求f(x)的解析式快点那!

Question

已知函数f(x)=f'(1)e的(x-1)次方-f(0)x+1/2x平方求f(x)的解析式
快点那!

Answer 1

f(x) =f'(1).e^(x-1) - f(0)x + (1/2)x^2
f'(x) = f'(1) .e^(x-1) - f(0) + x
put x=1
f'(1) = f'(1) -f(0) +1
f'(0)=1
put x=0
f'(0) = f'(1) .e^(-1) - f(0) + 0
f'(1) =2f'(0).e
= 2e
ie
f(x) =2e^(x) - x + (1/2)x^2