Question

函数f(x)=x^3+sinx+1, (x∈R), 若f(а)=2则f(-а)=?

Answer 1

由f(a)=a^3+sina+1=2
得a^3+sina=1
f(-a)=(-a)^3+sin(-a)+1
= - a^3-sina+1
= -(a^3+sina)+1
=0

Answer 2

f(а)=2,即x^3+sinx+1=2
x^3+sinx=1
f(-а)=（-x）^3+sin（-x）+1
=-x^3-sinx+1
=0