The range of functions fx x 2 x,x 1,3?

Derivation f'=1-2 x 2 so that it is greater than 0, and the solution is x> root number 2or x<-root number 2On this interval the function increases monotonically.

So the minimum value mn=f(root2)=2, f(3)=11 3Maximum=11 3

-3(x+a/3)^2+a^2/3+1

1. When x=-a 3<-1 3, i.e., a>1, x -1 3, 1 3 is monotonically decreasing.

So f(x)max=f(-1 3)=2a 3+2 3

f（x）min=f（1/3）=2/3-2a/3

That is, the value range is 2 3-2a 3, 2a 3+2 3

2. When x=-a 3>1 3, i.e., a<-1, x -1 3, 1 3 is monotonically increasing.

So f(x)max=f(1 3)=2 3-2a 3

f（x）min=f（-1/3）=2/3+2a/3

That is, the value range is 2 3-2a 3, 2a 3+2 3

3. When x=-a 3 -1 3,0, that is, a 0,1, so when x=-a 3, f(x)max=f(-a 3)=a 2 3+1

f（x）min=f（1/3）=2/3-2a/3

That is, the range is 2 3-2a 3, a 2 3+1

4. When x=-a 3 0,1 3, that is, a -1,0, so when x=-a 3, f(x)max=f(-a 3)=a 2 3+1

f（x）min=f（-1/3）=2/3+2a/3

That is, the range is 2 3-2a 3, a 2 3+1

In addition, there are products on the stationmaster's group **, cheap ***.

Hello function is y=(x-2) (x+1).

If. then y=(x-2) (x+1).

x+1-3）/（x+1）

1-3/（x+1）

3 (x+1)≠0

i.e. -3 (x+1)≠0

i.e. 1-3 (x+1)≠1

That is, y≠1, so the value key of the book destruction function is {y y≠0}

f(x)=x+ (2-x)2-x 0, which defines the stool grip field x 2f'ruler Qing (x) = 1-1 = x 7 4, f'(x) 0, f(x) monotonically increased; 7 at 4 x 2, f'(x) 0, f(x) monotonically minus x=7 4, the maximum ymax=7 4+ (2-7 4)=9 4 range (-infinity, 9 4).

Because of the definition domain of 2x.

is r, and the range is (0,+ macro talk Lee).

Therefore, the domain of f(x) is r, because the denominator cannot be 0 and is f(x)=1-2 (2 x+1), and the value of 2 x+1 is (1,+ we know that the range of 2 (2 x+1) is (0,2), thus.

The range of f(x) is (-1,1).

Summary. Hello, I am glad to answer for you: the range that belongs to [fx=x 2-2ax+1 =(x-a) 2+(1-a 2) is open upward, the axis of symmetry x=a, x=a has a minimum value of 1-a 2xe[<1, the axis of symmetry is in the interval [left, the minimum value of monotonically increasing f(1)=1-2a+1=2(1-a) maximum value f(3)=9-6a+1=2(5-3a) range [2(1-a),2(5-3a)] 1sa<2, fx=x -3 +x +3.

Hello, I'm glad to answer for you: the value range of [fx=x 2-2ax+1 =(x-a) 2+(1-a 2) opening up, the axis of symmetry x=a, x=a when there is a minimum value 1-a 2xe[<1 when Qi slows, the axis of symmetry is in the interval [left, the monotonic increase minimum value f(1)=1-2a+1=2(1-a) maximum value f(3)=9-6a+1=2(5-3a) value range [2(1-a),2(5-3a)] Kiwang 1sa<2, related information: The method of evaluating the range is as follows:

Direct method: starting from the range of independent variables, the value range is deduced; with method, find the maximum value and the minimum value; Observation method: For some simple functions, the value range of the function can be directly obtained according to the defined domain and correspondence.

1.Direct method: Brightness starts from the range of independent variables and deduces the value range.

2.Observation method: For some relatively simple functions, according to the definition range and correspondence, the auspicious key can be directly leaked to the value range of the function.

3.Matching method: (or the minimum value method) to find the maximum value and the minimum value, then the value range will come out.

Example: y=x2+2x+3xe[-1,2] formulas first, and obtains y=(x+1)2+1'.

ymin=（-1+1）2+2=2ymax=（2+1）2+2=11

Since the domain of 2 x is r, the range is (0,+ so the domain of f(x) is r, because the denominator cannot be 0 and is defined by f(x)=1-2 (2 x+1), and the range of 2 x+1 is (1,+ we know that the range of 2 (2 x+1) is (0,2), thus.

The range of f(x) is (-1,1).

Defining a domain is not equal to -1 2

The range is f(x) and is not equal to 1

This problem requires the absolute value to be removed to determine its range. Removing absolute values needs to be discussed in a categorical manner.

When x 2, f(x)=x 2+x-3 is a univariate quadratic function, has a minimum value at the axis of symmetry x=-b 2a=-1 2, decreases monotonically on (- 1 2), and increases monotonically at (-1 2,+ monotonically. And because x 2, the minimum value f(2)=3 is taken when x=2, so f(x) 3

When x<2, f(x)=x 2-x+1 is a unary quadratic function, has a minimum value at the axis of symmetry x=-b 2a=1 2, decreases monotonically on (- 1 2), and increases monotonically at (1 2,+ monotonically. And because x < 2, the minimum value f(1 2)=3 4 is taken when x = 1 2, so f(x) 3 4