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anonymous

  • one year ago

Find the x-coordinates where f '(x) = 0 for f(x) = 2x + sin(2x) in the interval [0, 2π]. so far I found f'(x)=2cos(2x)+2 cos(2x)=-1

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  1. myininaya
    • one year ago
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    so far good do you know how to solve: \[\cos(\theta)=-1 \text{ for } \theta \]

  2. anonymous
    • one year ago
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    no:(

  3. myininaya
    • one year ago
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    have you ever seen the unit circle before?

  4. anonymous
    • one year ago
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    oh yes

  5. myininaya
    • one year ago
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    Do x coordinates of the pairs represents cos these are the numbers we want to look at can you find when the x-coordinates on the unit circle will be -1?

  6. myininaya
    • one year ago
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    The x coordinates * (not do)

  7. anonymous
    • one year ago
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    so pie?

  8. myininaya
    • one year ago
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    pi is going to be one solution there is another solution we were solving cos(2x)=-1 on [0,2pi] but I replaced 2x with theta so we had 0<=x<=2pi and x is theta/2 so 0<=theta/2<=2pi multiply 2 on both sides we have 0<=theta<=4pi so we actually want to solve cos(theta)=-1 on [0,4pi] but that isn't too terrible we know pi is one solution in that interval but pi+2pi is another we wanted to solve for x not theta but we know the relationship between x and theta is given by 2x=theta so we have \[2x=\pi \text{ or also } 2x=\pi+2\pi\] simplify and solve for x

  9. anonymous
    • one year ago
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    so that would be 2x=3pi so x=pi/2 and x=3pi/2

  10. myininaya
    • one year ago
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    sounds great to me :)

  11. anonymous
    • one year ago
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    thanks again

  12. myininaya
    • one year ago
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    np

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