If the function f x is a monotonic function over the defined domain D, and there is an interval a,b

1) The f(x) expression is ambiguous. Take f(x)=x 2 as an example.

Let x 2 = x (x 0).

then x=0 or x=1

And f(0)=0, f(1)=1

Obviously, when x [0,1], f(x) [0,1], so the equidomain interval of f(x) is [0,1].

2) Note that g(x) is a subtraction function when x (-0].

If m 0, then g(x) 0

Obviously, g(x) cannot be a positive function on x (-0).

If m<0, then g(x) and x-axis intersect at two different points.

Let x 2 + m = 0, then the intersection of the negative half axis of the x axis is x = - (m) is obviously on the interval x [-m),0] g(x) [m,0] let - (m) = m, i.e. m = -1 (note m<0) indicates the existence of m = -1, such that when x [-1,0] g(x) [1,0] indicates that g(x) is a positive function over the interval x (-0).

1) It's relatively simple, just ask for it yourself.

Because the function g(x)=x2+m is a subtraction function on (- 0), when x [a,b], g(a)=b g(b)=a i.e. a2+m=b, b2+m=a, subtract the two equations to obtain a2-b2=b-a, i.e., b=-(a+1), substituting a2+m=b to get a2+a+m+1=0, by a b 0, and b=-(a+1).

-1 a - 1 2, so the equation for a a2+a+m+1=0 has a real solution in the interval (-1, - 1 2), and h(a)=a2+a+m+1, then h(-1) 0, h(- 1 2) 0, and m (-1,- 3 4) is solved

Therefore, the value range of m is: (-1, - 3 4).

The domain of the function f(x)=[ (2 x)] k is: ( 2], and it is decreasing within this domain.

The range of the function f(x) on [a,b] is: [ b, a] then: f(a) = a, f(b) = b

Get: [ 2 a)] k= a, [ 2 b)] k= b i.e.: [ 2 a)] a=k, [ 2 b)] b=k So, a and b are the two roots of the equation: [ 2 x)] x=k.

That is: the equation [ ( ( 2 x)] x=k has two unequal real roots in ( 2 ].

Let : (2 x)=t, then: t [0, and: x=2 t, then:

t＋(2－t²)=k

t t (k 2) = 0 in the interval [0, there are two unequal real roots in it.

Let g(t)=t t t (k 2), then:

g(0) 0, get: k 2

=1 4(k 2)>0, get: k<9 4 synthetic, get: 2 k<9 4

f(x)=2k+ (x+4) defines the domain as [-4,+ Obviously, f(x) is a monotonic increasing function within its defined domain! meet the requirements of (1);

And then according to the requirements of (2):

4≤a0；2) axis of symmetry x (2k+1) 2>-4;

3)f(-4)≥0

From (1): 12k 2-4k-17<0, (1-2 13) from (3): k 0

In summary: k [0,(1+2 13) 6].

f(x)=2k+(x+4) is a closed function, according to the title, x>=-4 is an increasing function, "the existence [a,b] is a subset of d so that f(x) in the range of x [a,b] is a sub-interval of d, a>=-4, the range f(x)>=4, in this formula, (b+4), in this equation we find an equation about the quadratic function of b, as long as b has a solution, it is true, b 2-4ac>=0, this formula finds the range of k, Then find the intersection, which is the range of values of k.

Solution: by a2-b2=b-a

Decompose the ballast factor(a-b)(a+b)=b-a

Extracting the common factor of the late travel hole a-b

Yield: (a-b)(a+b-1)=0

by a≠b code withering a+b-1=0

i.e. b=-(a+1).

Because the function g(x)=x2

m is a positive function on (-0), so a b 0, nucleus called.

So when x [a,b] and the function decreases monotonically, then g(a)=b, g(b) takes pretend=a, i.e., a2

m=b，b2

m=a, subtract the two formulas to get a2

b2b-a, i.e., b=-(a+1), is substituted for a2

m=b gets a2

a+m+1=0, by a b 0, and b = - (a + 1) to eliminate the foci, a - (a + 1) 0, ie.

a＜?a?1a+1＞0

a＜?1a＞?1 solution -1 a -1

Hence the equation a2 with respect to a

a+m+1=0 in the interval (-1, -1

There is a real number solution in it, and it is denoted h(a)=a2

a+m+1, then h(-1) 0,h(-1

0, i.e. 1-1+m+1 0 and 1

m+1 0 solution gives m -1 and m -3

Namely? 1＜m＜?3

So choose A

Answer: g(x)=1 m-1 x,m>0 is (0,+ is a positive function on [a,b] (0,+ such that g(x) is in the range of [a,b] so: b>a>0

Because g(x)=1 m-1 x is a monotonically increasing function, then:g(a)=1 m-1 a=a

g(b)=1/m-1/b=b

Subtract the two formulas: b-a=1 a-1 b=(b-a) (ab)>0 so: ab=1

Therefore: 1 m=a+1 a>=2 (a*1 a)=2 if and only if a=1 a, i.e., a=1, the minimum value of 2 is obtained

Because: a=1, b=1 a=1

Not eligible for b>a

So: 1 m>2

So: 0 So: m can be (0, 1, 2).

Because the function g(x)=x2+m is a positive function on (- 0), when x [a, b], g(a)=b g(b)=a i.e. a2+m=b, b2+m=a, subtract the two equations to get a2-b2=b-a, that is, b=-(a+1), and substitute a2+m=b to get a2+a+m+1=0, by a b 0, and b=-(a+1).

-1 a -12, so the equation for a is a2+a+m+1=0 in the interval (-1, -12) with a real solution, and h(a)=a2+a+m+1, then h(-1) 0,h(-1

2) 0, and 0, solution m (-1,-34).

So the answer is: (-1, -34).

k=2.As we can see, f(x) decreases monotonically on the defined domain r, and there must be f(x) [b,-a], so only f(b)=-b,f(a)=-a.Substituting k=2