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Grade 12 Calculus Problem - Differentiation and Division

  1. Aug 13, 2011 #1

    G-S

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    1. The problem statement, all variables and given/known data

    Let p be a polynomial function with p(a)= 0 =p'(a) for some real a. Which of the following must be true?

    A) p(x) is divisible by x+a
    B) p(x) is divisible by x^2+a^2
    C) p(x) is divisible by x^2-a^2
    D) p(x) is divisible by x^2+2ax+a^2
    E) p(x) is divisible by x^2-2ax+a^2

    2. Relevant equations



    3. The attempt at a solution

    I'm honestly beyond stumped for this one. Having a hard time finding out where to start on this problem.
     
  2. jcsd
  3. Aug 13, 2011 #2

    SammyS

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    What does it mean for a polynomial (or any function) if p(x) = 0 at x=a ?

    What does it mean for a polynomial, if p'(x) = 0 at x=a ?

    Put the answers together.
     
  4. Aug 13, 2011 #3

    vela

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    Hint: p(x) is a polynomial and p(a)=0 means a is a root of the polynomial. If you were to factor p(x), what factor do you know has to be there if a is a root?
     
  5. Aug 13, 2011 #4

    G-S

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    Thanks for the replies.
    So this is the conclusion I've come to so far,

    p(x)=0 at x=a
    p'(x)=0 at x=a
    a is a root of the polynomial

    So,
    x2-a2=0
    x2=a2
    x=a

    Therefore, C) must be true.

    Am I in the right direction?
     
  6. Aug 13, 2011 #5

    SammyS

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    No, unless you know that -a is also a root.

    What is true about a function if it's first derivative is zero at some value of x ?
     
  7. Aug 13, 2011 #6

    G-S

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    Sorry SammyS but the only thing that comes to mind is the function is a constant. Either that or the first step to find critical points.

    After looking at the problem again I've come to the conclusion that in order to fulfill the condition p(a)=0 only C) and E) can be true.

    Considering that we only know that a is a root (and not -a) E) seems like the only one that fulfills the conditions.

    0=x2-2ax+a2
    0=(x-a)2
    0=x-a
    x=a

    Taking stabs in the dark here.
     
  8. Aug 13, 2011 #7
    You have narrowed down the list of choices to just E, so E must be the answer. Is there anything more you need to know?
     
  9. Aug 13, 2011 #8

    G-S

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    I'm trying to get confirmation as to whether the answer and my thought process is correct and if not, how one would approach this question.
     
  10. Aug 13, 2011 #9

    SammyS

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    And, in general, why is it that you look for critical points?

    Also, what is the slope of a constant function? (I'm assuming that this is not a constant polynomial.)
     
  11. Aug 13, 2011 #10

    G-S

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    To find max/min values. And the slope of a constant function is 0.
     
  12. Aug 13, 2011 #11

    SammyS

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    More to the point, what is the slope of the line tangent to the polynomial at x = a, if p'(a) = 0 ?

    This is the question I should have asked in my previous post.
     
  13. Aug 13, 2011 #12

    G-S

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    The slope of the tangent line is 0 if p'(a)=0.
     
  14. Aug 13, 2011 #13

    SammyS

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    The behavior of a polynomial in the neighborhood of one of its zeros is due mainly to the factor which 'causes' the zero, and the multiplicity of that zero.

    So, in the neighborhood of x=a, the polynomial, p(x), behaves like ±(x - a)n, where n ≥ 2 .

    How do we know it's not like ±(x - a)1 ?
     
  15. Aug 13, 2011 #14

    G-S

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    Because it's not one of the options? Haha, I haven't got a clue.
     
  16. Aug 13, 2011 #15

    SammyS

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    The derivative of (x - a) ≠ 0 at x = 1 .
     
  17. Aug 14, 2011 #16

    HallsofIvy

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    I presume you mean at x= a.
     
  18. Aug 14, 2011 #17

    SammyS

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    Yes. Thanks for the correction.
     
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