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Math Help - X-Coordinates on Tangent Line

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    Member FalconPUNCH!'s Avatar
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    X-Coordinates on Tangent Line

    Find the x-coordinates of all points on the curve y = sin2x - 2sinx at which the tangent line is horizontal.

    I don't know what to do and where to begin.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    Find the x-coordinates of all points on the curve y = sin2x - 2sinx at which the tangent line is horizontal.

    I don't know what to do and where to begin.
    the tangent line is horizontal when the derivative is zero. find the derivative, set it equal to zero and solve for x
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    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by Jhevon View Post
    the tangent line is horizontal when the derivative is zero. find the derivative, set it equal to zero and solve for x
    Thanks I'll try that.
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    Member FalconPUNCH!'s Avatar
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    I got  y' = (cos(2)) (2x) - 2sin(x)cos(x) = 0 as the derivative and I pretty much don't know how to solve it for x.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    I got  y' = (cos(2)) (2x) - 2sin(x)cos(x) = 0 as the derivative and I pretty much don't know how to solve it for x.
    that's because your derivative is incorrect

    y' = 2 \cos 2x - 2 \cos x
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    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by Jhevon View Post
    that's because your derivative is incorrect

    y' = 2 \cos 2x - 2 \cos x
    could you please explain how you got that because every time I try I get a different answer.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    could you please explain how you got that because every time I try I get a different answer.
    for the first part we use the chain rule:

    \frac d{dx}f(g(x)) = f'(g(x)) \cdot g'(x)

    for \sin 2x: f(x) = \sin x and g(x) = 2x

    the other piece is trivial. \frac d{dx} \sin x = \cos x so \frac d{dx} -2 \sin x = -2 \frac d{dx} \sin x = -2 \cos x
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    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by Jhevon View Post
    for the first part we use the chain rule:

    \frac d{dx}f(g(x)) = f'(g(x)) \cdot g'(x)

    for \sin 2x: f(x) = \sin x and g(x) = 2x

    the other piece is trivial. \frac d{dx} \sin x = \cos x so \frac d{dx} -2 \sin x = -2 \frac d{dx} \sin x = -2 \cos x
    Oh I know what I did wrong. I forgot to take the -2 like this -2 \frac d{dx} sinx
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    Oh I know what I did wrong. I forgot to take the -2 like this -2 \frac d{dx} sinx
    ok. now set the derivative to zero and continue
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    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by Jhevon View Post
    ok. now set the derivative to zero and continue
    I got cos2x = cosx
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    I got cos2x = cosx
    are you forgetting your identities?

    \cos 2x - \cos x = 0

    \Rightarrow 2 \cos^2 x - 1 - \cos x = 0

    this is quadratic in cos(x). you can take it from here right?
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    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by Jhevon View Post
    ok. now set the derivative to zero and continue
    Quote Originally Posted by Jhevon View Post
    are you forgetting your identities?

    \cos 2x - \cos x = 0

    \Rightarrow 2 \cos^2 x - 1 - \cos x = 0

    this is quadratic in cos(x). you can take it from here right?
    To tell you the truth I can't remember any of the identities since I haven't used them in a year. I'll try and take it from there.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    To tell you the truth I can't remember any of the identities since I haven't used them in a year. I'll try and take it from there.
    you should look them up. for double angles you must know that:

    \cos 2x = \cos^2 x - \sin^2 x = 2 \cos^2 x - 1 = 1 - 2 \sin^2 x

    \sin 2x = 2 \sin x \cos x

    ok, continue. tell me if you have any problems
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    Quote Originally Posted by Jhevon View Post
    you should look them up. for double angles you must know that:

    \cos 2x = \cos^2 x - \sin^2 x = 2 \cos^2 x - 1 = 1 - 2 \sin^2 x

    \sin 2x = 2 \sin x \cos x

    ok, continue. tell me if you have any problems
    Yeah I don't think I can do it. I'm just going in circles. I was never good with functions
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    Yeah I don't think I can do it. I'm just going in circles. I was never good with functions
    are you taking me for a ride? or are you really trying? *suspicious smiley (we still need one of these)*

    2 \cos^2 x - \cos x - 1 = 0 ....this is a quadratic, see if you can factorize

    \Rightarrow (2 \cos x + 1)(\cos x - 1) = 0

    \Rightarrow 2 \cos x + 1 = 0 \mbox{ or } \cos x - 1 = 0

    \Rightarrow \cos x = - \frac 12 \mbox{ or } \cos x = 1

    what angles does this happen for?
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