已知函数f(x)=ax^3+bx^2+cx+d,有三个零点分别是0,1,2 f(x)在(-∞,x1]单增 [x1,x2]单减 [x2,+∞)单增 求x1^2+x2^2 __________错了.不是f(x)=ax^3+bx^2+cx+d 是f(x)=x^3+bx^2+cx+d
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![已知函数f(x)=ax^3+bx^2+cx+d,有三个零点分别是0,1,2 f(x)在(-∞,x1]单增 [x1,x2]单减 [x2,+∞)单增 求x1^2+x2^2 __________错了.不是f(x)=ax^3+bx^2+cx+d 是f(x)=x^3+bx^2+cx+d](/uploads/image/z/6075346-58-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dax%5E3%2Bbx%5E2%2Bcx%2Bd%2C%E6%9C%89%E4%B8%89%E4%B8%AA%E9%9B%B6%E7%82%B9%E5%88%86%E5%88%AB%E6%98%AF0%2C1%2C2+f%28x%29%E5%9C%A8%28-%E2%88%9E%2Cx1%5D%E5%8D%95%E5%A2%9E+%5Bx1%2Cx2%5D%E5%8D%95%E5%87%8F+%5Bx2%2C%2B%E2%88%9E%29%E5%8D%95%E5%A2%9E+%E6%B1%82x1%5E2%2Bx2%5E2+__________%E9%94%99%E4%BA%86.%E4%B8%8D%E6%98%AFf%28x%29%3Dax%5E3%2Bbx%5E2%2Bcx%2Bd+%E6%98%AFf%28x%29%3Dx%5E3%2Bbx%5E2%2Bcx%2Bd)
已知函数f(x)=ax^3+bx^2+cx+d,有三个零点分别是0,1,2 f(x)在(-∞,x1]单增 [x1,x2]单减 [x2,+∞)单增 求x1^2+x2^2 __________错了.不是f(x)=ax^3+bx^2+cx+d 是f(x)=x^3+bx^2+cx+d
已知函数f(x)=ax^3+bx^2+cx+d,有三个零点分别是0,1,2 f(x)在(-∞,x1]单增 [x1,x2]单减 [x2,+∞)单增 求x1^2+x2^2 __________
错了.不是f(x)=ax^3+bx^2+cx+d 是f(x)=x^3+bx^2+cx+d
已知函数f(x)=ax^3+bx^2+cx+d,有三个零点分别是0,1,2 f(x)在(-∞,x1]单增 [x1,x2]单减 [x2,+∞)单增 求x1^2+x2^2 __________错了.不是f(x)=ax^3+bx^2+cx+d 是f(x)=x^3+bx^2+cx+d
由题意,f(x)有三个解,可必可以分解因式,即
f(x)=x(x-1)(x-2)=x^3-3x^2+2x
f'(x)=3x^2-6x+2
令f'(x)=0,即3x^2-6x+2=0
设两根为x1,x2,由韦达定理,得
x1^2+x2^2=(x1+x2)^2-2x1x2
=(6/3)^2-2*2/3
=8/3
无解,f(x)本身为单增或者单减很明显可以化成
f(x)=a(x+m)^3+n它在横轴的解是唯一的不可能出现0,1,2三个零点
由函数有三个零点0,1,2可解得
f(x)=x^3-3x^2+2x 从而
f'(x)=3x^2-6x+2
令f'(x)=0,
设两根为x1,x2,由韦达定理,得
x1^2+x2^2=(x1+x2)^2-2x1x2
=(6/3)^2-2*2/3
=8/3