Given the function f (x) = a ^ x, G (x) = (a ^ 2x) + m, where m > 0, a > 0 and a is not equal to 1, when x belongs to [- 1,1], the sum of the maximum and minimum value of y = f (x) is 5 / 2 (1) Find the value of A; (2) If a > 1, note the function H (x) = g (x) - 2mf (x), and find the minimum value H (m) of H (x) when a belongs to [0,1]; (3) If a > 1 and the inequality | [f (x) - Mg (x)] / F (x) | is less than or equal to 1, then M is obtained when x belongs to [0,1]

Given the function f (x) = a ^ x, G (x) = (a ^ 2x) + m, where m > 0, a > 0 and a is not equal to 1, when x belongs to [- 1,1], the sum of the maximum and minimum value of y = f (x) is 5 / 2 (1) Find the value of A; (2) If a > 1, note the function H (x) = g (x) - 2mf (x), and find the minimum value H (m) of H (x) when a belongs to [0,1]; (3) If a > 1 and the inequality | [f (x) - Mg (x)] / F (x) | is less than or equal to 1, then M is obtained when x belongs to [0,1]

1. If a > 0, f (x) increases with the increase of X, then a ^ (- 1) + A ^ 1 = 5 / 2, i.e. 1 / A + a = 2.5
A = 2 or a = 0.5
2. If a > 1, then a = 2, H (x) = 2 ^ 2x + m-2m * 2 ^ x = (2 ^ x) ^ 2-2m * (2 ^ x) + M = (2 ^ x-m) ^ 2 + (M-M ^ 2)
The minimum value is M-M ^ 2 = 0
3. If a > 1, then a = 2 and y = 2 ^ x, then the inequality holds when y belongs to [1,2]
The inequality is - 1

Let a = 1 + 2x ^ 4, B = 2x ^ 3 + x ^ 2, X belongs to R, and X is not equal to 1, then the size relation of a and B is?
(2x ^ 3-x-1) / (x-1) how is this equal to (2x ^ 2 + 2x + 1) (x-1)?

A-B = 2x ^ 4-2x ^ 3-x ^ 2 + 1 = 2x ^ 3 (x-1) - (x + 1) (x-1) = (x-1) (2x ^ 3-x-1) = (x-1) (x ^ 3-x + x ^ 3-1) = (x-1) [x (x + 1) (x-1) + (x-1) (x ^ 2 + X + 1)] = (x-1) ^ 2 (x ^ 2 + X + 1) = (x-1) ^ 2 (2x ^ 2 + 2x + 1) x is not equal to 1, (x-1) ^ 2 > 02x ^ 2 + 2x + 1 = 2 (x ^ 2 + x) + 1 = 2 (x ^ 2 + X + X + 1 /

G (x) = 1-2x, f [g (x)] = 1-x & # 178; (x ≠ 0), f (1 / 2) is equal to,
When f [g (x)] = f (1-2x), why not replace X in 1-x & # 178 / X & # 178; with 1-2x

No, X in G (x) is also X in F [g (x)], the two are the same, but we can't change X in F [g (x)] to 1-2x! I hope you can understand my answer

Given that G (x) = 1-2x, f [g (x)] = 1-x & # 178; (x ≠ 0), what is f (1 / 2) equal to?

From G (x) = 1-2x = 1 / 2, we can get x = 1 / 4,
So f (1 / 2) = f (g (1 / 4)), we can substitute 1 / 4