Question

已知函数fx=|x-a|-9/x+a,x∈【1,6】,a∈R. （1）若a=6，写出函数fx的单调区间，并指出单调性

（2）若函数fx在【1，a】上单调，且存在x0∈【1，a】使fx0＞-2成立，求a的取值范围

Answer 1

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 <= x <= 6, both x and -9/x are increase ==> f(x) is increase

(2)
当 a <= x <= 6 时, f(x) = x-a - 9/x +a = x + (-9/x) is increase,
==> 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) <= 2a - 2(x(1/2)*(9/x)^(1/2)) = 2a - 6
==> 当 x 属于[1, a], a >= 3 时, f(x)的最大值 = f(3) = 2a - 6
a < 3 时, f(x)的最大值 f(a) = 2a - (a + 9/a) = a - 9/a < 0

So, 当 3 <= a < 6, f(x)的最大值的表达式M(a) = max{ 9/2, 2a-6}
当 1 < a < 3, M(a) = 9/2

9/2 = 2a - 6, a = 21/4 ==> 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 < 2a - 6 ==> M(a) = 2a - 6