It is known that a and B are real numbers, f (x) = x ^ 3 + ax, G (x) = x ^ 2 + BX, f '(x) and G' (x) are derivatives of F (x), G (x), if f '(x) g' (x) ≥ 0 in function area If f (x) and G (x) are monotonically consistent on interval I, then f (x) and G (x) are monotonically consistent on interval I 1. Let a > 0, if the monotonicity of functions f (x) and G (x) is consistent in the interval [- 1, + ∞), find the value range of real number B

It is known that a and B are real numbers, f (x) = x ^ 3 + ax, G (x) = x ^ 2 + BX, f '(x) and G' (x) are derivatives of F (x), G (x), if f '(x) g' (x) ≥ 0 in function area If f (x) and G (x) are monotonically consistent on interval I, then f (x) and G (x) are monotonically consistent on interval I 1. Let a > 0, if the monotonicity of functions f (x) and G (x) is consistent in the interval [- 1, + ∞), find the value range of real number B

f'(x)=3x²+a
g'(x)=2x+b
f'(x)g'(x)=(3x²+a)(2x+b)
If a > 0
Then 3x & # 178; + a ≥ 0 + a > 0
Consistent with monotonicity
On [- 1, + ∞)
g'(x)≥0
2x+b≥0
b≥2
If you think the explanation is not clear enough, I wish you:

Given the function f (x) = x ^ 3-ax ^ 2-A ^ 2x + 1 g (x) = 1-4x-ax ^ 2, where the real number a ≠ 0, if f (x) and G (x) are increasing functions in the interval (- A, - A + 2),
Find the value range of a

F '(x) = 3x ^ 2-2ax-a ^ 2 = (3x + a) (x-a) > = 0
a> 0, x > = a or X

If the distance between the maximum point and the minimum point of the function y = x ^ 2-2x in the interval [- 1, M] is 2 √ 5, then the value range of the real number m?
〔1,3〕
All the intervals on the question are closed, including those on the answer

This kind of problem is to take the extreme value of several points are found out; discussion
First of all, it can be determined that y = x ^ 2-2x is x = 1, so it is possible to take the extreme point x = - 1, m, 1
1. When M1 and M3
Ymin=f（1）=-1
Ymax=f(m)=m^2-2m
That is, (m ^ 2-2m) + 1 = 2 √ 5
I'm rather lazy. You can calculate the answer by yourself, O (∩)_ O ha!
I looked at the question again, and it seems that I have misunderstood the meaning of the question, but it doesn't matter, the idea is almost the same
1. When m = 1 and M3
Ymin=f（1）=-1
Ymax=f(m)=m^2-2m
That is ((m ^ 2-2m) + 1) ^ 2 + (m-1) ^ 2 = (2 √ 5) ^ 2
It's OK this time~

Given that the maximum value of the function y = x ^ 2-2x + 3 in the interval [0, M] is 3 and the minimum value is 2, what is the value range of the real number m?

y=(x-1)^2+2
When x = 1, the minimum value is 2
Y = x ^ 2-2x + 3 = 3, x = 0 or 2, combined with the title and the above, m ∈ [0,2]