It is known that the domain of the function f x is D, and f x satisfies both the following condition

1) f(x)=-x is subtracted on r, so the condition is satisfied, and when x [-1,1], the set of values of f(x) is also [-1,1], and the condition is satisfied.

2) Closed function, because the function is single-incremented, so when x takes the minimum value of the defined domain, f(x) should also take the minimum value and be equal, so the definition domain x》-2 is taken.

When x=-2, f(x)=-2, i.e., k=-2;The maximum value of the interval makes k+ (x+2)=x, and k=x- (x+2) is an increasing function, so the range of k is (-2, positive infinity).

Define the domain x>=-2

If -2<=a<=x<=b

Then the k+ (a+2)<=y<=k+ (b+2) range is also [a,b].

then a=k+ (a+2).

b=k+√(b+2)

So x=k+ (x+2) has two different solutions that are greater than or equal to -2.

x-k)^2=x+2

x^2-(2k+1)x+k^2-2=0

a>=-2,b>-2

a+b=2k+1,ab=k^2-2

a+2>=0,b+2>0

So a+2+b+2>0

2k+1+4>0

k>-5/2

a+2)(b+2)>=0

ab+2(a+b)+4>=0

k^2-2+4k+2+4>=0

k+2)^2>=0

Establish. The discriminant formula is greater than 0

2k+1)^2-4(k^2-2)>0

4k+1+8>0

k>-9/4

So k>-9 4

(1) y=-x 3 is the subtraction function on [a,b], that is, the larger the x, the smaller the f(x); The smaller the x, the greater the f(x).

f(a)=-a^3=b， f(b)=-b^3=a

f(b)/f(a)=a/b=-b^3/-a^3

a/b=±1

again a 3=b, a=-1, b=1

The interval is [ 1,1].

2), f (x) = 3 4-1 x 2, x (0, let f (x) = 3 4-1 x 2 0, get x (2 3) 3

x (2 3) 3, f(x) is an increment function on ((2 3) 3.

Let f (x) = 3 4-1 x 2 0 give 0 x (2 3) 3

f(x) is a subtractive function on (0, (2 3) 3).

f(x) is not a monotonic function on (0

f(x) is not a closed function on (0,

3) It is easy to know that f(x)=k+ (x+2) is [ 2, and the increase function on is obtained by (x+2) 0, and f(x) k (*

Let f(x)=k (x+2) satisfy the interval of [a,b].

Then f(a)=a, f(b)=b, from which we know.

The two roots of the equation f(x)=x are a, b, and a≠b

Collation equation f(x) = x.

x^2-(2k+1)x+k^2-2=0

(2k+1)^2-4(k^2-2)=4k+9

Let 0, solve k -9 4

x1=[(2k+1)-√4k+9)]/2，x2=[(2k+1)+√4k+9)]/2

From (*) we get x1 k, and we get -9 4 k -2

From (x+2) 0 we get x+2 0, i.e., x1 -2, and we get k -9 4

In summary, the function y=k+ (x+2) is a closed function, and the value range of k is -9 4 k -2

Solution: (1), easy to obtain: y=-x 3 is the subtraction function on [a,b].

f(a)=-a^3=b

f(b)=-b^3=a

f(b)/f(a)=a/b=-b^3/-a^3

a/b=±1

again a 3=b, a=-1, b=1

The interval is [ 1,1].

2), f (x) = 3 4-1 x 2, x (0, let f (x) = 3 4-1 x 2 0, get x (2 3) 3

x (2 3) 3, f(x) is an increment function on ((2 3) 3.

Let f (x) = 3 4-1 x 2 0 give 0 x (2 3) 3

f(x) is a subtractive function on (0, (2 3) 3).

f(x) is not a monotonic function on (0

f(x) is not a closed function on (0,

3) It is easy to know that f(x)=k+ (x+2) is [ 2, and the increase function on is obtained by (x+2) 0, and f(x) k (*

Let f(x)=k (x+2) satisfy the interval of [a,b].

Then f(a)=a, f(b)=b, from which we know.

The two roots of the equation f(x)=x are a, b, and a≠b

Collation equation f(x) = x.

x^2-(2k+1)x+k^2-2=0

(2k+1)^2-4(k^2-2)=4k+9

Let 0, solve k -9 4

x1=[(2k+1)-√4k+9)]/2，x2=[(2k+1)+√4k+9)]/2

From (*) we get x1 k, and we get -9 4 k -2

From (x+2) 0 we get x+2 0, i.e., x1 -2, and we get k -9 4

In summary, the function y=k+ (x+2) is a closed function, and the value range of k is -9 4 k -2

(1) y=-x is a subtractive function, [-1,1],[2,2],[3,3](2)f(x)=3 4x+1 x

f(x) =7 4x (x greater than 0) is a subtractive function, and if it is a closed function, it has f(a) 7, 4a=b——4ab=7

f（b）=7/4b=a——4ab=7

That is, as long as 4ab=7 is satisfied, for example, [1,7 4] so f(x)=3 4x+1 x(x is greater than 0) is a closed function (3)y=k+ (x+2) is an increasing function when x>=-2, and must meet a=k+ (a+2).

b=k+ (b+2)(-2=-2, so k<=x)x 2+k2-2kx-x-2=0

x^2-(2k+1)x+k^2-2=0

There are two unequal roots"b^2-4ac>0"

4k^2+4k+1-4k^2+8>0

k>-9/4

x=（2k+1-√4k+9）/2>=-2

k<=-2

So k (-9 4, -2].

The value of k is a bit dizzy, you can verify it yourself.

Solution: (1), y=-x is the subtraction function f(a)=-a =b on [a,b].

f(b)=-b³=a

a/b=±1

again a = b, a = -1, b = 1

The interval is [ 1,1].

2), f (x) = 3 4-1 x , x (0, let f (x) = 3 4-1 x 0, get x (2 3) root number 3 x (2 3) root number 3, f(x) is ((2 3) root number 3, the increase function on the root number.

Let f (x) = 3 4-1 x 0 get 0 x (2 3) root number 3 f(x) is the subtraction function on (0, (2 3) root number 3 ) f(x) is not a monotonic function on (0,

f(x) is not a closed function on (0,

3) It is easy to know that f(x)=k+root(x+2) is the increment function on [ 2, and f(x) k

Let f(x)=k the interval in which the root number (x+2) satisfies the condition be [a,b], then f(a)=a,f(b)=b, and we can see from this.

The two roots of the equation f(x)=x are a, b, and a≠b

Collation equation f(x) = x.

x²-(2k+1)x+k²-2=0

Discriminant equation 0 (the equation has two unequal real roots), solve the small root of the equation k -9 4 (find the root formula) k (according to the function value range), solve the small root of the -9 4 k -2 equation (find the root formula) -2 (according to the defined domain), solve the value range of the three ks above k -9 4 and take the intersection to get -9 4 k -2 sum up, the function y=k + root number (x+2) is a closed function, and the value range of k is -9 4 k -2

(1) According to the definition of closed function (mainly monotonicity), only the values at the two ends of the interval are required, so we get: -a 3=a, -b 3=b or -a 3=b, -b 3=a, [-1,1].

2) It is easy to know its monotony, just verify whether there is an interval [a, b], and the method is the same as (1).

3) The judgment process is similar to (2), and the difference may be that there is a discussion of the value of k. (Do the math yourself!) ）

|10 days and 22 hours until the end of the issue |Questioner: I'm bloody.

Monotonically increasing or monotonically decreasing within d 2The range on the existential interval [a,b] is [a,b], and f(x) is called a closed function. 1.

Find the three-way of the closing function y=-x that satisfies the interval 2 of condition 2To determine whether f(x)=3 4x+1 x (x is greater than 0) is a closed function, explain reason 3Determine whether x+2 under the function y=k + root number is a closed function, if so.

Find the range of values of k.

(1) -1 x 1 ,2)f(x)= x+1 x what does it mean?

3) In order for the range of f(x) on [a,b] to be [a,b], then the value of x f(x) cannot be greater than x, otherwise the f(b) of b in the defined domain is larger than the original b, and the range will be infinitely large.

k+ (x+2) x, the solution is k x - x+2), because x -2, k -2

(1) Because y=-x **********》y'=-3x 2 is by definition< =0 is a subtraction function.

Let the interval be [a,b]===》-b 3=a,-a 3=b, and use the substitution method to obtain a=-1, b=1

So in the interval [a,b]=[-1,1].

2）f(x)=¾x+1/x===>f'(x)=3 4-1 x 2, there are more than 0 and there are intervals less than 0 (please ask for it yourself), the professional does not meet the condition is not a closed function.

3) y=k+ (x+2)====》y'=1 (2 (x+2)) because x+2>0==x>-2,y'=1 (2 (x+2))>=0, is an increasing function.

He has to be satisfied

Because it is an increment function, the minimum value is taken on the minimum value, and the maximum value is the same, a=k+ (a+2), and the square of k is shifted to get (a-k) 2=a+2

Open the parentheses, because to satisfy , that is, the equation must have a solution, the discriminant formula of the equation "0

Needless to say, the first question, right? Very simple.

The definition field of 2 x is greater than or equal to 0The incremental proof is simple, and the second condition translates to y=k + the root number x and y=x have two different solutions in the interval where x is greater than or equal to 0.

k + root number x = x

Let the root x be tk+t=t2=quadratic function, t2-t-k, and there are two different positive roots in the interval between t greater than 0, so the discriminant formula greater than 0 gives k greater than -1 4

The sum of the two roots is greater than 0, and the product of the two roots is greater than 0.

Therefore, in summary, the range of k is greater than -1 and 4 is less than 0.

The test point of this question is: the range of functions.

Topic: Calculation questions.

Analysis: From the meaning of the title, it can be seen that f(x) is a monotonic increasing function in d, which is a "good function", so that the constructor f(x)=12x can be transformed into finding loga(ax+k)=12x has two heterogeneous positive roots, and the range of k can be found.

Answer: Solution: Since the function f(x)=loga(ax+k), (a 0,a≠1) is an increasing function in its defined domain, then if the function y=f(x) is a "good function", the equation f(x)=12x must have two different real roots, loga(ax+k)=12x ax+k=ax2 ax-ax2+k=0, and the equation t2-t+k=0 has two different positive roots, k (0,14)

Therefore, D. was chosenComments: This question examines the value range of the function, and the difficulty lies in the constructor, which is transformed into two functions with different binary intersections, which is solved by equations, which is a difficult problem.

With such a detailed increase in efficiency, the landlord is decisive!

Upstairs I think it's a bit problematic, f(x) [a,b] doesn't mean that the range of f(x) is [a,b], but that the range of f(x) belongs to [a,b].

So I think:

Define the domain x>=-2, and -2<=a<=x<=b;

It is easy to see that the function y=k+ (x+2) is monotonically increasing.

That condition is met.

Looking at the first condition, x [a,b], due to monotonic increase, then k+ (a+2)<=y<=k+ (b+2) and the range belongs to [a,b], which needs to be satisfied.

k+ (a+2)>=a and k+ (b+2)<=b;

The solution gives a- (a+2)<=k<=b- (b+2) where -2<=a<=b.