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求不等式恒成立的参数取值范围,是中学学习的难点之一,也是高考、竞赛的热点。下面就此问题的几种基本方法加以论述。一、利用一次函数的性质一次函数y=f(x)=ax+b在x∈[m,n]上恒大于零的充要条件是:(?)对于y=f(x)=ax+b恒小于零的条件亦可类似给出。例1若f(x)=(x-1)m~2-6xm+x+1在区间[0,1]上恒为正值,求实数m的取值范围。解:考查关于x的一次函数f(x)=(m~2-6m+1)x+1-m~2恒为正值的充要条件。
Constant inequality is inequality parameter range, is one of the difficulties in high school learning, but also college entrance examination, competition hot spots. Here are some basic methods to address this issue. First, the use of the nature of a linear function y = f (x) = ax + b in x ∈ [m, n] is always greater than zero on the necessary and sufficient conditions: (?) For y = f b constant less than zero conditions can be given similarly. Example 1 Find the range of the real number m if f (x) = (x-1) m ~ 2-6xm + x + 1 is constant in the interval [0,1]. Solution: A necessary and sufficient condition for a constant function f (x) = (m ~ 2-6m + 1) x + 1-m ~ 2 to be positive is obtained.