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Calculus - behaviour of functions - first derivative and the likes

  1. Nov 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Given P(x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real root of P'(x) = 0. If P(–1) < P(1), then in the interval [–1, 1]

    (1) P(–1) is not minimum but P(1) is the maximum of P
    (2) P(–1) is minimum but P(1) is not the maximum of P
    (3) Neither P(–1) is the minimum nor P(1) is the maximum of P
    (4) P(–1) is the minimum and P(1) is the maximum of P


    2. Relevant equations

    The answer is 1.

    3. The attempt at a solution

    • P(x) = x4 + ax3 + bx2 + cx + d
    • P′ (x) = 4x3 + 3ax2 + 2bx + c
    • P′ (0) = 0
    • c = 0

    • P′ (x) = 0 only at x = 0
    • P′ (x) is a cubic polynomial changing its sign from (–)ve to (+)ve and passing through O.
    • P′ (x) < 0 ∀ x < 0
    • P′ (x) > 0 ∀ x > 0
    • graph of P(x) is upward concave, where P′ (x) = 0

    Now P(–1) < P(1)
    ⇒ P(–1) cannot be minimum in [–1, 1] as minima in this interval is at x = 0.

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    The question is solved. That is not the problem. What I am wondering about right now is that such a logical question is too precious to be let go of without a discussion. Do you have any comments regarding either the geometrical interpretation or logical deduction of the answer?

    For me, the questions are getting monotonous by the day and I have almost lost the thrill of learning because now the questions are just too clear. However, I am sure you will have a lot to discuss about it, right?
     
  2. jcsd
  3. Nov 1, 2009 #2
    If anyone of you is a teacher - you must have had some sort of a frustrating experience on how the crystal clear logic doesn't even appeal to the students.
     
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