函数f(x)=px^3+qx+1(p,q∈R且为常数,x∈R,若f(a)=2,则f(-a)=?
人气:402 ℃ 时间:2019-10-26 22:37:31
解答
f(a)=pa³+qa+1=2
pa³+qa=1
f(-a)=p(-a)³+q(-a)+1
=-pa³-qa+1
=-(pa³+qa)+1
=-1+1
=0
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