How to draw the image of y = 1 / 2 SiNx

How to draw the image of y = 1 / 2 SiNx

Similar to the image of SiNx, the maximum and minimum values become 1 / 2 and - 1 / 2

The image of function y = 1 / X (x > 0) is drawn with the method of point tracing in list

X：              -4,       -3,      -2, -1,  1,  2,         3,4,Y=1／X：   -1／4...

Given the function y = 2sinx (- Π / 2 ≤ x ≤ 3 Π / 2), the area of the closed graph bounded by its image and the straight line y = - 2 is----

S=(2∏*4)/2=4∏

The area of the closed graph determined by the image of function (| Y-2 | / 3) + (| x + 7 / 5) = 1 is?

Draw a picture
To divide Y-2 greater than 0 and less than or equal to 0, and X + 7 greater than 0 and less than or equal to 0, remove the absolute value symbol, draw a drawing to determine the closed curve, with the figure area is very easy to find

The area of the closed graph enclosed by the image of function f (x) = x * 2 + 1 and the line y = x + 1 is quickly

The intersection of curve and straight line is the intersection of curve and straight line, y = x ^ 2 + 1y = x + 1, two types of simultaneous system x + 1 = x ^ 2 + 1 x = x ^ 2x = 0, x = 1y = 1, y = 1, y = 1, y = 2, intersection is (0,1) (1,2) use integral to get area s = ∫ (x + 1) - (x ^ 2 + 1) DX integral interval is (0,1) = ∫ x-x ^ 2 DX = ∫ x DX - ∫ x ^ 2DX = x ^ 2 / 2-x ^ 3 / 3 / 3 (0,1) = 1 / 2-2-x ^ 3 / 3 (0,1) = 1 / 2-2-2-2-1-1-1-1-1-1-1 / 3 = 1 / 6

How to get the image of y = 2cosx from the image of function y = SiNx

y=sinx→y=sin(x+π/2)=cosx→y=2cosx
First, the image of y = SiNx is shifted to the left by π / 2 units, and then the abscissa of the image is unchanged, and the ordinate is changed to twice of the original, and y = 2cosx is obtained

To get the image of the function y = √ 2cosx, just change the function y = √ 2sinx (2x + tt / 4),

Function y = √ 2sinx (2x + tt / 4)
=√2sin[π/2+(2x-π/4)]
=√2cos(2x-π/4)
=√2cos[2(x-π/8)]
In the image of y = √ 2cosx, the ordinate of each point is unchanged, and the abscissa is reduced to 1 / 2 times of the original, and the image of y = √ 2cos2x is obtained
The image of y = √ 2cos [2 (x - π / 8)] is obtained by shifting the image of y = √ 2cos2x to the right by π / 8 units
Then the image of y = √ 2cos [2 (x - π / 8)] is shifted to the left by π / 8 units to get the image of function y = √ 2cos2x
In the image of y = √ 2cos2x, the image with y = √ 2cosx is obtained by extending the abscissa of each point to twice the original ordinate
In other words, the image of y = √ 2sinx (2x + tt / 4) needs to be shifted to the left by π / 8 units, and then the ordinate of each point in the image is unchanged, and the abscissa is extended to twice the original value to obtain the image of y = √ 2cosx

What is the minimum value of the function y = - 2sinx + √ 2cosx?

Let me tell you a conclusion,
Y=Asinx+Bcosx
The maximum value of must be ± √ (a 2 + B 2)

Find the function y = LG (2sinx-1) + √ (1 + 2cosx) Define domain

The definition domain of X is (π / 6 + 2K π, 2K π + 2 π / 3]

The image of the function y = 1-sin3x is drawn by the five point method RT

List:
3x:0 pi/2 pi 3pi/2 2pi
x:0 pi/6 pi/3 pi/2 2pi/3
y:1 0 1 2 1
Use the data of the second line to trace the points to get the desired image