Question

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

Answer 1

so you need to solve f'=0 in the interval [0,2pi]
first find f'

Answer 2

2cos(2x)+2 equals f'

Answer 3

\[2\cos(2x)+2=0 \\ 2\cos(2x)=-2 \\ \cos(2x)=-1\]

Answer 4

solve that equation on the interval 0 to 2pi

Answer 5

i had this earlier but i could not figure out how to get rid of the 2x like it was

Answer 6

let u=2x
so if x is between 0 and 2pi
and x=u/2 then u is between 0 and 4pi
so solve cos(u)=-1 on the interval [0,4pi]

Answer 7

cos(u)=-1 when u=pi
can you find another value?

Answer 8

If u equals pi then do you have to add the one and set it equal to zeros to solve?

Answer 9

i was asking you know another value between 0 and 4pi
such that cos(u)=-1

Answer 10

I gave you one value for u that satisfies the equation

Answer 11

cos(u)=-1 gives us the solutions as u=pi or u=3pi in the interval [0,4pi]

Answer 12

since u=2x
then we actually have to solve 2x=pi and solve 2x=3pi

Answer 13

when you solve those two equations for x you are done

Answer 14

do i just do cos(2pi) and cos(2*3pi)?

Answer 15

no you just solve 2x=pi for x
and solve 2x=3pi for x

Answer 16

divide both sides by 2

Answer 17

the interval was 0 to 2pi, so would i just have pi

Answer 18

no the interval for x was 0 to 2pi
but we let u=2x so that means we had u/2 is between 0 and 2pi
so that means u is between 0 and 4pi
so we solved cos(u)=-1 on the interval [0,4pi]
giving us u=pi and u=3pi
but we had u=2x
so we have 2x=pi and 2x=3pi
solving both of these equations gives you solutions for x only between 0 and 2pi

Answer 19

isn't x=pi/2 and x=3pi/2 both in the interval 0 to 2pi?

Answer 20

\[\cos(2x)=-1 ; 0 \le x \le 2 \pi \\ \\ \text{ if we let } u=2x \text{ then we have: } \\ \cos(u)=-1; 0 \le \frac{u}{2} \le 2\pi \\ \text{ but that inequality actually says } 0 \le u \le 4\pi (\text{ if you multiply both sides by 2}) \]