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Thread: Differentiation Problem!!

  1. #1
    Member looi76's Avatar
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    Differentiation Problem!!

    Question:
    Find the coordinates of the stationary points on the graph of the following functions, and find weather these points are maxima or minima.
    $\displaystyle (a) 2x^3+3x^2-72x+5$

    Attempt:
    $\displaystyle f(x) = 2x^3+3x^2-72x+5$
    $\displaystyle f'(x) = 6x^2+6x-72$
    $\displaystyle f'(x) = 0$

    $\displaystyle 6x^2+6x-72 = 0$
    $\displaystyle a=1, b=1, c=-72$
    $\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

    $\displaystyle x = \frac{-1 \pm \sqrt{1^2-4\times1\times-72}}{2\times1}$

    $\displaystyle x = \frac{-1 \pm 17}{2}$

    $\displaystyle x = \frac{-1 + 17}{2}$ $\displaystyle x=8$

    $\displaystyle x = \frac{-1 - 17}{2}$ $\displaystyle x=-9$

    $\displaystyle f'(2) = 6x^2+6x-72$
    $\displaystyle f'(2) = 6(2)^2+6(2)-72$
    $\displaystyle f'(2) = -36$

    $\displaystyle f'(12) = 6x^2+6x-72$
    $\displaystyle f'(12) = 6(12)^2+6(12)-72$
    $\displaystyle f'(12) = 864$

    $\displaystyle x \leq 8$ ; minima

    Where did I go wrong?
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  2. #2
    Eater of Worlds
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    It would appear your error is in the use of the quadratic formula.

    Your quadratic is $\displaystyle f'(x)=x^{2}+x-12=0$
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  3. #3
    Member looi76's Avatar
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    Question:
    Find the coordinates of the stationary points on the graph of the following functions, and find weather these points are maxima or minima.
    $\displaystyle (a) 2x^3+3x^2-72x+5$

    Attempt:
    $\displaystyle f(x) = 2x^3+3x^2-72x+5$
    $\displaystyle f'(x) = 6x^2+6x-72$
    $\displaystyle f'(x) = 0$

    $\displaystyle 6x^2+6x-72 = 0$
    $\displaystyle a=1, b=1, c=-12$
    $\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

    $\displaystyle x = \frac{-1 \pm \sqrt{1^2-4\times1\times-12}}{2\times1}$

    $\displaystyle x = \frac{-1 \pm 7}{2}$

    $\displaystyle x = \frac{-1 + 7}{2}$ $\displaystyle x=3$

    $\displaystyle x = \frac{-1 - 17}{2}$ $\displaystyle x=-4$

    $\displaystyle f'(3) = 6x^2+6x-72$
    $\displaystyle f'(3) = 6(3)^2+6(3)-72$
    $\displaystyle f'(3) = 0$

    $\displaystyle f'(-4) = 6x^2+6x-72$
    $\displaystyle f'(-4) = 6(-4)^2+6(-4)-72$
    $\displaystyle f'(-4) = 0$

    I don't know what to do next... answer in the text book is $\displaystyle (-4,213)$$\displaystyle ,maximum;$ $\displaystyle (3,-130)$$\displaystyle ,minimum$
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  4. #4
    Eater of Worlds
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    Factor your quadratic or use the formula.

    $\displaystyle x^{2}+x-12=(x-3)(x+4)$

    Now, plug thjose values back into f(x) to get the y values.

    You had it, but just used the wrong values on the formula.
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