函数(1)已知函数f(x)=(x
解:(1)。 x属于1,6的闭区间, x-1 >= 0
f(x) = (x-1) - 9/x + 1 = x - 9/x = x + (-9/x)
当1 f(x) is increase
(2)
当 a 当 x 属于[a, 6]时, f(x)的最大值=f(6) = 6-9/6 = 27/6 = 9/2
当 x 属于[1, a]时, f(x) = a-x - 9/x + a = 2a - (x + 9/x) 当 x 属于[1, a], a >= 3 时, f(x)的最大值 = f(3) = 2a - 6
a 当a属于[3,21/4] 时, 9/2 > 2a-6, M(a) =...全部
解:(1)。 x属于1,6的闭区间, x-1 >= 0
f(x) = (x-1) - 9/x + 1 = x - 9/x = x + (-9/x)
当1 f(x) is increase
(2)
当 a 当 x 属于[a, 6]时, f(x)的最大值=f(6) = 6-9/6 = 27/6 = 9/2
当 x 属于[1, a]时, f(x) = a-x - 9/x + a = 2a - (x + 9/x) 当 x 属于[1, a], a >= 3 时, f(x)的最大值 = f(3) = 2a - 6
a 当a属于[3,21/4] 时, 9/2 > 2a-6, M(a) = 9/2。
So, 当a属于 (1,21/4] 时 ==> M(a) = 9/2
当a属于 (21/4, 6) 时, 9/2 M(a) = 2a - 6
来源: 。收起