f x log takes 1 2 as the base x 2 ax a 13 on ,2 as the increment function, find the range of the

f(x)=log, with 1 2 as the base x 2-ax+a+13 as the increment function on (2), let g(x)=x 2-ax+a+13

Then according to the question g(x)=x 2-ax+a+13 is a subtraction function on (2), then a 2>=2 a>=4

f(x)=-log2(x 2-ax+a+13), which is an increasing function on (2), so x 2-ax+a+13 is a decreasing function on (2), so the axis of symmetry is greater than or equal to 2, i.e., -(-a) 2>=2, so a>=4

And x 2-ax+a+13 must be greater than 0 on (2).

Therefore, x=2 is required to be greater than or equal to 0, 4-2a+a+13>=0,a<=17, so a belongs to [4,17].

Let g(x)=x 2-ax+a+13, obviously g(x) must be greater than 0 because the logarithmic function with a base of 1 2 is a subtraction function.

Therefore, this problem is equivalent to finding the range of the value of a when g(x) is a subtraction function on (2).

Since g(x) is open upward, as long as the axis of symmetry is greater than or equal to 2.

i.e. a 2 > = 2

A >= 4

To satisfy g(x)>0 on (2), g(x)>0 is sufficient when x=2.

i.e. 4-2a+a+13>0

Got a < 17

Take the intersection of the two ranges, and get it.

The value range of a is [4,17].

f(x) is an additive function on [2,4].

Then: y=-ax +2x+3 is a subtraction function on [2,4], and y(min)>0, i.e.: y(4)>0;

1) When a=0, y=2x+3, obviously not satisfied, give up;

2) When a<0, y is a parabola with an opening upward, the axis of symmetry is x=1 a, 1 a<0, and y is increasing on [2,4] and rounded;

3) When A>0, Y is a parabola with an opening downward, and the axis of symmetry is x=1 a, and it should be decreasing on [2,4], then: 1 A 4, get: 00, that is: -16A+11>0, get: A<11 16 So, 0 in summary, the value range of the real number A is: 0 I wish you a a happiness! I hope it can help you, if you don't understand, please ask, I wish you progress! o(∩_o

Is 1 2 the bottom?

If yes:

First of all, it is agreed that the logarithm of x will be based on m, which will be denoted as: log[m]a solution:

f(x)=log[1 2](-ax +2x+3)f(x)=[ln(-ax +2x+3)] [ln(1 2)]f(x)=[ln(-ax +2x+3)] (-ln2)f(x)=(-1 ln2)ln(-ax +2x+3)f'(x)=(-1 ln2)(-2ax+2) (-ax +2x+3)f'(x)=(2 ln2)(ax-1) (-ax +2x+3) because: f(x) is an increment function, so: f'(x) 0, i.e.: (2 ln2)(a-1) (-ax +2x+3) 0

Defined by logarithms, we know: -ax +2x+3 0.........1) So: AX-1 0.........2) From (1):

a＜(2x+3)/x²……3) From (2): a 1 x .........4) Because :x [2,4], so:

From (3): a 11 16

From (4): a 1 2

i.e.: a [1 2, 11 16].

The whole is subtracted from a<0 or a>=1 2

Yes, if the function is an increasing function in that range, then make sure that ax square -2x+4 is a subtraction function in that range. According to this range, we can know that the function image opening is upward, a is greater than 0, and the axis of symmetry is greater than or equal to 3, but there should be another problem, for the logarithmic function, the formula of the ax square should be greater than zero,,, I don't understand, see for yourself.

The function f(x)=log<1 2>(x -ax-a), is an increasing function on (- 1 2), because log<1 2>u is a decreasing function, so u=x 2-ax-a>0, on (- 1 2) is a subtraction function, the axis of symmetry of the parabola u=x 2-ax-a x=a 2>=-1 2, and u(-1 2)=1 4-a 2>=0, a>=-1 and a<=1 2,-1<=a<=1 2, is what is sought.

First, the true number of the logarithmic function is greater than 0, i.e., x 2-ax-a>0

Secondly, if 1 2 is the base, then the function is a subtraction function in the effective range, so the function is reduced to f(x)=--log2(x 2-ax-a); , f(x) is a subtractive function within the valid range.

To sum up, (-1 2) must be in the range of x 2-ax-a>0 to solve x, and the value range of a can be obtained by comparing the two ranges.

According to the analysis of the topic, it is as follows:

The opening is upwards and is greater than 0 in the [- 3] range;

2.Since log(1 2)x is a subtraction function (the base is less than 1), g(x) is also a subtraction function in the [- 3] interval;

In the [- 3] interval is not equal to 0;

Here's how to solve it: g(x) is evergreen to 0 in the field of real numbers

b^2-4ac=4-16a≤0

Solution A 1 4

and 3 -b 2a = 1 a

Solution: a 1 3 (because a>0).

So 1 4 a 1 3

g(x) has two intersections with the x-axis, g(3) 0

0, solution a<1 4

Substituting x=3 into g(x)>0 gives 9a-6+4>0 to get a>2 9

Therefore, in 2 9 sum, the value range of a is (2 9, 1 3].

ax^2-2x+4>0

The base number 1 2<1, f(x) is a subtractive function that decreases as ax 2-2x+4 increases. The function is incrementing in (-3), then in (-3), ax 2-2x+4 is monotonically decreasing. a>0

The axis of symmetry x=1 a>=3 a<=1 3

x=3 brings in 9a-6+4>0 a>2 9

The value of a can be (2 9, 1 3).

f(x)=log1 2(ax 2-2x+4) log1 2 x is a subtraction function in the defined domain In order for the function to be an increasing function in (— 3), then x must also be a subtraction function in (— 3) ax 2-2x+4 is a subtraction function in (— 3) Then the axis of symmetry 2 2a must be less than or equal to 3 Then the range of a can be calculated.

ax 2-2x+4 (axis of symmetry x=2 a) (—3) is the subtraction function, both 2 a<=3, in =4-16a<0, a>=2 3

2-ax>0 makes sense.

x belongs to [1,2[.]

a<2 x, and 2 x belongs to [1,2].

So a belongs to (0,1).

That is, when x belongs to [1,2], 2-ax>0

a is the base of the logarithm, then a > 0

So u=2-ax is a subtraction function of 2-2ax>0 when x=2, i.e., 2-4a>0 a<1 2

The value range of the real number a is 0