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anonymous

  • 5 years ago

what are the local minimum an maximum pints of this equation -1.28x4 + -33.36x3 + -323.41x2 + -1384.63x + -2206.60 ?

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  1. anonymous
    • 5 years ago
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    hi are you there? if this is a y or f(x) then differentiate it then equaled it to zero and find its roots, that rot will be your max or min point..just plug it into your original equation....

  2. anonymous
    • 5 years ago
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    yeah hehe. kay thanks (:

  3. anonymous
    • 5 years ago
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    your welcome

  4. anonymous
    • 5 years ago
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    the derivative would be like y'=-5.12x^3 -100.8x^2 -646.82X -1384.63,,....NOW FIND ITS ROOTS OR ZEROES

  5. anonymous
    • 5 years ago
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    TO FIND THE ROOTS HERE I THNK ITS EASIER TO GRAPH IT...OR USE SYNTHETIC DIVISION

  6. anonymous
    • 5 years ago
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    doing synthetic division now...big numbers ):

  7. anonymous
    • 5 years ago
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    lets try graphing the y'=-5.12x^3 -100.8x^2 -646.82X -1384.63,,...

  8. anonymous
    • 5 years ago
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    can you tell what the asymptotic behavior with this equation is?

  9. anonymous
    • 5 years ago
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    ok this is quartic equation,, so you have ti look for the family of curves of all the quartic equations behaviour....its like a sine wave but not quite as good as sine wave it wil touch the grap at x, 4 times in the graph

  10. anonymous
    • 5 years ago
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    it will have 4 roots, and it will touch the x axis four times

  11. anonymous
    • 5 years ago
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    can i post the graph of the equation?

  12. anonymous
    • 5 years ago
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    did you graph it yet? the derivsative graph?

  13. anonymous
    • 5 years ago
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    well it is a parent function picture so i took a picture with the shape of a quartic function and plotted points on the graph

  14. anonymous
    • 5 years ago
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    ok get its roots and plug it in the original equation

  15. anonymous
    • 5 years ago
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    are you familiar with newtons method of approximation?

  16. anonymous
    • 5 years ago
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    if you have a calculus book its in there also its in the web

  17. anonymous
    • 5 years ago
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    no whats that?

  18. anonymous
    • 5 years ago
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    newtons method of aprox is good in finding he rots of equations..

  19. anonymous
    • 5 years ago
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    oh well what do you have to do?

  20. anonymous
    • 5 years ago
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    this is the graph

  21. anonymous
    • 5 years ago
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    http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/approx/newton.html

  22. anonymous
    • 5 years ago
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    http://en.wikipedia.org/wiki/Newton%27s_method

  23. anonymous
    • 5 years ago
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    y'=-5.12x^3 -100.8x^2 -646.82X -1384.63,,....NOW FIND ITS ROOTS OR ZEROES ..to find its roots use newton method of approx.here you wil find 3 roots,then this 3 roots you will then plug them into your origimal equation to find you max y or min y

  24. anonymous
    • 5 years ago
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    did you see the graph?

  25. anonymous
    • 5 years ago
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    it didnot download in my laptop bec my c is full..here i can help u with newtons method,did you read the website i gave u?

  26. anonymous
    • 5 years ago
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    I read the wikipedia one

  27. anonymous
    • 5 years ago
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    newtons formula is x2=xi-[f(x1)/f'(x1)]

  28. anonymous
    • 5 years ago
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    how do I get the actual coordinates?

  29. anonymous
    • 5 years ago
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    here your f(x)=5.12x^3-100.8x^2-646.82x-1384.63 now use derivative to find f'(x)?

  30. anonymous
    • 5 years ago
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    f'(x)=15.36x^2-201.6x-646.82 now you can use newtons method formula to find the roots of f(x)=5.12x^3-100.8x^2-646.82x-1384.63

  31. anonymous
    • 5 years ago
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    newton formula is x2=x1-[f(x1)/f ' (x)] start with x1 like -8, -9, -7, hope you are using a good calculator with this calculation

  32. anonymous
    • 5 years ago
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    using x2=x1- [(5.12x^2-100.8x^2-646.82x-1384.63)/(15.36x^2-201.6x-646.82)] i got a root of x=25.13527 now you can plug it on you original equation

  33. anonymous
    • 5 years ago
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    x2=x1- [(5.12x^3-100.8x^2-646.82x-1384.63)/(15.36x^2-201.6x-646.82)] im sorry that was 5.12x^3

  34. anonymous
    • 5 years ago
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    just plug the root x =25.13527 here, f(x)=-1.28x4 + -33.36x3 + -323.41x2 + -1384.63x + -2206.60

  35. anonymous
    • 5 years ago
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    and that should get me my minimum/maximum?

  36. anonymous
    • 5 years ago
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    f(25.13527)=-1.28x^4 + -33.36x^3 + -323.41x^2 + -1384.63x + -2206.60= -789369.6612 yes this is one of the minimum y..........you will have to find 3 more roots......

  37. anonymous
    • 5 years ago
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    is this your college subject calculus?

  38. anonymous
    • 5 years ago
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    im in high school pre calculus!

  39. anonymous
    • 5 years ago
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    i think in this problem they want you to learn how to graph this function, and to learn how to get the roots of the first derivative.....then to find its max or min y you will have to use the root to find the min n the max

  40. anonymous
    • 5 years ago
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    i wish i could look at the graph you want to show me, is it the graph of y=f(x)=-1.28x^4 + -33.36x^3 + -323.41x^2 + -1384.63x + -2206.60..or the graph of y'=f '(x)=5.12x^3-100.8x^2-646.82x-1384.63 ..trace it at x=0, then you can find the y axis which is max or min

  41. anonymous
    • 5 years ago
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    its the first one

  42. anonymous
    • 5 years ago
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    yes only the first one bec x^4 is quartic therefore it must have 4 roots and 4 max or min y...correction on the first root i check it again on my calculator it is x=25.14043077

  43. anonymous
    • 5 years ago
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    ook i have another question... how do i know which one is the positive/negative/complex roots?

  44. anonymous
    • 5 years ago
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    so when you plug this x=25.14043077 to the orig eq f(25.043077)= -1282838.114

  45. anonymous
    • 5 years ago
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    ypu notice that this is neg 1.million which is hard to graph,,therefore you will need to use the newton method formula..thats the conclusion here,,,you will be using the newton formula again and guess again for the roots

  46. anonymous
    • 5 years ago
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    oh well thanks soo much for your help i really appreciate it!

  47. anonymous
    • 5 years ago
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    hope you will follow the procedure again and again to find the other 3 roots....good luck ....is this in college subject? or ap calculus in high school?

  48. anonymous
    • 5 years ago
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    umm i go to an option school, and we dont really have ap classes, but there is advanced math...thats what i take

  49. anonymous
    • 5 years ago
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    oh ok....whews this is really for college problem hehehe thats really advance calculus

  50. anonymous
    • 5 years ago
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    ook i have another question... how do i know which one is the positive/negative/complex roots? if you remember the quadratic formula x= [-b+-sqrt(b^2-4ac)/2a]...the equation is y=ax^2+bx+c

  51. anonymous
    • 5 years ago
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    positive complex root is like x=(2+sqrt-4)/2 = (2+2i)/2= 1+i did you noticed the plus or positive sign? that a positive complex root..for conjugate or neg complex root its x=1-i

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