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已知函数f(x)=x^2-mx+m-1.当x∈在[2,4]时,f(x)≥-1恒成立,求m取值范围.
人气:194 ℃ 时间:2020-06-02 14:42:09
解答
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
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