The function f x x 2 ax a 3 if x 2 and 2 f x 0 is constant, find the range a

f(x)=x 2+ax-a+3=(x+a2) -a 4-a+3, i.e., f(x) is a parabola with an open phase and an axis of symmetry x=-a2.

When -a 2 -2, that is, [-2,2] is on the right side of the parabolic axis of symmetry, a single increase, f(-2) is the minimum value, f(-2)=4-2a-a+3=7-3a, when 7-3a 0, f(x) 0 is constant on [-2,2], there is no solution.

When -a 2 2, i.e., [-2,2] is on the left side of the parabolic axis of symmetry, single subtraction, f(2) is the minimum value, f(2)=4+2a-a+3=a+7, when a+7 0, f(x) 0 is constant on [-2,2], and the solution set is -7 a -4

When -2<-a 2<2, the minimum value is obtained at the parabolic vertex, f(a 2)=-a 4-a+3=-(a 2+1) +4, when -(a 2+1) +4 0, f(x) 0 is constant on [-2,2], and the range of the solution set is -7 a 2

You can also find a derivative to do it, and I personally think that the combination of numbers and shapes is simpler

Solution: The highest term coefficient of f(x) is not 0, so f(x) is a quadratic function.

If the discriminant equation of the equation f(x)=0 δ 0, then x r,f(x) 0 is constant.

At this point, a 4 (3 a) 0, i.e. a +4a 12 0, i.e. (a + 6) (a 2) 0

a∈【﹣6，2】

If the discriminant equation of the equation f(x)=0 δ 0, then f(x)=0 has two x1,x2 and x1= ( a δx2= ( a+ δ then f(x) 0 when x (x1, x2).

Therefore, (x1,x2) has no intersection with [ 2,2].

At this point x1 2, or x2 2

Substituting δ=a +4a 12 into it, it can be obtained after a series of lengthy calculations such as shifting, squarering, shifting, and merging similar terms.

a∈【﹣7，﹣6）∪（2，7/3】

In summary, the value range of a is [ 7, 7 3].

If you need a more detailed process, I can help you write about it.

The function f(x)=x 2+ax+3 is known, when x r, f(x) a is constant, f(x)=x 2+ax+3=(x+a 2) 2-a 2 4+3, because (x+a 2) 2 0, so f(x) a 2 4+3;

It is known that when x r, f(x) a is always trapped early, so -a 2 4+3 > cover finch = a, a 2+4a-12

f(x)=x 2+ax+3-a=(x+a know-file2) 2+3-a-a 2 4

x [-2,2], f(x) 0 is established.

A 2 2, A -4.

f(2)=4+2a+3-a=7+a 0,a -7a -7a 2 -2,a 4,f(-2)=4-2a+3-a=7-3a 0,a 7 3

Hitching = a 2-4(3-a) = a 2 + 4a-12 = (a + 6) (a-2) 0

6 A 2So, the value range of a: [-7,2].

The function f(x)=x 2+ax+3 axis of symmetry x=-a 2, according to the meaning of the problem When -a 2 -2, when x [-2,2], the minimum value of f(x) a is: f(-2)=4-2a+3 a, there is no solution.

When -2 -a 2 2 and when x [-2,2], the minimum value of f(x) a i.e.: f(-a 2) a, we get -4 a 2

When -a 2 2 and when x [-2,2], the minimum value of f(x) a is: f(2)=4+2a+3 a, resulting in -7 a -4

In summary: -7 a 2

x [-2,2], the maximum value of x is 2, and the minimum value is -2Then:

1. When x=-2, substitute f(x) to get 4-2a+3-a=7-3a, and require f(x) 2 to become constant immediately: 7-3a 2, get a 5 3.

2. When x=2, substituting f(x) to get 4+2a+3-a=7+a, requiring f(x) 2 to become constant immediately: 7+a 2, get a -5.

Therefore, the value range of a is: -5 a 5 3

Friend, you have a question with your answer, the following is my answer, I hope it will help you!

f(x)=x^2+ax+3-a

x+a 2) 2 +3-a-a 2 4 vertex coordinates [-a 2,(3-a-a 2 4)] because when x [-2,2], f 0 is constant.

Discussion 1, When -a 2<=-2 (a>=4)f minimum = f(-2)=4-2a+3-a>=0 is calculated as a<=7 3 contradiction, round off.

2, when -2<-a 2<2 (-4=0 is -6<=a<=2 combined to get -4=2 (a<=-4)f minimum = f(2)=4+2a+3-a>=0 is calculated to be a>=-7 merged to get -7 a -4

f(x)=x²+2x-3a

f(x)+2a≥0

i.e.: x +2x-a 0

A x +2x against the x [-2,2] constant holds, the constant bend is small is that the left a is smaller than the right side without the bottom of the side, x +2x=(x+1) -1 minimum is -1, a -1