Given the function f (x) = x ^ 2 + ax + 3-a. when x belongs to [- 2,2], f (x) is greater than or equal to 2, find the range of A Why can't we use the method of finding the most value

Given the function f (x) = x ^ 2 + ax + 3-a. when x belongs to [- 2,2], f (x) is greater than or equal to 2, find the range of A Why can't we use the method of finding the most value

We can use the method of finding the most value. We should classify it like this
f(x)=x²+ax+3-a
Axis of symmetry x = - A / 2
(1) If - A / 24
f(x)min=f(-2)=7-3a≥2 a≤ 5/3
To sum up, there is no intersection, and this situation does not exist
(2) If - 2 ≤ - A / 2

Given the function f (x) = ax ^ 2 + AX-1, if f (x) is less than or equal to 0, the value range of a is obtained

First of all, the quadratic function must be open downward, otherwise there must be a part greater than 0
Second, if the function is always less than zero, it means that the function has no intersection with the x-axis (opening downward), that is, the function = 0 has no real root
Thirdly, if the function is equal to 0, then there is only one intersection point between the function and the X axis, which is the extreme value and the maximum value of the function
According to these three things, we can list the equation, the judgment formula of the direction of the opening, the judgment formula of the number of roots. If you forget, you can look it up in the book
According to the derivative theorem, we can know that the extremum is the value of zero first derivative of the function, and the extremum of the quadratic function with downward opening is its maximum value. If the maximum value of the function is on the x-axis, then there is only one intersection point between the function and X, that is, the maximum value itself. If the maximum value is below the x-axis, then there is no intersection point between the secondary function and x-axis, that is, there is no real root If f '(x) = 2aX + A, the extremum is f' (x) = 2aX + a = 0;
Then the X coordinate of the extreme value is x = - 1 / 2
Then f (- 1 / 2) = A / 4-A / 2-1 = - A / 4-1
So f (- 1 / 2)

If f (x) is equal to AX + INX and f (x) is less than or equal to 0, the range of a is obtained

It's not right. It's similar to the tangent to y = LNX through the origin. The slope of the tangent is 1 / E, so a
A satisfying condition