设f(x)＝e∧x－x＋cosx－2f＇(x)＝e∧x－sinx－1再次求导,f“(x)＝e∧x－cosx∵x＞0,∴e∧x＞1,0＜cosx＜1,∴e∧x－cosx＞0,即f“(x)＞0∴f＇(x)递增,又f＇(0)＝e∧0－sin0-1＝0∴f＇(x)＞0,∴f(x)在(0,正无穷)递增,...