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Find the points on the graph

  1. May 29, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the points on the graph of y = x3ex where the tangent line is horizontal.
    (Your answer should include x-values and y-values)


    2. Relevant equations

    Product rule

    fg'+gf'


    3. The attempt at a solution

    (x3)(ex)+(ex)(3x2)

    x3ex+3x2ex= y'

    x3ex+3x2ex= 0

    x2ex(x+3) = 0

    x = -3

    Going back to original equation, and plugging in x:

    y = -33e-3

    y = -27e-3

    y ≈-1.34

    Not sure if I did this correctly.
     
  2. jcsd
  3. May 29, 2013 #2

    Mark44

    Staff: Mentor

    Looks fine, but you should provide a summary of what you found - something like this: The only point with a horizontal tangent is (-3, -27e-3), or approximately (-3, -1.34).
     
  4. May 29, 2013 #3
    Okay. Also, with

    x2ex(x+3) = 0

    x = -3 but does it also = 0? Graphing it, it looks like I could include that value as well.
     
  5. May 29, 2013 #4

    Mark44

    Staff: Mentor

    Sorry, I overlooked that factor of x2, so yes, your function has a horizontal tangent at (0, 0).
     
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