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sandy524 Group Title

find the derivative of f(x) = cos(x) − sin^2(x) on [0,2pi]

  • one year ago
  • one year ago

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  1. sandy524 Group Title
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    i now know that its -sin(x)(2cos(x)+1)

    • one year ago
  2. sandy524 Group Title
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    can yu guys help me find the critical points

    • one year ago
  3. myininaya Group Title
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    Are you sure it ask you to find f'(x) on [a,b]?

    • one year ago
  4. myininaya Group Title
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    If so, you did that. You are done.

    • one year ago
  5. sandy524 Group Title
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    yes. may you help me find the critical points of this now

    • one year ago
  6. myininaya Group Title
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    Set f'(x)=0 and solve for x to find the critical numbers.

    • one year ago
  7. myininaya Group Title
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    You only need to solve this on the interval [0,2 pi]

    • one year ago
  8. sandy524 Group Title
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    can we do it together please

    • one year ago
  9. myininaya Group Title
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    You are already factored f'(x) Now set both factors equal to 0.

    • one year ago
  10. sandy524 Group Title
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    how? can we work out the problem

    • one year ago
  11. sandy524 Group Title
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    step by step please

    • one year ago
  12. myininaya Group Title
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    Yes. I'm walking you through it. You have -sin(x)*(1+2cos(x))=0 Set both factors equal to 0.

    • one year ago
  13. sandy524 Group Title
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    ok

    • one year ago
  14. myininaya Group Title
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    Just like if you have a*b=0 then either a=0 or b=0 or both=0

    • one year ago
  15. sandy524 Group Title
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    http://www.wolframalpha.com/input/?i=-sin%28x%29%282cos%28x%29%2B1%29%3D0

    • one year ago
  16. sandy524 Group Title
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    this is all i have

    • one year ago
  17. sandy524 Group Title
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    im confused about the results

    • one year ago
  18. myininaya Group Title
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    Use your unit circle. Keep in mind you only have to look between 0 and 2pi

    • one year ago
  19. myininaya Group Title
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    You have -sin(x)=0 or 1+2cos(x)=0 Solve both of these for the trig function. Solve the first one for sin(x) and solve the second one for cos(x). Tell me what you get after doing this.

    • one year ago
  20. sandy524 Group Title
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    -sinx=pin

    • one year ago
  21. sandy524 Group Title
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    \[\pi n\]

    • one year ago
  22. myininaya Group Title
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    Well you are trying to solve -sin(x)=0 for x. You could divide both sides by -1 since -1 will never be 0. Try solving this: sin(x)=0 Then we will come back to 1+2cos(x)=0

    • one year ago
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