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Math Help - finding derivatives of a function

  1. #1
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    finding derivatives of a function

    Alright I have a function y = sin2x-2sinx and I have to find x-coordinates to all the points on the graph that have a tangent line that is horizontal, or zero. So I'm guessing chain rule along with the derivatives of sin which I can do but how to find the points where it is zero? Find the derivative of this function then plug in zero? Thanks for any help.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by UMStudent View Post
    Alright I have a function y = sin2x-2sinx and I have to find x-coordinates to all the points on the graph that have a tangent line that is horizontal, or zero. So I'm guessing chain rule along with the derivatives of sin which I can do but how to find the points where it is zero? Find the derivative of this function then plug in zero? Thanks for any help.
    ok, so what you do is find the derivative and set the expression equal to zero and solve for x. these x values are the coordinates you seek
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by UMStudent View Post
    Alright I have a function y = sin2x-2sinx and I have to find x-coordinates to all the points on the graph that have a tangent line that is horizontal, or zero. So I'm guessing chain rule along with the derivatives of sin which I can do but how to find the points where it is zero? Find the derivative of this function then plug in zero? Thanks for any help.
    in case you wanna check your answers when done.

    y = sin2x - 2sinx
    => y' = 2cos2x - 2cosx

    for horizontal slope, set y' = 0
    => 2cos2x - 2cosx = 0
    => cos2x - cosx = 0
    => cos^2x - sin^2x - cosx = 0
    => cos^2x - (1 - cos^2x) - cosx = 0
    => 2cos^2x - cosx - 1 = 0
    let y = cosx
    => 2y^2 - y - 1 = 0
    => (2y + 1)(y - 1) = 0
    => y = -1/2 or y = 1
    => cosx = -1/2 or cosx = 1
    so x = cos^-1(-1/2) or x = cos^-1(1) .............cos^-1 is cosine inverse. plug this into your calculator to find a value and remember you have to add + kpi for all solutions
    Last edited by Jhevon; March 8th 2007 at 06:39 PM.
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    Ok so I kind of followed your work there. Kinda fell off at => cos2x - cosx = 0
    => cos^2x - sin^2x - cosx = 0. But then at the end if I plug in the values you got and trace them on the original graph of y = sin2x - 2sinx and the points don't match up with where a logical horizontal tangent appears on the graph. I'm probably just doing something wrong but what I can guess is horizontal tangents at 1.11111 and -1.111111 then at 4 other points but inverse cosine of 1 is 0 which on the graph isn't a max or min value. I'm pretty lost I think.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by UMStudent View Post
    Ok so I kind of followed your work there. Kinda fell off at => cos2x - cosx = 0
    => cos^2x - sin^2x - cosx = 0. But then at the end if I plug in the values you got and trace them on the original graph of y = sin2x - 2sinx and the points don't match up with where a logical horizontal tangent appears on the graph. I'm probably just doing something wrong but what I can guess is horizontal tangents at 1.11111 and -1.111111 then at 4 other points but inverse cosine of 1 is 0 which on the graph isn't a max or min value. I'm pretty lost I think.
    i divided both sides of the equation by 2 to get cos2x - cosx = 0
    cos2x = cos^2x - sin^2x is a trig identity.

    it turns out that the values gained for the cos^-1(1) solutions are inflection points. you can check this with the second derivative. it's not wrong, the derivative is zero at inflection points as well, they are technically critical points
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by UMStudent View Post
    Ok so I kind of followed your work there. Kinda fell off at => cos2x - cosx = 0
    => cos^2x - sin^2x - cosx = 0. But then at the end if I plug in the values you got and trace them on the original graph of y = sin2x - 2sinx and the points don't match up with where a logical horizontal tangent appears on the graph. I'm probably just doing something wrong but what I can guess is horizontal tangents at 1.11111 and -1.111111 then at 4 other points but inverse cosine of 1 is 0 which on the graph isn't a max or min value. I'm pretty lost I think.
    i dont know what graph you're looking at, but i plotted sin2x - 2sinx in maple and it seems to confirms my results. r u sure the function wasn't sin^2x - 2sinx?
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  7. #7
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    No I posted the original problem correctly. I'm using my TI-86 to graph the function and by the looks of it I should have 6 horizontal tangents. And when I bring up the menu on the calculator to trace and I enter cos^-1 (1) its 0 and 0 is on a curve not a min or max where the tangent line would be horiztonal. And my calculator is in radians mode so I know thats not the problem.
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  8. #8
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    Quote Originally Posted by UMStudent View Post
    No I posted the original problem correctly. I'm using my TI-86 to graph the function and by the looks of it I should have 6 horizontal tangents. And when I bring up the menu on the calculator to trace and I enter cos^-1 (1) its 0 and 0 is on a curve not a min or max where the tangent line would be horiztonal. And my calculator is in radians mode so I know thats not the problem.
    see the graph at

    graph of y = sin(2x) - 2 sin(x) /
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