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Math Help - Challenging Calculus question

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    Challenging Calculus question

    Suppose that the tangent at a point P on the curve y=x^3 intersects the curve again at a point Q. Show that the slope of the tangent at Q is four times the slope of the tangent at P.

    I have no idea where to start except to take the derivative of y=x^3
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  2. #2
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    Quote Originally Posted by pluvam View Post
    Suppose that the tangent at a point P on the curve y=x^3 intersects the curve again at a point Q. Show that the slope of the tangent at Q is four times the slope of the tangent at P.

    I have no idea where to start except to take the derivative of y=x^3
    I like to draw a picture to help me see what is going on.

    Challenging Calculus question-graph.jpg

    So you know the derivative is

    y'=3x^2

    So the slope at the point (a,a^3) is

    m_a=3a^2

    So the equation of the tangent line is

    y-a^3=3a^2(x-a) \iff y=3a^2x-2a^3

    but we know that the tangent line intersects the curse again at some point (b,b^3)

    So we know that

    y=3a^2b-2a^3 and that y=b^3

    So setting the two above equations equal gives

    b^3=3a^2b-2a^3 \iff b^3-3a^2b+2a^3=0

    This is cubic in b which can make it difficult to solve but know that this is trivially true when a=b so b-a must be a factor of the equation by long division we get
    (b-a)(b^2+ab-2a^2) this can be factored to gives

    (b-a)(b-a)(b+2a) so the only non trivial root is

    b=-2a

    Now just plug the points into the derivative to get

    y'(a)=\textcolor{red}{3a^2};y'(b)=y'(-2a)=3(-2a)^2=12a^2=4\textcolor{red}{(3a^2)}
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    Thank you so much
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