Knowing f x x 2 m x 2, x 1, 2, find the minimum value of the function f x

You should have learned derivatives! It's easiest to use the derivative, find the derivative of f(x), and get 2x+m. In the case of x [-1,2], the range of the derivative is [m-2,m+4], so the positive and negative of the derivative cannot be judged, and it needs to be discussed.

1. When m<-4, the derivative is negative, and the original function is single subtracted in the interval of x, so the minimum value of the function is 2m+6 when x=22. When m [-4,2], the derivative is zero at x=-m 2, and by analysis, the extreme value is the minimum value, which is the minimum value of the original function in the interval, so the minimum value is 2-m 2 4;3. When m>2, the derivative is positive, the original function is single-increased, and the function obtains a minimum value of x=-1, which is 3-m.

If you use the properties of the binary function to do it, you can also discuss it, first formulate the original function, and then see which interval its symmetry axis falls in, and use the image of the binary function to intuitively find out the minimum value. But to paraphrase our teacher, this approach is not authoritative...

The axis of symmetry of the function is x=-m 2

When -m 2<=-1(m>=2) f(x)min=f(-1)=3-m

When -1<-m 2<2 (-4=2(m<=-4) f(x)min=f(2)=6+2m

Well, I think it's necessary to bring x=-1 2 in according to the value of m (m=0, greater than 0 or less than 0) to get 3+m and 6-2m

Then just do it according to the value of m.

It's just my thoughts)

Function f(x)=x2

2ax+2=（x-a）2

2-a2When a -1, the minimum value of f(x) on [-1,1] is Qizheng f(-1)=2a+3;Quietly repentant.

When -1 a 1, the minimum value of f(x) on [-1,1] is f(a)=2-a2

When a 1, the minimum value of f(x) on [-1,1] is f(1)=3-2a limb.

You draw |2x+1|,|x-2|The minimum value of f(x)=max is the intersection of the two brightening functions, that is, when it is positive 2x+1=2-x, that is, when x=1 3, then you can now calculate the answer.

2x+1|=|x-2|=5/3

First, find the axis of symmetry, which is a

Discuss the scope of a talk and make peace.

First, when a is less than or equal to -1, then fx increases monotonically between [-1,1], so when x is -1, there is a minimum value of 3 2a

Second, when a is between (-1,1), the function takes the minimum value when x is a, which is -a 2 2

Third, when a is greater than or equal to 1, fx decreases monotonically in the defined domain, so when x is 1, the minimum value of the closed argument is 3-2a

In summary, the minimum value of fx is ......, write a segment letter with a state mark.

Answer: When a is [1,2].

f(a) is the minimum.

f（a）=-a^2+1

When a is greater than 2.

f(2) is the smallest.

f（2）=5-4a

When a is less than 1.

f(1) is the smallest.

f（1）=2-2a

Or you can do it with a derivative, and also discuss the value of a.

f(x)=(x-a)^2-a^2

fmin=f(a)=-a^2

If it is judged that the number of excavated beams is in the interval, that is, -1=1, fmin=f(1)=1-2a

If a is on the left side of the interval, i.e., a<-1, fmin=f(-1)=1+2a