The concept of a math function value range

1）.y=x²-4x+6,x∈[1，5）

y=(x-2)²+2

y min = 2 y max = (5-2) + 2 = 11

So the range is [2,11].

2）.y=2-root (-x +4x).

y=2-√[4-(x-2)²]

Y max = 2 y min = 2-2 = 0

So the range is [0,2].

3）.y=1/（1+x²）

y max = 1 When x approaches infinity, y approaches 0

So the range is (0,1].

4）.y=x + root (1-2x).

Let 1-2x=t

then x=(1-t) 2

So y=(1-t) 2+t=(1 2)(-t +2t+1)(1 2)[-t-1) +2].

Y max = 1t approaching infinity and y approaching infinity.

So the range is (- 1].

The range of x-1 is -1,1

f[x] is defined by [0,2].

Ask me if it's one topic or two.

If it's a question.

I have to give up 0y is 1+x/1, and y is [,is infinite) I am a senior in high school and have just graduated. You have to trust me.

The 1st floor is very good, but the 2nd question is wrong.

Let t=-x 2+4x

Since t needs to be squared, t>=0, and t=-(x-2) 2+4<=4So, 0<=t<=4,0<=2- t<=2, that is, the range of the function is: [0,2].

The vertex x=2 is in the interval, and when x=2, the function has a minimum value ymin=2 so that x=1 y=1-4+6=3

Let x=5 y=25-20+6=11

The range of the function is [2,11).

The square term is constant and non-negative, (x +4x) 0

y=2-√(x²+4x)≤2-0=2

The range of the function is (- 2].

The square term is constant and non-negative, x 0 1+x 1

0<1/(1+x²)≤1

The range of the function is (0,1].

-1 2)(1-2x)+1-2x)+1 2=(-1 2)[ 1-2x)-1] +1 (1-2x)Constant non-negative, when (1-2x)=1, i.e., x=0, the function has a maximum value ymax=1

The value range of the function is (- 1].

（1）.y=x²-4x+6,x∈[1，5）y=(x-2)²+2

y min=2 ymax=(5-2) +2=11, so the range is [2, 11].

2）.y=2-root (-x +4x).

y=2-√[4-(x-2)²]

Y max = 2 y min = 2-2 = 0

So the range is [0,2].

3）.y=1/（1+x²）

y max = 1 When x approaches infinity, y approaches 0

So the range is (0,1].

4）.y=x + root (1-2x).

Let 1-2x=t then x=(1-t) 2, so y=(1-t) 2+t=(1 2)(-t +2t+1)=(1 2)[-t-1) +2].

ymax=1 t tends to infinity, y tends to -infinity, so the range is (- 1].

Let t=-x 2+4x

Since t needs to be squared, t>=0, and t=-(x-2) 2+4<=4

So, 0<=t<=4,0<=2- t<=2, i.e., the range of the function is: [0,2].

1 The range of all x values has been implied to us, and it is clear that the first x cannot be equal to 3, and the second x is the first of any real numbers: y=[2(x-3)+7] (x-3)=2+7 (x-3) Since 7 (x-3) is not equal to 0, the range is (negative infinity-2) and (2-positive infinity), i.e., it cannot be 2

Second Path: Same as First Path: y=(x 2+1-2) (x 2+1)=1-2 (x 2+1).

Since 2 (x 2+1) is greater than 0, the range is (minus infinity-1) i.e. less than 1

Is there a range of values for x? Otherwise, you can't do it.

Solution: (1) The domain is defined as r, so kx 2+4kx+4 0 is constant, and when k=0, 4 0 is true.

When k ≠ 0, k 0 and =(4k) 2-4k 4 0 are required to solve: 0 k 1, so the range of k is [0,1].

2) Define the domain as r, then kx 2+4kx+4≠0 is constant, and when k=0, 4≠0, is true.

When k ≠ 0, 0, i.e. (4k) 2-4k 4 0, the solution is: 0 k 1 so the range of k is [0,1).

yx^2+y=ax+b

yx^2-ax+(y-b)=0

If there is a solution to this equation about x, the discriminant formula is not less than 0

So a 2-4y(y-b)>=0

4y^2-4by-a^2<=0

The range is [-1,4], i.e. the set of solutions of this inequality is -1<=y<=4, so the corresponding equation 4y2-4by-a, the roots of 2=0 are -1 and 4, so by Veda's theorem.

1+4=-(4b)/4=b

1*4=-a^2/4

a=4, b=3 or a=-4, b=3

I hope mine is helpful to you, o(o!

1.There seems to be a clerical error in the question.

5/[2(x-1)^2 + 1]

The range of the denominator is [1, positive infinity) When x=1, the range of 1y is (0,5].

3.This question is directly divided into four quadrants (+ means that the value of the function is positive, and - means negative)1)sin+,cos+,tan+,cot+,y=42)sin+,cos-,tan-,cot-,y=-23)sin-,cos-,tan+,cot+,y=04)sin-,cos+,tan-,cot-,y=-2Of course, 0, 2, ,3 2 involve the meaninglessness of the function, so we will not discuss it.

As a reminder, the function y=|x|x is a symbolic function, when x>0, y=1, x<0, y=-1, x=0 is meaningless.

4.First of all, to ensure that the formula under the root number is not less than 0, then the value range of x [0,4]x 2+4x is [0,4], and the open root number is [0,2] and then subtract with 2 to get the final value range of [0,2].

1 + 1/(2^x+1)]

The range of 2 x+1 is (1, positive infinity).

The range of 1 (2 x+1) is (0,1).

The value range of y is (1,2).

infinite, since the analytic formula has been given, and the range is also given.

So, you just need to determine the definition domain.

If the value range is [0,1].

y=-x^2+1

Obviously, x [-1 0], [0,1] are satisfied.

Obviously, [-1,0],[1,1 2][-1,1 6][-1,1 7] are all satisfied.

So y=-x 2+1,x [-1,0],[1,1 2][-1,1 6][-1,1 7].

Isn't that the function you've already decided??? How can you still ask how many such functions there are.

You didn't write the title wrong.

The function has been determined, and the number of defined fields can be determined according to the value range.

Depending on the defined domain, different functions can be derived.