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Coordinates at which 2 tangent lines of a function pass through a single point

  1. Nov 11, 2007 #1
    Here is what the question asks:

    Find the coordinate of all points in the graph of y=3e^x - x^3 at which the tangent line passes through the point (1,0).


    I am told that the 2 points are (-.872, -1.027) and (2.275, 13.657). Any help at all that you can give will be very much appreciated.

    I was thinking of letting the 2 coordinates be (X1, Y1) and (X2, Y2), calculating the slope of the line through (1,0) and one of these points, and setting this slope equal to the derivative of y=3e^x - x^3 ......but I seem to be going no where.

  2. jcsd
  3. Nov 11, 2007 #2


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    Yes, the slope of the derivative would be the slope of the tangent at what point on the line?
  4. Nov 11, 2007 #3


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    What's the equation for the tangent line at the point (a,f(a))?

    By the way. This final expression is nasty to solve. I think you'll need a computer to evaluate the answers numerically in the end.
  5. Nov 11, 2007 #4
    Alright -- the final equation i get is 6e^x + 2x^3 - 3x^2 - 3xe^x = 0

    The 2 answers that the book gives work in this equation......but how the hell would I solve this?

    If I were to set 6e^x + 2x^3 - 3x^2 - 3xe^x equal to y in my calculator and graph it, is there a way for the calculator to tell you the x-intercepts?

    EDIT - Never mind - I figure it out! Thanks for helping out.
    Last edited: Nov 11, 2007
  6. Nov 11, 2007 #5


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    Newton's Method?
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