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Finding max. and min. values analytically

  1. Dec 6, 2009 #1
    1. The problem statement, all variables and given/known data
    y = 2x^2-14x


    2. Relevant equations



    3. The attempt at a solution
    y' = 4x - 14
    4x - 14 = 0
    x = 14/4
    f(14/4) = 2(14/4)^2 - 14 (14/4) = -49/2

    I don't know how to continue from there. ((14/4), (-49/2)) is my critical point?
     
  2. jcsd
  3. Dec 6, 2009 #2

    there are a couple of steps involved with finding local min/max's:

    1) differentiate
    2) find f'(x) = 0
    3) take values larger and smaller than x | f'(x) = 0 to see if it is a min/max [note: you must do this because you may have f'(x) = 0 without being a min/max]
    min: \ /
    max : / \
    meaning, values less than x | f'(x) = 0 will have negative derivatives if it is a minimum value.. et c .
     
    Last edited: Dec 6, 2009
  4. Dec 6, 2009 #3

    Mark44

    Staff: Mentor

    You can also use the second derivative to determine a local maximum or local minimum. At any critical point c, if f''(c) > 0, there is a local minimum at (c, f(c)). If f''(c) < 0, there is a local maximum at (c, f(c)).

    For your particular function, you could also notice that its graph is a parabola that opens upward. The only critical point is at the vertex of the parabola.

    BTW you should simplify 14/4 to 7/2.
     
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