有这样一道测试题:若函数f(x)=x3-12x在区间(-∞,a]上存在反函数,求a的最大值.同学们的解法大致有以下三种:解法1:∵f(x)=x3-12x∴f′(x)=3x2-12,∴由f′(x)0,得x∈(-∞,-2)∪(2,+∞);由f′(x)0,得x∈(-2,2).∴函数f(x)=x3-12x的单调增区间为(-∞,-2]、[2,+∞),单调减区间为[ There is such a test question: If the function f(x) = x3-12x has an inverse function in the interval (-∞, a), find the maximum value of a. The students’ solutions are roughly the following three: Solution 1: ∵f (x)=x3-12x∴f′(x)=3x2-12, ∴ from f′(x)0, get x∈(-∞,-2)∪(2,+∞); from f′(x ) 0, get x ∈ (-2, 2). The monotonically increasing interval of the ∴ function f(x) = x3-12x is (-∞, -2), [2, +∞), and the interval of monotonic reduction is [