The graph of the function f (x) = ︱ x + 3 - ︱ x-3 ︱ is drawn and observed. The monotone increasing interval and monotone decreasing interval are pointed out

The graph of the function f (x) = ︱ x + 3 - ︱ x-3 ︱ is drawn and observed. The monotone increasing interval and monotone decreasing interval are pointed out

x> At 3, f (x) = x + 3 - (x-3) = 6; it does not increase or decrease;
-When 3 ≤ x ≤ 3, f (x) = x + 3 + x-3 = 2x is an increasing function;
When x is less than - 3, f (x) = - (x + 3) + x-3 = - 6 is neither increased nor decreased;
Therefore, the increasing interval of F (x) is [- 3,3]; f (x) has no decreasing interval;

It is known that the function f (x) is an odd function defined on R. when x is greater than or equal to 0, f (x) = x (x + 1). The image of function f (x) is drawn and the function analysis is obtained

F (x) = x (x + 1), when x

Draw the image of the function y = 5x-9, and answer the following questions according to the image: (1) when x takes what value, the function value y equals 0? (2) what value does x take (1) What value does x take, the function value y equals 0? (2) when x takes what value, the function value y equals to - 2? (3) what is the relationship between the above two questions and the unary linear equation 5x-9 = 0, 5x-9 = - 2?

(1) When y = 0, 5x-9 = 0
X = 9 / 5
(2) When y = - 2, 5x-9 = - 2
X = 7 / 5
(3) The solutions of the latter two equations are the points on the graph of the previous function

Draw the graph of the following function and write the monotone interval of the function (1) y = - x ^ 2 + 2 | x | + 1 (2) y = | - x ^ 2 + 2x + 3| There is no need to draw an image It's mainly about how to turn and choose between images after making absolute values

The first one is even function, as long as you draw > 0, and then symmetry about y axis, you can draw (1) y = - x ^ 2 + 2|x| + 1
The second one is to draw - x ^ 2 + 2x + 3, and then turn up y < 0
In fact, if an absolute value is added to the second one, the negative value will be changed into a positive value, as long as it is symmetrical about the x-axis, but the character > O is still the original character

Draw the graph of the function y = | 2x + 1 | and write the monotone interval

Method 1: classified discussion
When x "- 1 / 2, y = 2x + 1, draw the image
When x ＂ - 1 / 2, y = - 2x-1, draw the image
Monotonically increasing interval is [- 1 / 2, positive infinite)
The monotone decreasing interval is (negative infinity, - 1 / 2]
Method 2
Draw all the images of y = 2x + 1, and then turn up the image below the X axis
Monotonicity is invariable

The graph of the function y = - x? + 2x is drawn, and the monotone increasing interval of the function is found

y'=-2x+2=2(1-x)
Easy to see when x

The sum of π - 3 and π - 3 of π - 3 is obtained How to write a big problem? I can draw pictures, but how to write the theoretical part of the big problem?

The title should be x ∈ (- π / 3, π / 3). The person who writes the wrong questions is very careless. How can a careless person learn mathematics well? Mathematics is rigorous!
For such a function, we first transform it into the standard formula, that is, y = - 2Sin (2x - π / 3), and write a monotone interval. According to the meaning of the title, when 2x - π / 3 ∈ [2K π - π / 22K π + π / 2], it is a decreasing function; when 2x - π / 3 ∈ [2K π + π / 2K π + 3 π / 2], it is an increasing function
When x ∈ [K π - π / 12, K π + 5 π / 12], we try to take the value of K and make it fall into the range of (- π / 3, π / 3), and we get that x ∈ [- π / 12 π / 3) is a decreasing function
When x ∈ [K π + 5 π / 12 K π + 11 π / 12], we try to get the value of K and make it fall into the range of (- π / 3, π / 3). We get that x ∈ (- π / 3 - π / 12] is an increasing function
Therefore, the monotone decreasing interval of the function is [- π / 12 π / 3], and the monotone increasing interval is (- π / 3 - π / 12]
When x = - π / 12, - π / 3 and π / 3, the values are 2,0, - 2Sin π / 3, respectively, so the value range of the function is (- root 32]

Given the function FX = 2Sin (2x - π / 6), find the value range of function FX on the interval [- π / 12, π / 2]

x∈[-π/12,π/2]
2x∈[-π/6,π]
2x-π/6∈[-π/3,5π/6]
sin(2x-π/6)∈[-√3/2,1]
2sin(2x-π/6)∈[-√3,2]
The range is [- √ 3,2]

Given the function y = 2Sin (x - π / 3) + 3, find the monotone interval of value domain

Maximum y = 2 * 1 + 3 = 5 minimum y = 2 * (- 1) + 3 = 1
Let 2K π - π / 2 ≤ X - π / 3 ≤ 2K π + π / 2 to obtain 2K π - π / 6 ≤ x ≤ 2K π + 5 π / 6, and K is an integer
Let 2K π + π / 2 ≤ X - π / 3 ≤ 2K π + 3 π / 2 to obtain 2K π + 5 π / 6 ≤ x ≤ 2K π + 11 π / 6, and K is an integer
Therefore, the value range of the function y = 2Sin (x - π / 3) + 3 is [1,5]
The monotone increasing interval is [2K π - π / 6,2k π + 5 π / 6], and K is an integer;
The monotone decreasing interval is [2K π + 5 π / 6,2k π + 11 π / 6], and K is an integer

The monotone increasing interval of the function y = Log1 / 2 [2Sin (2x + π / 3) - 1] is to find the detailed solution,

Due to 1 / 20
From π / 2 + 2K π