Find the x-coordinates where f '(x) = 0 for f(x) = 2x - sin(2x) in the interval [0, 2π].

The x-coordinates where f '(x) = 0 for f(x) = 2x - sin(2x) in the interval [0, 2π] are: x=0, π, and 2π.

Step-by-step explanation:

f(x)=2x-sin(2x)

f'(x)=[f(x)]'

f'(x)=[2x-sin(2x)]'

f'(x)=(2x)'-[sin(2x)]'

f'(x)=2-cos(2x) (2x)'

f'(x)=2-cos(2x) (2)

f'(x)=2-2 cos(2x)

f'(x)=0, then:

2-2 cos(2x)=0

Solving for x. First solving for cos(2x): Subtracting 2 from both sides of the equation:

2-2 cos(2x)-2=0-2

-2 cos(2x)=-2

Dividing both sides of the equation by -2:

-2 cos(2x) / (-2)=-2 / (-2)

cos(2x)=1

If cos(2x)=1, then:

2x=2nπ

Dividing both sides of the equation by 2:

2x/2=2nπ/2

x=nπ

If n=-1→x=-1π→x=-π, and -π is not in the interval [0,2π], then x=-π is not in the solution.

If n=0→x=0π→x=0, and 0 is in the interval [0,2π], then x=0 is in the solution.

If n=1→x=1π→x=π, and π is in the interval [0,2π], then x=π is in the solution.

If n=2→x=2π, and 2π is in the interval [0,2π], then x=2π is in the solution.

If n=3→x=3π, and 3π is not in the interval [0,2π], then x=2π is not in the solution.

Solution: x=0, π, and 2π.

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

source:

include a picture of the drawing in order for me to

step-by-step explanation: