Given the function f (x) = x ^ 3-x decreasing on (0, a) and increasing on [a, positive infinity), try to find the value of A

Given the function f (x) = x ^ 3-x decreasing on (0, a) and increasing on [a, positive infinity), try to find the value of A

The derivative of the original function: F '(x) = 3x ^ 2-1 is less than zero in the interval of one third of the positive and negative radicals, so the original function decreases monotonically in this interval. Therefore, according to the meaning of the problem, we can calculate a = 1 / ㄏ 3
We can't find a place to root

The known function f * x) = x ^ 3-x decreases on (0, a], increases on x greater than or equal to a, and calculates the value of A
Please give details of the process

f'(x)=3x^2-1
From F '(x) > 0: X √ 3 / 3
That is, f (x) increases on X √ 3 / 3
From F '(x)

Given the function f (x) = - x3-bx2-5cx, it decreases monotonically on (- ∞, 0) and increases monotonically on [0, 6]. (1) find the value of real number C; (2) find the value range of B

(1) F ′ (x) = - 3x2-2bx-5c ∵ function f (x) = - x3-bx2-5cx decreases monotonically on (- ∞, 0] and increases monotonically on [0, 6]; {function has a minimum at x = 0,} f ′ (0) = - 5C = 0, C = 0 (2) ∵ f ′ (x) = - 3x2-2bx ≥ 0 is constant on [0, 6], that is, 2b ≤ - 3x is constant on [0, 6]