若函数f(x)=ax^3+bx^2+cx+d满足f(0)=f(x1)=f(x2)=0,(0
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若函数f(x)=ax^3+bx^2+cx+d满足f(0)=f(x1)=f(x2)=0,(0
若函数f(x)=ax^3+bx^2+cx+d满足f(0)=f(x1)=f(x2)=0,(0
若函数f(x)=ax^3+bx^2+cx+d满足f(0)=f(x1)=f(x2)=0,(0
f(0)=0 得出d=0 则f(x)=ax^3+bx^2+cx 则f(x)=ax^2+bx+c 两根x1 x2
【x2,……)上单调递增 所以a大于0 x1+x2>0 x1*x2>0所以b
f(0)=0 ==>d=0
f(1)=0 ==>a+b+c=0
f(2)=0 ==>8a+4b+2c=0 ==>4a+2b+c=0
==>3a+b=0
==> f(x)=-b/3*x^3+bx^2-2b/3*x
由题==>b!=0
二次函数对称轴为x=2/3
要满足题意需要a>0 ==> -b/3>0
==> b<0
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