Question

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

Answer 1

@Luigi0210 can u plz help on this 1? :) its not a word problem

Answer 2

@pooja195

Answer 3

take the derivative set it equal to zero and solve
it is not a word problem

Answer 4

okay so i take the derivative of 2x-sin(2x),
can u plz show me those steps?
i like to see it all written out it helps me understand better.

Answer 5

do you know what the derivative of \(2x\) is?
`

Answer 6

yes 2

Answer 7

Idk the derivative of -sin(2x)

Answer 8

ok that was the first step
how about the derivative of \(-\sin(2x)\)?

Answer 9

-2cos(2x)

Answer 10

?:)

Answer 11

@satellite73

Answer 12

@welshfella @pooja195 @Luigi0210 plz help!!!!!! :(

Answer 13

yes you have it

Answer 14

solve
\[2-2\cos(2x)=0\] and you are done

Answer 15

What do you mean solve? what about the interval [0,2pi]?

Answer 16

what do you mean "what do you mean"
solve for \(x\) there is no other way i can think of to say it

Answer 17

start with
\[2=2\cos(2x)\] so
\[\cos(2x)=1\]

Answer 18

then solve that one more or less in your head for the first value

Answer 19

do we divide both sides by cos?

Answer 20

so it'd be 1/cos = 2x or can we not separate them

Answer 21

to get
\[2x=\frac{1}{\text{cos}}\]?

Answer 22

yes :)

Answer 23

@perl

Answer 24

a bit confused right?

Answer 25

yes just at the solving for x part

Answer 26

\[f(x)=x+2\\
x=\frac{x+2}{f}\]

Answer 27

so i have:
cos(2x)=1

Answer 28

cosine is a function
you are trying to solve \[\cos(2x)=1\] think of a number whose cosine is 1

Answer 29

0?

Answer 30

yes

Answer 31

and if \(2x=0\) then \(x=0\)
now we get another one

Answer 32

\[\cos(2\pi)=1\] also, so if
\[2x=2\pi\] then
\[x=\pi\]

Answer 33

those are the two in your interval from zero to two pi

Answer 34

ohh so the coordinates are 0 & pi?

Answer 35

the first coordinates are, yes

Answer 36

lol there's more???????

Answer 37

@perl @perl @perl can u plz help ? :)