y=ax^5+bx^3+cx+d x=0 y=-3 , x=-5 y=9 , x=5 y=?

x=0
y=0+0+0+d=-3
d=-3
所以y=ax^5+bx^3+cx-3

x=-5
y=a*(-5)^5+b*(-5)^3+c*(-5)-3
=-a*5^5-b*5^3-5c-3
=-(a*5^5+b*5^3+5c)-3=9
所以
a*5^5+b*5^3+5c=-12

所以
x=5
y=a*5^5+b*5^3+5c-3=-12-3=-15

当x=0，原式变为d=-3
即y=ax^5+bx^3+cx-3
当x=-5时，y=9，那么亦即
g(x)=ax^5+bx^3+cx在x=-5时其值为9+3=12
易知这是个奇函数，所以
g(5)=-g(-5)=-12
∴f(5)=-12-3=-15

x=0,y=d=-3
x=-5,y=ax^5+bx^3+cx-3=-a*5^5-b*5^3-5c-3=9
a*5^5+b*5^3+5c=-12

x=5,y=a*5^5+b*5^3+5c-3=-12-3=-15