设函数f (x)=x3-4x+a,0<a<2.若f (x)的三个零点为x1,x2,x3,且x1<x2<x3,则 ()A.x1>-1 B.x2<0 C.x2>0 D.x3>2 应该选哪一个?具体思路
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![设函数f (x)=x3-4x+a,0<a<2.若f (x)的三个零点为x1,x2,x3,且x1<x2<x3,则 ()A.x1>-1 B.x2<0 C.x2>0 D.x3>2 应该选哪一个?具体思路](/uploads/image/z/2085830-62-0.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f+%28x%29%EF%BC%9Dx3%EF%BC%8D4x%EF%BC%8Ba%2C0%EF%BC%9Ca%EF%BC%9C2%EF%BC%8E%E8%8B%A5f+%28x%29%E7%9A%84%E4%B8%89%E4%B8%AA%E9%9B%B6%E7%82%B9%E4%B8%BAx1%2Cx2%2Cx3%2C%E4%B8%94x1%EF%BC%9Cx2%EF%BC%9Cx3%2C%E5%88%99+%EF%BC%88%EF%BC%89A%EF%BC%8Ex1%EF%BC%9E%EF%BC%8D1++++++++B%EF%BC%8Ex2%EF%BC%9C0+++++++++C%EF%BC%8Ex2%EF%BC%9E0++++++++++D%EF%BC%8Ex3%EF%BC%9E2+%E5%BA%94%E8%AF%A5%E9%80%89%E5%93%AA%E4%B8%80%E4%B8%AA%3F%E5%85%B7%E4%BD%93%E6%80%9D%E8%B7%AF)
设函数f (x)=x3-4x+a,0<a<2.若f (x)的三个零点为x1,x2,x3,且x1<x2<x3,则 ()A.x1>-1 B.x2<0 C.x2>0 D.x3>2 应该选哪一个?具体思路
设函数f (x)=x3-4x+a,0<a<2.若f (x)的三个零点为x1,x2,x3,且x1<x2<x3,则 ()
A.x1>-1 B.x2<0 C.x2>0 D.x3>2
应该选哪一个?具体思路
设函数f (x)=x3-4x+a,0<a<2.若f (x)的三个零点为x1,x2,x3,且x1<x2<x3,则 ()A.x1>-1 B.x2<0 C.x2>0 D.x3>2 应该选哪一个?具体思路
f'(x)=3x²-4
令f'(x)≥0
3x²-4≥0
3x²≥4
x≥2/√3或x≤-2/√3
即函数在区间(-∞,-2/√3]上单调递增;在区间[-2/√3,2/√3]上单调递减;在[2/√3,+∞)上单调递增.
f(-1)=-1+4+a=a+3 30,即f(x)在区间(2,+∞)上无零点,D错.
综上,C是正确的,选C.
(x-x1)(x-x2)(x-x3)=0
(x^2 -x1x -x2x +x1x2)(x-x3)=0
(x^2 -x1x -x2x +x1x2)x-(x^2 -x1x -x2x +x1x2)x3=0
x^3 -x1x^2 - x2x^2 -x3x^2+x1x2x +x1x3x+x2x3x-x1x2x3=0
所以 x1+x2+x3=0, x1x2+x2x3+x1x3=-4, x1x2x3=-a<0
由1,3两式可知,三个零点 1负2正,正数为x3, 其它2个为负,选B