The monotonicity of the function is known to be discussed

The monotonicity of a function can also be referred to as the addition or decrease of a function.

When the independent variable of the function f(x) increases (or decreases) within its defined interval, and the value of the function f(x) also increases (or decreases), the function is said to be monotonionic in that interval.

The monotonicity of this function has nothing to do with adding or not adding 2, so it is equivalent to discussing the monotonicity of f(x)=a"x+1".

The monotonicity of the function is then judged based on the sign of a. a=0 is a function that does not increase or decrease; When a is not equal to 0, the image of this function can be obtained by translating the image of f(x)=a"x" one unit to the left, then we can know that when a>0, the monotony decreases from negative infinity to -1, and increases monotonically from -1 to positive infinity, and the monotonicity at a<0 can be obtained.

Solution: f(x)=(x+1+2) (x+1)=1+2 (x+1)x+1 is a monotonic increase function over the interval (-1, +00).

1 (x+1) is a monotonically subtracting function over the interval (-1, +00).

1+2 (x+1) is a monotonically subtracting function over the interval (-1, +00).

That is, f(x) is a monotonically subtracting function over the interval (-1, +00).

x+1 is monotonically increasing on the interval (-2, -1).

1 (x+1) is a monotonically subtracting function on the interval (-2, -1).

1+2 (x+1) is a monotonically subtracting function over the interval (-2, -1).

That is, f(x) is a monotonically decreasing function over the interval (-2, -1).

When x=-1, the function has no solution.

In summary: f(x) is a monotonically decreasing function over the intervals (-2, -1) and (-1, +00).

Note that it should be used in the middle of the two intervals here"with"(not an intersection), cannot be changed to"and"

f（x）=（x+3）/（x+1） =1+ 2/(x+1)

If you look at the equation, you can see that the inverse function shifts one unit to the left.

2, -1) minus, (-1, +00) minus.

The hook function is an odd function, at x>0, monotonically decreasing at (0,1) and increasing monotonically at [1,+.

x<0 is sufficient.

The proof can be proved by the definition of the monotonicity of the function.

When x<1, monotonically decreases. When x>1, it increases monotonically.

f(x) = y = 3 -2f(x)

In [a,b] arbitrarily take x1 < x2, then:

f(x2) -f(x1) = 3 - 2f(x2) -3 - 2f(x1))

2f(x1) -2f(x2)

Since f(x) is a subtraction function, so: 2f(x1) -2f(x2) >0;

So y = 3 -2f(x) is an increment function.

It's already okay to do it upstairs; Do this kind of question directly with the definition method of proving monotonicity, and try it out a hundred times! Study hard!

There are many ways to solve function monotonicity.

1 Properties of basic elementary functions.

Many functions are composite functions (combinations of different or similar basic elementary functions), so they can be solved according to such characteristics: in the same interval, if the monotonicity of each composite part of the composite function is the same, then the monotonicity of the composite function is increased, and vice versa, such as the fifth.

Solution 5: The cube of y=x is an increasing function when a>0 the original function is an increasing function and vice versa is a decreasing function.

2. Finding the derivative This is a commonly used and more versatile method, except for functions that cannot find the derivative, such as 1, , ,

3.It is a bit of a hassle to draw a picture based on an image, but it can be solved by a categorical discussion method to draw a rough image.

All monotonicity problems are solved by setting the x1 x2 x1, x1, x2 function definition domains f(x1) f(x2).

If the value is greater than 0, the monotonic increase is less than 0, and the monotonic decrease.

1 Derivative f (x)=-2x+1 When the derivative is greater than 0, x 1 2 is the increasing interval of the original function, then when x 1 2 the original function is a monotonically decreasing function.

2 This function is monotonically increasing at x 0.

3 Simplifying Finding the Derivative Function After that, it is the same as in question 1.

4 Derivative function Make the derivative function greater than 0 Evaluate This interval is the increasing interval of the original function, and vice versa is the decreasing interval (in fact, it is a method for solving 1 problem, which is the general method of this kind of problem).

5 Derivative Function y =3ax This is a quadratic equation with parametric variables, which can be combined with images The interval of the derivative function greater than 0 is the increasing interval of the original function, but the interval of the function less than 0 is the subtracting interval of the original function, which you can find yourself.

6 I don't know why there should be other conditions.

It's easy to use the method of derivation!

First, the analytic formula is found, then the derivative function is obtained, the size of and is discussed, and then the size of and is discussed separately, and the monotonic interval of the function is obtained according to the sign of the derivative function.

Solution: Yes, at that time, the function was an increasing function on top;

At that time, the function was a subtraction function on top and an increase function on top;

Yes, at that time, the function was subtractive on top;

At that time, the function was a subtraction function on top and an increase function on top;

This question mainly examines the monotonicity of piecewise functions, the relationship between the positive and negative derivatives of the derivative function and the monotonicity of the original function, that is, when the derivative function is greater than the monotonic increase of the original function, when the derivative function is less than the monotonic decrease of the original function, and the mathematical ideas of classification discussion, it belongs to the mid-range questions.

The method is shown in the figure below, please check it carefully, and I wish you a happy study:

The reciprocal of f(x) is always at 0, and x is meaningless at 0, so f(x) increases monotonically in the intervals (-infinity, 0) and (0, +infinity).