Question

函数f(x)=(x^2+1)\(bx+c)是奇函数且f(1)=2求f(x)

Answer 1

f(x)=(x^2+1)(bx+c)

f(1)=(1+1)(b+c)=2

f(-1)=(1+1)(-b+c)=-2

b=1 c=0

f(x)=(x^2+1)x