How to draw the function image of y = - 2x

How to draw the function image of y = - 2x

Set a for loop, starting from 0, such as (0,0), (1, - 2), (2, - 4) It's OK to link such points into lines

Draw an image of the function y = | X-1 | + | 4-2x | Draw it out

1. When 2 < x, y = X-1 - (4-2x) = 3x-52.1 < x < 2 y = X-1 + 4-2x = - x + 33. X > 1 y = - (x-1) + 4-2x = - 3x + 3 when 2 < x, y = X-1 - (4-2x) = 3x-52

If the function f (x) = AB, where a = (2cosx, cosx + SiNx), B = (SiNx, cosx SiNx) If the function f (x) = vector a * vector B, where vector a = (2cosx, cosx + SiNx), vector b = (SiNx, cosx SiNx) 1. The image symmetry center and axis equation of F (x) are obtained 2. For any x belonging to [0, PI / 2], if f (x) is less than m ^ 2 + m + radical 2-2, the value range of real number m is found

(1) The vector a = (2cosx, cosx + SiNx), vector b = (SiNx, cosx SiNx)
And the function f (x) = vector a * vector B,
So,
f(x)=2sinxcosx+(cosx+sinx)(cosx-sinx)
=2sinxcosx+cos²x-sin²x
=sin2x+cos2x
=√2sin(2x+π/4)
The image symmetry center of F (x) (- π / 8 + K π / 2,0), K ∈ Z
The image symmetry axis equation of F (x) is x = π / 8 + K π / 2, K ∈ Z
(2) For any x ∈ [0, π / 2], f (x) ＜ M 2 + m + √ 2-2 holds,
By X ∈ [0, π / 2], 2x + π / 4 ∈ [π / 4, π],
f(x)max=f(π/8)=√2
According to the title, f (x) < m 2 + m + √ 2-2 always holds,
Therefore, √ 2 ＜ M 2 + m + √ 2-2
m²+m-2＞0
M ∈ (- ∞, - 2) ∪ (1, + ∞) is obtained

How can 1 / SiNx * cosx be simplified by multiplying it by 2. Is it 2 / 2 SiNx * cosx

This simplification means that there are as few different function names as possible, and it is better not to have denominators
1/(sinxcosx）
=2/(2sinxcosx）
=2/sin2x
=csc2x

How to simplify SiNx / (1-cosx) I've read the solution, but I can't understand it. It's better for you to write the symbols on the paper and take them to my computer

A:
Double angle formula: sin2a = 2sinacosa, cos2a = 2cos? A-1 = 1-2sin? A
sinx/(1-cosx)
=2sin(x/2)cos(x/2)/[2sin²(x/2)]
=cot(x/2)

How to simplify (1-cosx) / SiNx

(1-cosx)/sinx=sinx(1-cosx)/sin²x =sinx(1-cosx)/(1-cos²x) =sinx(1-cosx)/(1+cosx)(1-cosx)=sinx/(1+cosx)=2sin(x/2)cos(x/2)/2cos²(x/2) =sin(x/2)/cos(x/2)=tan(x/2)

How to simplify SiNx + cosx?

sinx+cosx
=√2（√2/2*sinx+√2/2*cosx）
=√2(sinπ/4sinx+cosπ/4cosx)
=√2cos(x-π/4)

How to simplify y = SiNx + cosx

√2/2y=√2/2sinx+√2/2cosx
√2/2y=cos(π/4)sinx+sin(π/4)cosx
√2/2y=sin(x+π/4)
y= √2sin(x+π/4)

How to simplify SiNx + cosx? The best formula and derivation process have

sinx+cosx
=√2（√2/2sinx+√2/2cosx)
=√2[cos(π/4)sinx+sin(π/4)cosx]
=√2sin(π/4+x)

(sinx+cosx)/(2sinx-cosx)=

（tanx+1)/(2tanx-1)