已知函数f(x)=x²-mx+m-1

问题描述:

已知函数f(x)=x²-mx+m-1
(1)当x∈[2,4]时,f(x)≥-1恒成立,求实数m的取值范围.
(2)是否存在整数a,b(a>b),使得不等式a≤f(x)≤b的解集为{x|a≤x≤b}?若存在,求出a,b的值;若不存在,说明理由.

f(x)=x²-mx+m-1≥-1
则:
x²-mx+m≥0
(x-1)m≤x²
因为:1≤x-1≤3
则:
m≤x²/(x-1)=[(x-1)²+2(x-1)+1]/(x-1)=[(x-1)+1/(x-1)]+2
因为:x-1≥0,则:(x-1)+1/(x-1)≥2
即:(x-1)+1/(x-1)的最小值是2
因:m≤【[(x-1)+1/(x-1)+2]】最小值,则:m≤4
(2)
x^2-mx+m-1
=(x-m/2)^2+m-1-(m/2)^2
=(x-m/2)^2-(m/2-1)^2
=(x-1)(x-m+1)
a≤f(x)≤b
a≤x≤b存在
f(x)=0
x=1 x=m+1
m>=0
a=1 b=m+1
m