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sandy524

  • 2 years ago

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

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  1. sandy524
    • 2 years ago
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    i now know that its -sin(x)(2cos(x)+1)

  2. sandy524
    • 2 years ago
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    can yu guys help me find the critical points

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

  4. myininaya
    • 2 years ago
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    If so, you did that. You are done.

  5. sandy524
    • 2 years ago
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    yes. may you help me find the critical points of this now

  6. myininaya
    • 2 years ago
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    Set f'(x)=0 and solve for x to find the critical numbers.

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

  8. sandy524
    • 2 years ago
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    can we do it together please

  9. myininaya
    • 2 years ago
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    You are already factored f'(x) Now set both factors equal to 0.

  10. sandy524
    • 2 years ago
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    how? can we work out the problem

  11. sandy524
    • 2 years ago
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    step by step please

  12. myininaya
    • 2 years ago
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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.

  13. sandy524
    • 2 years ago
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    ok

  14. myininaya
    • 2 years ago
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    Just like if you have a*b=0 then either a=0 or b=0 or both=0

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

  16. sandy524
    • 2 years ago
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    this is all i have

  17. sandy524
    • 2 years ago
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    im confused about the results

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

  19. myininaya
    • 2 years ago
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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.

  20. sandy524
    • 2 years ago
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    -sinx=pin

  21. sandy524
    • 2 years ago
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    \[\pi n\]

  22. myininaya
    • 2 years ago
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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

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