Question

函数f(x)ax五次方+bx三次方+cx+8,且f(-2)=10,则f(2)的答案是多少

Answer 1

∵f(x)=ax^5+bx^3+cx+8,f(-2)=10∴f(-2)= -32a -8b -2c + 8 =10∴-32a -8b -2c =2 32a + 8b + 2c=-2 ∴f(2)=32a + 8b + 2c + 8= -2 +8 = 6