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在一次学生测验中出现了这样一道题:若对任意的t∈(-2,2),函数f(x)=x2+(t-6)x+11-2t的值恒大于零,试求x的取值范围。分析:若以x为主元,则是关于x的二次函数,可以分类讨论解答,这样虽然能解答出来,但过程较为烦琐。若变换一下思路以参数t为主元,将它整理成关于t的一次函数.
In a student test, a question appears: If for any t ∈ (-2,2), the value of the function f (x) = x2 + (t-6) x + 11-2t is always greater than zero, The range of values. Analysis: If x is the main element, it is about the quadratic function of x, can be classified to discuss the solution, although this can be answered, but the process is more cumbersome. If you change the idea of the parameter t as the main element, it is sorted into a function on t.