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Math Help - polynomial caculus

  1. #1
    Member Rimas's Avatar
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    polynomial caculus

    Determine a,b,c and d so that the graph of

    y= ax^3+bx^2+cx+d

    has a point of inflection at the origin and a relative maximum at the point (2,4)
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  2. #2
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    y = ax^3 + bx^2 + cx + d

    so \frac{dy}{dx} = 3ax^2 + 2bx + c

    and \frac{d^2y}{dx^2} = 6ax + 2b

    For a point of inflection, \frac{d^2y}{dx^2} = 0.

    So 6ax + 2b = 0 at x = 0 (origin)

    \therefore b = 0.

    Now at (2,4) \frac{dy}{dx} = 0 as we have a local maximum, and we must make sure \frac{d^2y}{dx^2} < 0 so it is a max and not a min.

    Substituting in the values:
    <br />
12a + c = 0 (b is gone as it equals zero)

    \Rightarrow c = -12a

    Now, we know that y = ax^3 - 12ax + d

    We know that at x = 0, y = 0, so you can see d = 0.

    Now we get y = ax^3 - 12ax.

    You know that at x = 2, y = 4. Simply substitute these values into the equation to find a. Then you can find c.
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  3. #3
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    Hello, Rimas!

    Determine a,b,c,d so that the graph of: . f(x) \:=\:ax^3+bx^2+cx+d

    has a point of inflection at the origin and a relative maximum at the point (2, 4)
    Since the origin (0,0) is on the graph: . f(0) = 0
    . . We have: . a\!\cdot0^3 + b\!\cdot\!0^2 + c\!\cdot\!0 + d \:=\:0 \quad\Rightarrow\quad\boxed{d \:=\:0}
    The function is: . f(x) \:=\:ax^3+bx^2 + cx


    The first derivative is: . f'(x) \:=\:3ax^2 + 2bx + c
    The second derivative is: . f''(x) \:=\:6ax + 2b


    Since the origin is an inflection point: . f''(0) \,=\,0
    . . We have: . f''(0) \:=\:6a\!\cdot\!0 + 2b \:=\:0\quad\Rightarrow\quad\boxed{b \:=\:0}
    The function is: . f(x) \:=\:ax^2 + cx


    Since (2, 4) is on the graph: . f(2) \,=\,4
    . . f(2) \:=\:8a + 2c \:=\:4 \quad\Rightarrow\quad c \:=\:2 - 4a .[1]

    Since (2, 4) is a critical point: . f'(2) \,=\,0
    . . We have: . f'(2) \:=\:3a\!\cdot\!2^2 + c \:=\:0 \quad\Rightarrow\quad c \:=\:-12a .[2]

    Equate [1] and [2]: . 2 - 4a \:=\:-12a \quad\Rightarrow\quad\boxed{ a \:=\:\text{-}\tfrac{1}{4}}

    Substitute into [2]: . c \:=\:-12\left(\text{-}\tfrac{1}{4}\right) \quad\Rightarrow\quad\boxed{ c \:=\:3}


    Therefore: . f(x) \:=\:-\frac{1}{4}x^3 + 3x

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  4. #4
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    Quote Originally Posted by Rimas View Post
    Determine a,b,c and d so that the graph of

    y= ax^3+bx^2+cx+d

    has a point of inflection at the origin and a relative maximum at the point (2,4)
    since the curve passes through the origin, d = 0.

    y = ax^3 + bx^2 + cx

    curve also passes through the point (2,4) ...

    4 = 8a + 4b + 2c

    2 = 4a + 2b + c


    y' = 3ax^2 + 2bx + c

    since y has a relative max at (2,4) ...

    y' = 0 at x = 2

    0 = 12a + 4b + c


    y'' = 6ax + 2b

    y'' = 0 at x = 0

    0 = 2b

    b = 0

    so ... now you have the two equations

    12a + c = 0

    4a + c = 2

    finish up by solving for a and c.
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