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Finding x-coordinates

  1. May 15, 2009 #1
    1. The problem statement, all variables and given/known data
    hi everyone ,, got this question and I need the idea today if possible:

    Find the x-coordinate of the point on the graph of y = x^3 where
    the tangent line is parallel to the secant line that cuts the curve at x = -1 and x = 1.

    2. Relevant equations


    3. The attempt at a solution
    i got the derivative and then the tangent line equaled m=3 and then i got this equation:
    y=3x+2 so my x coordinates are (1,5) (-1,-1) ,, and I think it's wrong because (-1,-1) is not parallel ,, so how to get it and then what should i do to get the coordinates ,,
     
    Last edited: May 15, 2009
  2. jcsd
  3. May 15, 2009 #2

    Mark44

    Staff: Mentor

    Nope. Your line has slope m = 3, which is not equal to the slope of the secant line. What are the coordinates of the points on the graph of y = x^3 when x = 1 and when x = -1?

    There are two points of the graph of y = x^3 with tangent lines that are parallel to the given secant line.
     
  4. May 15, 2009 #3
    yea,, got it ,, m =1 not 3 ,, and then i should y`=3x^2
    3x^2=1 then I'll get the coordinates ,, x=+-sqrt(1/3) ,, thanks :D
     
  5. May 15, 2009 #4

    Mark44

    Staff: Mentor

    Keep in mind that 3x^2 = 1 has two solutions.

    I hope that you have drawn a graph of the function you're working with...
     
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