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Math Help - Help with finding Critical Numbers

  1. #1
    Member FalconPUNCH!'s Avatar
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    Help with finding Critical Numbers

    Ok so the equation is
    f(x) = cos^2(x) - 2sin(x)

    0 < or equal to x < or equal to 2pi

    I get the derivative which is

    f'(x) = -sin^2(x) - 2cos(x)

    I was never good with Trigonometric identities so I need help with finding the critical numbers.
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  2. #2
    Super Member
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    Quote Originally Posted by FalconPUNCH! View Post
    Ok so the equation is
    f(x) = cos^2(x) - 2sin(x)

    0 < or equal to x < or equal to 2pi

    I get the derivative which is

    f'(x) = -sin^2(x) - 2cos(x)

    I was never good with Trigonometric identities so I need help with finding the critical numbers.
    Hello,

    if
    f(x)=\cos^2(x)-2\sin(x)\ ,~0 \leq x \leq 2\pi

    then the first derivative (use chain rule!) is

    f'(x)=2\cos(x)\cdot (-\sin(x)) - 2\cos(x)=-2\cos(x)(\sin(x)+1)

    f'(x)=0~\implies~-2\cos(x)=0~\vee~\sin(x)+1=0

    I leave the rest for you.
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  3. #3
    Member FalconPUNCH!'s Avatar
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    Thanks I can do the rest but what does the V sign stand for?
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  4. #4
    Super Member
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    Quote Originally Posted by FalconPUNCH! View Post
    Thanks I can do the rest but what does the V sign stand for?
    Hi,

    \vee means or

    and

    \wedge means and
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