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anonymous

  • 5 years ago

please help to find cubic equation when f' (x) =0 at(-1;-3,5) f'(x)<0 for x<-1 and x>2 f'(x)>0 for -1<x<2 f(0) =0 f(-2)=0 and f(4)=0

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  1. anonymous
    • 5 years ago
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    use cardano's method given below

  2. anonymous
    • 5 years ago
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    ok you know your derivative is negative then positive then negative, so it is a parabola facing down. the zeros are -1 and 2 so it looks like \[f'(x)=a(x+1)(x-2)\] where a is some negative number. could even be -1 who knows.

  3. amistre64
    • 5 years ago
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    is the derivative the cubic? it has 3 zeros...

  4. amistre64
    • 5 years ago
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    f' = 0 at -1,-3,5 ?

  5. anonymous
    • 5 years ago
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    \[f'(x)=ax^2-ax-2a\] and therefore \[f(x) = \frac{ax^3}{3}-\frac{ax^2}{2}-2ax+c\]

  6. anonymous
    • 5 years ago
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    good morning.

  7. amistre64
    • 5 years ago
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    howdy :)

  8. anonymous
    • 5 years ago
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    i assume this is looking for the cubic so the derivative is quadratic. with zeros at -1 and 2

  9. amistre64
    • 5 years ago
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    "please help to find cubic equation when f' (x) =0 at(-1;-3,5)" ??

  10. anonymous
    • 5 years ago
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    i am assuming you are looking for the cubic function whose derivative is zero at -1

  11. anonymous
    • 5 years ago
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    in other words \[f(x)=a_3x^3+a_2x^2+a_1x+a_0\]

  12. anonymous
    • 5 years ago
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    \[f(-1)=-3.5\] \[f(0)=0\] which of course means \[a_0=0\]

  13. amistre64
    • 5 years ago
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    a quartic f(x) can have 3 zeros if its a touch and turn, coupled with a thru and thru

  14. amistre64
    • 5 years ago
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    that would have 3 f' zeros then

  15. anonymous
    • 5 years ago
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    it is a cubic, and they have given us the zeros. the original function is a cubic. the derivative is quadratic. the zeros of the original function are 0, -2 and 4, which means it is \[f(x)=a(x+2)(x-4)x\]

  16. amistre64
    • 5 years ago
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  17. anonymous
    • 5 years ago
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    ho ho , that is a 4th degree poly. this one is sposed to be cubic!

  18. amistre64
    • 5 years ago
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    how does a cubic have 3 zeros in the derivative? unless the original question is in error....

  19. anonymous
    • 5 years ago
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    the derivative only has two zeros: -1 and 2

  20. amistre64
    • 5 years ago
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    "please help to find cubic equation when f' (x) =0 at(-1;-3,5" from the original statement ... I find this to either be an error, or the intended question...

  21. anonymous
    • 5 years ago
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    it is a parabola facing down, negative until -1, positive between -1 and 2, then negative

  22. amistre64
    • 5 years ago
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    f' = 0 at -1,-1,5 is my best interp of it.... right or wrong lol

  23. anonymous
    • 5 years ago
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    no it is not error. what it means is that f'(-1) = 0 and f(-1)=-3.5

  24. amistre64
    • 5 years ago
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    ooohhh....... that helps, thnx :)

  25. anonymous
    • 5 years ago
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    now let me see if i can finish this. before the day is over. lol

  26. anonymous
    • 5 years ago
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    ok where were we before we were interrupted?

  27. anonymous
    • 5 years ago
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    oh yes, you have a cubic polynomial with zeros at -2,0, and 4 so it must look like \[f(x)=a(x+2)(x-4)x\] where a is some negative number. negative because from the sign of the derivative we know that this polynomial is decreasing then increasing then decreasing. so we just need to find a

  28. anonymous
    • 5 years ago
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    The question says that f'(x) is 0 at -1,-3 and 5. But f'(-1)=0 yields a=0!

  29. anonymous
    • 5 years ago
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    i guess i cannot read. i read it as \[f'(-1)=0\] at \[(-1,-3.5)\]

  30. anonymous
    • 5 years ago
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    Yes but \[f^\prime(-1)=0 \implies a=0\]

  31. anonymous
    • 5 years ago
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    actually i get that too. damn

  32. anonymous
    • 5 years ago
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    the original quest. says find the equation of the cubic function when f'(x) = 0 at (-1;-3.5) then f'(x)<0 for x<-1 and x>2 f'(x)>0 for -1<x<2 f(0) =0 f(-2)=0 and f(4)=0

  33. anonymous
    • 5 years ago
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    in fact it get \[-6a=-8a\]

  34. anonymous
    • 5 years ago
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    @satellite73 that means a=0.

  35. anonymous
    • 5 years ago
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    think too many constraint to solve this one. yes, know it means a=0 that was what was bothering me.

  36. amistre64
    • 5 years ago
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    if a = 0, just use a different letter lol

  37. anonymous
    • 5 years ago
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    lets go slow. do we agree that \[f(x)=a(x+2)(x-4)x\] ?

  38. anonymous
    • 5 years ago
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    because f is cubic and the roots are given.

  39. anonymous
    • 5 years ago
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    The cubic is \[f(x)=ax(x+2)(x-4)\].

  40. anonymous
    • 5 years ago
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    ok so far so good. and we agree that a < 0 yes?

  41. anonymous
    • 5 years ago
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    thx a mil for the awesome insight, any chance i would be able to graph this equation.....

  42. anonymous
    • 5 years ago
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    why?

  43. anonymous
    • 5 years ago
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    because of the behavior of the derivative: negative then positive then negative means f is decreasing then increasing then decreasing. so a < 0

  44. anonymous
    • 5 years ago
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    to show max an min values and intercepts

  45. anonymous
    • 5 years ago
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    and we also agree that \[f(x)=ax^3-2ax^2-8ax\]

  46. anonymous
    • 5 years ago
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    i know what they want you to graph, but i think this problem is crap. could be wrong. they want you to graph a cubic that is decreasing until -1, then increasing until 2, then decreasing. and they want it to cross the x axis at -2, 0, and 4.

  47. anonymous
    • 5 years ago
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    -4 that is

  48. anonymous
    • 5 years ago
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    no it says 4

  49. anonymous
    • 5 years ago
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    f(4)=0

  50. amistre64
    • 5 years ago
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    when f'(-1) = 0 and f(-3.5)=0 f'(x) = - ; (-inf,-1) U (2,inf) f'(x) = +; (-1,2) <.......-3.5..................-1.....0............2............> + 0 + 0 - - 0 + ????????

  51. amistre64
    • 5 years ago
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    little backwards, but still i quandry

  52. anonymous
    • 5 years ago
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    I think -1,-3.5 and the f'(x) being 0 date are wrong.

  53. anonymous
    • 5 years ago
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    it says f'(-1)=0 it changes direction at (-1,-3.5)

  54. amistre64
    • 5 years ago
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    f'(x)=0 could also indicate an inflection point... right?

  55. anonymous
    • 5 years ago
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    it is entirely possible that this thing is over determined. you have not only the zeros of the cubic, but also the zeros of its derivative. i will try it later because i gotta go, but i think there is a contradiction here.

  56. anonymous
    • 5 years ago
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    no it is not an inflection point because the derivative changes sign there.

  57. amistre64
    • 5 years ago
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    if the slope is the same on both sides of f' = 0, that indicates inflection i believe

  58. amistre64
    • 5 years ago
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    the signs dont change at -3.5

  59. anonymous
    • 5 years ago
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    derivative changes sign. that is what it says.

  60. anonymous
    • 5 years ago
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    is inflexion point noy determind by f''?

  61. amistre64
    • 5 years ago
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    in inflection point is made sure by f''; but an inflection point can still have a 0 slope in f'

  62. anonymous
    • 5 years ago
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    says f'(x)<0 for x<-1 f'(x)>0 for -1<x<2 f"(x)< 0 for x>2

  63. anonymous
    • 5 years ago
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    derivative changes sign so f changes direction from decreasing to increasing. still i cannot come up with a. will try later.

  64. amistre64
    • 5 years ago
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    "when f'(x) = 0 at (-1;-3.5) then f'(x)<0 for x<-1 and x>2 f'(x)>0 for -1<x<2"

  65. anonymous
    • 5 years ago
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    -3.5 is an inflection point..

  66. anonymous
    • 5 years ago
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    thx for all the help, egan helmie, south africa grade 12 -2011-provincial assesment task

  67. amistre64
    • 5 years ago
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    I see f(x) looking somthing like this if i interpret your info correctly

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  68. anonymous
    • 5 years ago
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    oh no amistre, this is not a cubic polynomial. i interpret the question to be for a cubic polynomial. cubics do not look like this. also the second derivative of a cubic polynomial is a line, so only one inflection point, where the line crosses the x - axis (i.e. f''(x)=0). i am thinking this question was made up without sufficient thought. you cannot specify the zeros of a cubic polynomial, its value at some point, AND the zeros of the derivative.

  69. anonymous
    • 5 years ago
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    in fact i assert that there is no polynomial of degree 3 with zeros at -2, 0 and 4 for which f'(-1)=0. let me see if i fall flat on my face with a (rather simple) proof. by the factor theorem we know that \[f(x)=ax(x+2)(x-4)=ax^3-2ax^2-8ax\] thus \[f'(x)=3ax^2-4ax-8a\] and \[f'(-1)=3a+4a-8a\] the only way for this to be 0 is for \[a=0\] which is a contradiction.

  70. amistre64
    • 5 years ago
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    i trust your interpretations and its method; I dont trust that the given information is rigid enough to come to a solid conclusion tho ;)

  71. anonymous
    • 5 years ago
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    it is always possible i misinterpreted the question. but it is also possible some math teacher said "gee i will give them a problem to see if they can graph it" without thinking through whether such a thing exists, and that is the last second i am going to think about this.

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