The function f (x) defined on the interval (- 1,1) is a decreasing function, and f (1-A)

The function f (x) defined on the interval (- 1,1) is a decreasing function, and f (1-A)

Because the function f (x) defined on the interval (- 1,1) is a decreasing function, and f (1-A) 1-A > A ^ 2-1 > - 1
The solution is 1 > a > 0

In this paper, the function f (x) = x2 + 10x-a + 3 is known. When x ∈ [- 2, + ∞), f (x) ≥ 0 is constant. The value range of real number a is obtained

Since the equation of the symmetry axis of the image of the function f (x) = x2 + 10x-a + 3 is x = - 5, the function increases monotonically on [- 2, + ∞), so the minimum value of the function is f (- 2) = - 13-a. because when x ∈ [- 2, + ∞), f (x) ≥ 0 is constant, and a ≤ - 13-A ≥ 0 is obtained

Let f (x) = 1 / 3x3-4x + a function (1) find the extreme value of F (x) (2) find that the equation f (x) = 0 has only one real root, and find the value of real number

1. Let f (x) = 1 / 3x3-4x + A, f '(x) = x ^ 2-4, f' (x) = 0, x = - 2 or x = 2, x = - 2 function has maximum value 16 / 3 + A, x = 2 function has minimum value - 16 / 3 + A, 2

If f (x) = 2 ^ x + m and the focus of X axis are between 1 and 2, then the value range of M? I don't understand,

When the intersection point with X axis is (1,0), there is: 2 ^ 1 + M = 0, M = - 2
When the intersection is (2,0), there is: 2 ^ 2 + M = 0, M = - 4
So, the range of M is: - 4