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Math Help - Differentiation help

  1. #1
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    Differentiation help

    I have a problem with this exercise:

    "The curve with equation: y = ax^3 + bx^2 + cx + d has a local maximum point where x=-3 and a local minimum point where x=-1. If the curve passes through the points (0;4) and (1;20), find the equation for y in terms of x"

    I get that "d = 4"; "a + b + c=16"; "27a - 6b + c" and "3a - 2b + c" both are equal to 0; and that "2b - 18a is less than zero" while "2b - 6a is greater than zero"

    But I'm stuck, and can't seem to find the values for a;b and c.

    Please help.
    Thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by johnnyboy View Post
    I have a problem with this exercise:

    "The curve with equation: y = ax^3 + bx^2 + cx + d has a local maximum point where x=-3 and a local minimum point where x=-1. If the curve passes through the points (0;4) and (1;20), find the equation for y in terms of x"

    I get that "d = 4"; "a + b + c=16"; "27a - 6b + c" and "3a - 2b + c" both are equal to 0; and that "2b - 18a is less than zero" while "2b - 6a is greater than zero"

    But I'm stuck, and can't seem to find the values for a;b and c.

    Please help.
    Thanks

    we know that y passes through (0,4) and (1,20)
    y = ax^3 + bx^2 + cx + d

    => 4 = a(0)^3 + b(0)^2 + c(0) + d
    => d = 4

    => 20 = a(1)^3 + b(1)^2 + c(1) + d
    => 20 = a + b + c + 4
    => 16 = a + b + c
    that is, a + b + c = 16
    call this equation (1)

    so a + b + c = 16 .....................(1)


    we also know that the derivatve gives the slope at any value of x right? so let's start there

    y = ax^3 + bx^2 + cx + d
    => y' = 3ax^2 + 2bx + c
    we have extrema occuring at the points where y' = 0

    set y' = 0
    => 3ax^2 + 2bx + c = 0

    we know that x = -1 and x = -3, so we have the simultaneous equations:

    3a(-1)^2 + 2b(-1) + c = 0 ....................(2)
    3a(-3)^2 + 2b(-3) + c = 0 ....................(3)

    3a - 2b + c = 0 ....................(2)
    27a - 6b + c = 0 ..................(3)

    so we know d, to find a,b, and c we must solve the system:

    a + b + c = 16 .....................(1)
    3a - 2b + c = 0 ....................(2)
    27a - 6b + c = 0 ..................(3)

    => 2a - 3b = -16 ...................(4) = (2) - (1)
    => 24a - 4b = 0 .....................(5) = (3) - (2)

    => 24a - 36b = -192 ...............(4)*12
    => 24a - 4b = 0 .....................(5)

    => 32b = 192 .........................(5) - (4)
    => b = 6

    but 2a - 3b = -16
    => 2a - 18 = -16
    => 2a = 2
    => a = 1

    but a + b + c = 16
    => 1 + 6 + c = 16
    => c = 16 - 7 = 9

    so we have a = 1, b = 6, c = 9, d = 4

    so the curve is y = x^3 + 6x^2 + 9x + 4
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  3. #3
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    thanks a lot. It seems so obvious now
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