y log1 3 X2 6X 5 .

The defined field is -x 2+6x-5>0

x^2-6x+5<0

x-1)(x-5)<0

1. The outer function is a subtraction function.

Therefore, the increase interval of the inner function is the subtract interval of the composite function.

Knowing from the image.

1, 5) is the image "0 time.

x-3)^2-4<0

The axis of symmetry is x=3

So the subtracted interval of the composite function is (1,3).

f(max)=+infinity f(min)=f(3)=log(1 3)(4) The compound function increment interval is (3,5).

f(min)=log(1/3)(4)

f(max)=+infinity.

So the range is [log(1 3)(4),+infinite).

y=log(1/3)(-x²+6x-5)=-log3(-x²+6x-5)

Domain. 1,5) value range.

2log3(2),+

Monotically increasing interval (3,5).

Monotically decreasing interval (1,3).

Solution: The domain is defined by the function of -x 2+6x-5>0 as follows: (1,5) Let t(x)=-x 2+6x-5, then y=log(1 3)tWhen 1 x1 x2 3, t(1)y2>log(1 3)4;

When 3 x1 x2 5, t(3) t1 t2 t(5), i.e. 4 t1 t2 0, log(1 3)4 y1 y2

Therefore, the monotonic increase interval of the function is [3,5] and the monotonically decreasing interval is (1,3).

The range is log(1, 3)4,

-x +6x-5) 0 x 6x 5 0 1 x 5 The axis of symmetry is x 3

The logarithmic function with base is decreasing.

The increase interval of the true number 1,3] is the subtraction interval of the original function.

The decreasing interval of the true number [3,5 is the increasing interval of the original function.

The subtract interval of the function is 1,3], and the increase interval of the function is [3,5

This is a land scraping composite function.

Let u=7+6x-x 2(-1,7), y=log4u, where chunjian is log4u in (0,+ is the increment function, and the definition domain is in (-1,7), so log4u in (0,7) is the increment function.

And because u=7+6x-x 2, the axis of symmetry is x=3, so at (-1,3) is the increasing function and at (3,7) is the decreasing function.

Therefore, according to the same increase and difference subtraction, y=log4 (7+6x-x 2) is an increase function at (-1,3) and a subtraction function at (3,7).

Here's how, please refer to:

If it helps,.

Solution: Function y log1 2 (x 2-3x 5), let u x 2-3x 5

then u (x-3 2) 2+11 4 11 4 so the domain of the original function is x r;

When x 3 2, u is the subtraction function, then y log1 2 (x 2-3x 5) is the increasing function;

When x 3 2, u is the increment function, then y log1 2 (x 2-3x 5) is the subtraction function.

First, consider defining the domain, i.e., x 2-4x+5 0 (x-5) times (x+1) 0 to get x>5 or x

Let x 2-5x+6=u

then y=log2(u).

The base number is 2, so when you increment, y increments; When you decrement, y decreases.

u=x 2-5x+6, is a parabola with an opening upward, the axis of symmetry shouting as a straight line x=5 2, when x=5 2, u increments.

Because u>0 (to be an integer), i.e., x 2-5x+6> nuclear banquet 0, the solution is x3, so the increase interval of the function that matches Zheng Shi u is x3

So the function y=log2(x 2-5x+6) monotonically increases the interval (- 2); The monotonic reduction interval is (3, ).

This is a composite function.

Let u=7+6x-x 2(-1,7), y=log4u, where log4u is at (0,+ is the number of trembling finches of the increasing function, and the definition domain is in (-1,7), so log4u is the increasing function at (0,7).

And because u=7+6x-x 2, the axis of symmetry is x=3, so at (-1,3) is the increasing function, and at (3,7) is the subtraction function.

Therefore, according to the same increase and difference subtraction, y=log4 (7+6x-x 2) is the increasing function at (-1,3) and subtracting at (3,7).

s=-x2+4x+5=-(x-2) 2+9>0 -10 range: [-2log2(3),+infinity)

Single minus: (-1,2】

Single increase: [2, 5).

value range [-log2(9),+infinity).

The monotonic increase interval is [3,5].

The monotonic reduction interval is (-1,3].

Find the monotonic interval of y=log 1 2 (x -5x-6).

Solution: Define the domain: from x -5x-6 = (x+1)(x-6) > 0, the domain is defined as x<-1, or x>6

y=log‹1/2›u，u=x²-5x-6=(x-5/2)²-25/4-6=(x-5/2)²-49/4.

y is a subtraction function with respect to u; u is a quadratic function with respect to x, which is a parabola with an opening facing up, the vertices of which are (5 2, -49 4);

According to the principle of same increase and difference decrease, when x (-1) u decreases monotonically, so y increases monotonically in this interval; When x (6,+, u increases monotonically, so y decreases monotonically in this interval.

That is, the single increment interval of y=log 1 2 (x -5x-6) is (- 1); The single minus interval is (6,+

(x+1)(x-6)>0

x<-1;x>6

y=log1/2[(x+1)(x-6)]

(x+1)(x-6) decreases monotonically when x<-1, and (x+1)(x-6) increases monotonically when x>6.

So the monotonically increasing interval of y=log1 2[(x+1)(x-6)] is (- 1); The monotonically decreasing interval is (6,+

f(x)=log1 2(x) is monotonically decreasing at (0, positive infinity).

k(x)=x-5x-6 decreases monotonically at (negative infinity, 5 2); In [5 2, positive infinity) monotonically increases.

According to: subtraction increases and decreases, and increases and decreases decreases.

At (0,5 2) monotonically increasing, at [5 2,positive infinity) monotonically decreasing.

According to the definition, -x 2+2x+8>0 is -2, and the increase and decrease of the quadratic function is related to the axis of symmetry, and the axis of symmetry in this problem is x=1 Therefore, the -21 composite function: the composite function is a function within a number. Let the definition domain of y=f(u) be du, the value range be mu, and the definition domain of the function u=g(x) be dx and the value range be mx, then for any x in dx, pass u; There is a uniquely determined value of y corresponding to it, so that the relationship between the variable x and y is formed by the variable u is denoted as:

y=f[g(x)], this function is called a composite function, where x is called the independent variable, u is the intermediate variable, and y is the dependent variable (i.e., the function).

First, zeros and negative numbers have no logarithm, -x 2+2x+8 0 again: -x 2+2x+8 = -(x-1) +9 9 0 -x 2+2x+8 9

Base 1 3 1

When x 2 + 2x + 8 = 9, the minimum value ymin = log1 3(9) = -2

value range [-2, +infinity).

g(x) = -x 2+2x+8=-(x-1) +9=-(x+2)(x-4), the opening is downward, the axis of symmetry x=1, the zero point x1=-2, x2=4

y=f(x)=log(1 3)[g(x)] base 1 y=f(x)) monotonically decreasing interval (-2,1), monotonically increasing interval (1,4).