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Thread: 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

    $\displaystyle y= ax^3+bx^2+cx+d$

    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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    $\displaystyle y = ax^3 + bx^2 + cx + d$

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

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

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

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

    $\displaystyle \therefore b = 0$.

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

    Substituting in the values:
    $\displaystyle
    12a + c = 0$ (b is gone as it equals zero)

    $\displaystyle \Rightarrow c = -12a$

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

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

    Now we get $\displaystyle y = ax^3 - 12ax$.

    You know that at $\displaystyle x = 2$, $\displaystyle y = 4$. 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 $\displaystyle a,b,c,d$ so that the graph of: .$\displaystyle f(x) \:=\:ax^3+bx^2+cx+d$

    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: .$\displaystyle f(0) = 0$
    . . We have: .$\displaystyle a\!\cdot0^3 + b\!\cdot\!0^2 + c\!\cdot\!0 + d \:=\:0 \quad\Rightarrow\quad\boxed{d \:=\:0}$
    The function is: .$\displaystyle f(x) \:=\:ax^3+bx^2 + cx$


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


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


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

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

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

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


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

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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

    $\displaystyle y= ax^3+bx^2+cx+d$

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

    $\displaystyle y = ax^3 + bx^2 + cx$

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

    $\displaystyle 4 = 8a + 4b + 2c$

    $\displaystyle 2 = 4a + 2b + c$


    $\displaystyle y' = 3ax^2 + 2bx + c$

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

    $\displaystyle y' = 0$ at $\displaystyle x = 2$

    $\displaystyle 0 = 12a + 4b + c$


    $\displaystyle y'' = 6ax + 2b$

    $\displaystyle y'' = 0$ at $\displaystyle x = 0$

    $\displaystyle 0 = 2b$

    $\displaystyle b = 0$

    so ... now you have the two equations

    $\displaystyle 12a + c = 0$

    $\displaystyle 4a + c = 2$

    finish up by solving for a and c.
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