(2010•台州二模)已知函数f(x)=13x3−12x2−23(t−1)2x.
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(2010•台州二模)已知函数f(x)=
x
1 |
3 |
![(2010•台州二模)已知函数f(x)=13x3−12x2−23(t−1)2x.](/uploads/image/z/4465667-11-7.jpg?t=%EF%BC%882010%E2%80%A2%E5%8F%B0%E5%B7%9E%E4%BA%8C%E6%A8%A1%EF%BC%89%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%EF%BC%9D13x3%E2%88%9212x2%E2%88%9223%28t%E2%88%921%292x%EF%BC%8E)
(I)当t=1时,记h(x)=f(x+a)+b=
1
3(x+a)3−
1
2(x+a)2+b
则h′(x)=x2+(2a-1)x+a2-a
∵h(x)为奇函数
∴h(0)=
1
3a3−
1
2a2+b=0(1)------(3分)
且 h′(x)为偶函数 即2a-1=0(2)------(5分)
由(1)、(2)解得:a=
1
2,b=
1
12.------(7分)
(II)f/(x)=x2−x−
2
3(t−1)2
令f′(x)=0解得:x1=
1−
1+
8
3(t−1)2
2,x2=
1+
1+
8
3(t−1)2
2------(9分)
(i)当1<t<4时,则有-2<x1<x2<t
∴f′(x)在(-2,x1)和(x2,t)为正,在(x1,x2)为负
∴f(x)在(-2,x1)和(x2,t)上递增,在(x1,x2)上递减
此时,x1=
1−
1+
8
3(t−1)2
2为极大值点,x2=
1+
1
3(x+a)3−
1
2(x+a)2+b
则h′(x)=x2+(2a-1)x+a2-a
∵h(x)为奇函数
∴h(0)=
1
3a3−
1
2a2+b=0(1)------(3分)
且 h′(x)为偶函数 即2a-1=0(2)------(5分)
由(1)、(2)解得:a=
1
2,b=
1
12.------(7分)
(II)f/(x)=x2−x−
2
3(t−1)2
令f′(x)=0解得:x1=
1−
1+
8
3(t−1)2
2,x2=
1+
1+
8
3(t−1)2
2------(9分)
(i)当1<t<4时,则有-2<x1<x2<t
∴f′(x)在(-2,x1)和(x2,t)为正,在(x1,x2)为负
∴f(x)在(-2,x1)和(x2,t)上递增,在(x1,x2)上递减
此时,x1=
1−
1+
8
3(t−1)2
2为极大值点,x2=
1+
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