已知二次函数f(x)=ax^2+bx+c(a,b,c∈R,a≠0),f(-2)=f(0)=0,f(x)的最小值为-1.

求(1)函数f(x)的解析式。
(2)设g(x)=f(-x)-mf(x)+1,若g(x)在[-1,1]上是减函数,求实数m的取值范围。
(3)设函数h(x)=log2[n-f(x)],若此函数在定义域范围内不存在零点,求实数n的取值范围。

解:

(1)

f(-2)=f(0)=0

∴可设f(x)=a(x+2)x,对称轴x=-1,顶点纵坐标是f(-1)=-a=-1,得a=1,

∴f(x)=x²+2x,

(2)

g(x)
=x²-2x-mx²-2mx+1
=(1-m)x²-2(1+m)x+1

当m=1时,g(x)=-4x+1,满足题意,

当m>1时,需(1+m)/(1-m)<-1,恒成立,

当m<1时,需(1+m)/(1-m)>1,解得0<m<1,

综上,m的取值范围是[0, ∞),

(3)

h(x)=lg2(-x²-2x+n),

定义域是-x²-2x+n>0,有解的条件是n>-1,

-x²-2x+n最大值是n+1,

不存在零点说明n+1<1,即n<0,

所以-1<n<0,

完毕!

谢谢!
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第1个回答  2010-07-16
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第2个回答  2010-07-16
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