When x ∈ [0,2], the function f (x) = ax ^ 2 + 4 (A-1) x-3 gets the maximum value when x = 2, then the value range of A

When x ∈ [0,2], the function f (x) = ax ^ 2 + 4 (A-1) x-3 gets the maximum value when x = 2, then the value range of A

f'(x)=2ax+4(a-1)
Because f (x) has a maximum at x = 2
So f (x) increases monotonically in [0,2], so the minimum value of F '(x) is greater than 0
If a > 0, f (0) = 4 (A-1) > 0, a > 1
If A0 a > 1 / 2
In conclusion, a > 1

12. When x ∈ [0,2], the function f (x) = ax ^ 2 + 4 (A-1) x-3 obtains the maximum value when x = 2, then the value range of a is

First find the derivative of F (x) = 2aX + 4a-4
Let f (x) derivative = 2aX + 4a-4 = 0
We get x = (2 / a) - 2
The results show that f (2) > F {(2 / a) - 2}
f（2）>f（0）
Finally, find out the answer

When x ∈ [0,2], the function f (x) = ax ^ 2 + 4 (A-1) - 3 gets the maximum value when x = 2, then the value range of a is

Substituting x = 2 into the function, we get 12a-11
Because x = 2 is the maximum (so take any number, as long as it is within the interval, it is less than x = 2)
Then f (x) = - 33 / 2 when x = 0

It is known that f (x) is an even function and an increasing function on [0, + ∝). If f (AX + 1) ≤ f (X-2) is constant on X ε [&# 189;, 1], the value range of real number a is obtained

It can be obtained that | ax + 1 | X-2|
First of all, we can get that if X-2 ＜ 0, then | ax + 1 | ≤ 2-x, we can get (a + 1) * x ≤ 3 and - 3 ≤ (A-1) * x, let x = 1 and x = 1=
&#It's OK to solve the inequality