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Finding points on a curve where the tangent slope = 6

  1. Oct 31, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the x values of the points on the curve f(x) = (x^3) + (3x^2) + 3x + 6 where the tangent has a slope equal to m = 6

    2. Relevant equations

    3. The attempt at a solution

    This question was on an online quiz for my into calculus course... can't seem to wrap my head around it.

    I differentiated:
    f'(x) = 3(x^2) + 6x + 3

    Set it equal to 6:

    3(x^2) + 6x + 3 = 6

    Then attempted to factor and solve for x, in different ways:

    3(x^2) + 3x + 3x + 3 = 6
    3x(x+1) + 3(x+1) = 6
    (3x+3)(x+1) = 6
    3x+3=6
    x=(6-3)/3
    x=1
    x+1=6
    x=5

    I also tried:
    3(x^2) + 6x + 3 = 6
    3((x^2) + 2x + 1) = 6
    3(x+1)(x+1) = 6
    (x+1)(x+1) = 6/3
    x+1 = 2
    x = 1

    Needless to say I got the problem wrong... the correct answers were given: -1 + sqrt(2), and -1 - sqrt(2).

    I would appreciate any kind of help or pointers... I just can't seem to get this straight. What am I missing??
     
  2. jcsd
  3. Oct 31, 2009 #2
    I have no idea what you are doing with that factoring! You can only factor when one side is set equal to zero. (Remember we can solve by factoring because ab=0 implies that a=0 or b=0 ... but ab=6 will NOT imply a=6 or b=6!)

    You have:

    6 = 3x^2 +6x +3

    Set it equal to zero, like this:

    0 = 3x^2 +6x - 3

    But this will not factor... so, use the quadratic formula.
     
  4. Oct 31, 2009 #3
    Thank you, I honestly had no idea what I was doing either. Appreciate the help.
     
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