Question

求函数Y=(ax^2+x+1)\(x+1),(x≥3且a﹥0）的最小值拜托各位了 3Q

Answer 1

f(x) = (ax + x + 1)/(x + 1) = ax/(x + 1) + 1 令f'(x)的分子为零,即 2ax(x + 1) - ax = 0 (x + 1) = 1 x = -2,或x = 0 就是说,f(x)在x = -2和x = 0时分别有极值点,经判断,在x = -2时,f(x)有极大值,在x = 0时,f(x)有极小值.题目要求x≥ 3,那么在极值点右侧的f(3)就是该函数的最小值,即 f(x)min = f(3) = (9/4)a + 1