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Homework Help: Extrema on high degree polynomial

  1. Mar 13, 2010 #1
    1. The problem statement, all variables and given/known data

    Find any extrema, points of inflection, asymptotes, and symmetry for function.

    2. Relevant equations

    f(x) = (x^5-10x^3+9x) / ( x^4 - 16)

    3. The attempt at a solution


    Extrema: I took the first derivative by using the Quotient Rule, and got

    (x^8 + 10x^6 - 107x^4 + 480x^2 - 144) / ( x^4 - 16)^2

    I know that to find an extrema, I need to determine the critical numbers. Which are when f ' is equal to 0 or is undefined. I determined that "2" makes f ' undefined, but it also is not defined in the original function, f(x), so that is not a critical number. But I cannot for the life of me figure out how to factor the numerator when set to 0.

    I tried to graph f ' , and it seems like x = 0 is a critical number, but when i plug it into the numerator it gives me -144...I feel like I am missing something, can someone please help me figure out how to determine the critical numbers please? I think I'm having more algebra issues than calculus.

    Then there's the possibility that I took the wrong first derivative. If someone could check me on that, I would be thankful.
     
  2. jcsd
  3. Mar 13, 2010 #2

    Mark44

    Staff: Mentor

    Your derivative is incorrect. I get
    f'(x) = (5x^8 - 4x^7 + 10x^6 - 107x^4 + 480x^2 - 144)/(x^4 - 16)^2

    Edit: My mistake. My brain misfired when I added 5 and 3 and got 7. It should be
    f'(x) = (x^8 + 10x^6 - 107x^4 + 480x^2 - 144)/(x^4 - 16)^2
    f is undefined at x = 2 and x = -2, which you can see by factoring the denominator in f(x).
    It might be helpful to sketch a graph of the function first. That way you could get an idea of approximately where the minima and maxima are. The numerator of your function factors easily, making it easy to find the five x-intercepts for the graph of the function. The denominator also factors easily. Since there are no factors in common between the numerator and denominator, there will be four vertical asymptotes.

    If you divide the numerator by the denominater, you get x + a proper rational function, which means that there is a slant asymptote (i.e., the graph of the function eventually approaches the graph of y = x).

    To find the zeroes of the numerator of f'(x), you'll probably need to use the rational root theorem. In a polynomial anxn + ... + a1x + a0 = 0, any rational roots p/q are such that p divides a0 and q divides an. For your problem, the polynomial you're trying to factor is 5x^8 - 4x^7 + 10x^6 - 107x^4 + 480x^2 - 144, so p has to divide 144 and q has to divide 5. Fortunutely 5 has factors only of +/-1 and +/-5. If you drew a graph of y = f(x), there will be many potential candidates that you won't need to check.


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    Last edited: Mar 13, 2010
  4. Mar 13, 2010 #3

    phyzguy

    User Avatar
    Science Advisor

    It looks like you took the first derivative correctly. Try plotting out f and f' to identify the extrema, points of inflection, etc.
     
  5. Mar 13, 2010 #4
    For what it's worth, I got the same first derivative as you did.

    y'=[(x^4-16)(5x^4-30x^2+9) - (x^5-10x^3+9x)(4x^3)]/(x^4-16)^2

    =(5x^8-30x^6-71x^4+480x^2-144-4x^8+40x^6-36x^4)/(x^4-16)^2

    =(x^8+10x^6-107x^4+480x^2-144)/(x^4-16)^2
     
  6. Mar 13, 2010 #5

    Mark44

    Staff: Mentor

    phyzguy and JOhnJDC, you are correct.
     
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