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anonymous

  • 5 years ago

determine the location of each local extrenum of the function f(X)= x^4/4 - 8X^3/3 +19X^2/2 -12X +2

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  1. anonymous
    • 5 years ago
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    take the derivative, set = 0, and solve. The derivative is a cubic polynomial, not easy to find the roots, but you can see by inspection that 1 is a solution. Factor (x-1) from your derivative so get \[f'(x)=(x-1)(x^2-7x+12 )= (x-1)(x-4)(x-3))\] therefore zeros are -1,3 and 4. -1 and 4 are relative min, as you can see if you draw a picture of a fourth degree polynomial with positive leading coefficient, and 3 is a relative max

  2. anonymous
    • 5 years ago
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    oops my typo. zeros are 1, 3 and 4, not -1!

  3. anonymous
    • 5 years ago
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    that's my problem i'm having trouble setting it to f' how did u get (x-1)(x^2-7x+12)??

  4. anonymous
    • 5 years ago
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    that's my problem i'm having trouble setting it to f' how did u get (x-1)(x^2-7x+12)??

  5. anonymous
    • 5 years ago
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    Find the derivative of each term. After that factor, like in college algebra or pre-cal.

  6. anonymous
    • 5 years ago
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    ok

  7. anonymous
    • 5 years ago
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    I got it by cheating. It is hard to find the zeros of a cubic polynomial but I figured that this problem was easy so I guessed 1. I checked that f'(1) = 0. Now "factor theorem" says if r is a zero of a polynomial p(x) then p(x) = (x-r)q(x) so I knew that f'(x) = (x-1) (something) I found the "something" by synthetic division, but you can do long division as well. http://www.youtube.com/watch?v=bZoMz1Cy1T4 Once you have a quadratic you can always find the zeros using the formula but this one factored easily.

  8. anonymous
    • 5 years ago
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    ok thanx:)

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