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Math Help - Differentiation and Geometry

  1. #1
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    Differentiation and Geometry

    Hi, I ve been trying to solve this question for days, but my answer is not the same as in the book; could someone help, plz?!!

    Q: Find the point(s) on the curve: y=x^3+2x+2 whose tangent line is parallel to this line: 3x-y=2

    Thanksssss
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Slope of the line 3x-y=2 is 3.

    y=x^3+2x+2

    \frac{dy}{dx}=3x^2+2

    \frac{dy}{dx}=3

    So, 3x^2+2=3

    3x^2=1

    x=\pm\frac{1}{\sqrt{3}}=\pm0.58

    The points on the curve: y=x^3+2x+2 whose tangent line is parallel to this line: 3x-y=2 are (0.58, 3.35) and (-0.58, 0.645).
    Last edited by alexmahone; July 3rd 2009 at 10:58 PM.
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  3. #3
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    based on this, the points will be where x= +/- (1/sqrt 3)??!!
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  4. #4
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    Quote Originally Posted by fadeer View Post
    based on this, the points will be where x= +/- (1/sqrt 3)??!!
    Yes.
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  5. #5
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    Thankz

    This means the text book is wrong then.
    Tnx very very much..
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  6. #6
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    Quote Originally Posted by fadeer View Post
    This means the text book is wrong then.
    Tnx very very much..
    What does the textbook say?
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  7. #7
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    It says:

    x= +/- (1/sqrt2)
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  8. #8
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    Quote Originally Posted by fadeer View Post
    It says:

    x= +/- (1/sqrt2)
    Are you sure you gave us the right question (like not mistyping anything)?
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  9. #9
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    sure

    oh yes I am sure, this book has a lot of mistakes, this was not the first nor the last mistake.
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