Finding the monotone interval of a function

A common method of finding function monotone interval

• The monotone interval of the function is consistent with the method of determining monotonicity

1 Image method: if \ (f (x) \) is given in the form of an image, or \ (f (x) \) the image is easy to make, it can be the visual nature of the image to write its monotonic interval.

\ (\fbox{example 1}\)"2018 Tianjin simulation adaptation"

For an image of a known function \ (y=f (x) (x\in R) \) , the function \ (f (x) \) has a monotonic interval of _____________.

Analysis: The graph shows that the function \ (f (x) \ ) is monotonically decreasing on the interval \ ((-\infty,0) \) and \ ([\cfrac{1}{2},+\infty) \) ,

monotonically increasing on the interval \ ([0,\cfrac{1}{2}]\) ,

"Reviews": ① learn to read pictures, interpretation of images, is to change the trend of the same (only rise or fall only) that part of the image, to the \ (x\) axis projection, the resulting interval is a monotone interval.

② This method can solve the monotonicity of many simple functions in high school, such as basic elementary function, once, two times function, etc.

2 Definition method: First, define the domain, then use the monotonicity definition.

\ (\fbox{case 2}\)"The topic is self-proposed, this method is seldom used"

Using the definition method, the monotone interval of function \ (f (x) =x-\cfrac{1}{x}\) is obtained.

Analysis: Define domain as \ ((-\infty,0) \cup (0,+\infty) \);

Take x_1<x_2\in \ (0,+\infty) \),

Then \ (f (x_1)-F (x_2) =x_1-\cfrac{1}{x_1}-(x_2-\cfrac{1}{x_2}) \)

\ (= (x_1-x_2)-(\cfrac{1}{x_1}-\cfrac{1}{x_2}) \)

\ (= (x_1-x_2)-\cfrac{x_2-x_1}{x_1x_2}\)

\ (= (x_1-x_2) +\cfrac{x_1-x_2}{x_1x_2}\)

\ (= (x_1-x_2) (1+\cfrac{1}{x_1x_2}) <0\)

i.e. \ (f (x_1) <f (x_2) \),

Therefore, the function \ (f (x) =x-\cfrac{1}{x}\) is monotonically increasing on the interval \ ((0,+\infty) \) ;

Similarly, it can be proved that the function \ (f (x) =x-\cfrac{1}{x}\) is monotonically increasing on the interval \ ((-\infty,0) \) ;

[or by using \ (f (x) \) as the odd function, you can prove the monotonically incrementing on the interval \ ((-\infty,0) \) ]

"Reviews": ① with the above topic, if the difference in the interval \ ((0,+\infty) \) \ (f (x_1)-f (x_2) \) is not certain to be positive or negative,

It is necessary to find a new point, the above interval refinement, such as the above interval \ ((0,+\infty) \) refinement to \ ((0,x_0) \ ) and \ ( (x_0,+\infty) \),

Then we judge the positive and negative of \ (f (x_1)-f (x_2) \ ) on the interval \ ((0,x_0) \) and Interval \ ((x_0,+\infty) \) to determine the monotone interval.

② Note the use of the parity of functions effectively, simplifying the proof.

3 The monotone interval is obtained by using the monotonicity of the known function, that is, the difference or the compound function which is transformed into a known function.

\ (\fbox{example 3}\)"pending Edit"

4 derivative Method: The monotone interval of the positive and negative definite function using the derivative value. "The main method used in high school, Tong FA"

See blog: Strategies for determining the monotonicity of functions by derivative method

Second, to find the compound function =\ (f (g (x)) \)The steps of the monotone interval:

(1) Determine the definition field of the function.

(2) The composite function is decomposed into basic elementary functions \ (y=f (u) \),\ (u=g (x) \).

(3) The monotone intervals of the two functions are determined respectively.

(4) If these two functions are reduced together, then \ (Y=f (g (x)) \) is the increment function, and if one is increased, then \ (Y=f (g (x)) \) is the subtraction function, that is, the "same increase and decrease".

\ (\fbox{example 5}\)"Finding the monotonicity of a compound function"

Known function \ (f (x) =log_2 (x^2-3x+2) \), the monotone interval is obtained.

Analysis: Make \ (u=x^2-3x+2\),

The original compound function is split into outer function \ (y=f (u) =log_2u\) and inner function \ (u=x^2-3x+2\)

by \ (u=x^2-3x+2>0\), Solution \ (x\in (-\infty,1) \cup (2,+\infty) \),

That is, the definition domain for this compound function is \ (x\in (-\infty,1) \cup (2,+\infty) \).

In order to study its monotonicity, we must first define the domain priority principle within the scope of the above defined domain.

Then by \ (u=x^2-3x+2= (x-\cfrac{3}{2}) ^2-\cfrac{1}{4}\),

The inner function \ (u (x) \ ) is monotonically decreasing on the interval \ ((-\infty,1) \ ), monotonically incrementing on the interval \ ((2,+\infty) \) ,

The outer function \ (y=f (u) =log_2u\) is monotonically incremented on \ ((0,+\infty) \) ,

Therefore, the compound function \ (f (x) \ ) is monotonically decreasing on the interval \ ((-\infty,1) \ ) and monotonically increasing on the interval \ ((2,+\infty) \) .

\ (\fbox{example 6}\)"to find the monotone interval of the compound function" "2018 Tianjin Simulation"

For an image of a known function \ (y=f (x) (x\in R) \) , the function \ (g (x) =f (Log_ax) (0<a<1) \) has a monotonically decreasing interval of ""

\ (A, [0,\cfrac{1}{2}]\) \ (\hspace{4em}\) \ (B, [\sqrt{a},1]\)

\ (C, (-\infty,0) \cup[\cfrac{1}{2},+\infty) \) \ (\hspace{4em}\) \ (D, [\sqrt{a},\sqrt{a+1}]\)

Analysis: By the figure, the outer function \ (f (x) \) is monotonically decreasing on the interval \ ((-\infty,0) \) and \ ([\cfrac{1}{2},+\infty) \) ,

monotonically increasing on the interval \ ([0,\cfrac{1}{2}]\) ,

Also \ (0<a<1\) , the inner function \ (y=log_ax\) on the interval \ ((0,+\infty) \) on the monotonically decreasing,

Therefore, to make the compound function function \ (g (x) =f (Log_ax) (0<a<1) \) monotonically decreasing,

You need \ (log_ax\in [0,\cfrac{1}{2}]\), i.e. \ (0\leq log_ax\leq \cfrac{1}{2}\),

Solution (x\in [\sqrt{a},1]\), so select B.

Finding the monotone interval of a function