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Thread: Examining a Curve

  1. #1
    MHF Contributor Jason76's Avatar
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    Examining a Curve

    All of this is wrong. Here is the actual question: "Find the local maximum and minimum values of f using both the First and Second Derivative Tests." Please look thru this to see what is mistaken.

    $\displaystyle f(x) = -8x^{3} + 12x^{2} + 8$

    $\displaystyle f'(x) = -24x^{2} + 24x$

    $\displaystyle f''(x) = -48x + 24$

    Find Critical Numbers:

    $\displaystyle f'(x) = -24x^{2} + 24x$

    $\displaystyle -24x^{2} + 24x = 0$

    $\displaystyle -24x^{2} = - 24$

    $\displaystyle x^{2} = 1$

    $\displaystyle \sqrt{x^{2}} = \sqrt{1}$

    $\displaystyle x = \pm \sqrt{1}$

    $\displaystyle x = -1$

    $\displaystyle x = 1$

    Find Min Max

    $\displaystyle f(x) = -8x^{3} + 12x^{2} + 8$

    $\displaystyle f(-1) = -8(-1)^{3} + 12(-1)^{2} + 8 = 28$

    $\displaystyle f(1) = -8(1)^{3} + 12(1)^{2} + 8 = 12$
    Last edited by Jason76; Oct 29th 2013 at 12:34 AM.
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  2. #2
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    Re: Examining a Curve

    Hey Jason76.

    There is nothing wrong with your first answer.
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    Re: Examining a Curve

    Quote Originally Posted by Jason76 View Post
    All of this is wrong. Here is the actual question: "Find the local maximum and minimum values of f using both the First and Second Derivative Tests." Please look thru this to see what is mistaken.

    $\displaystyle f(x) = -8x^{3} + 12x^{2} + 8$

    $\displaystyle f'(x) = -24x^{2} + 24x$

    $\displaystyle f''(x) = -48x + 24$

    Find Critical Numbers:

    $\displaystyle f'(x) = -24x^{2} + 24x$

    $\displaystyle -24x^{2} + 24x = 0$

    $\displaystyle -24x^{2} = - 24$
    The last line here is missing an x on the right hand side.

    You can also write $\displaystyle f'(x)$ as:

    $\displaystyle f'(x) = 24x(1-x)$

    This is zero when $\displaystyle x=0$ or $\displaystyle x=1$
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  4. #4
    MHF Contributor Jason76's Avatar
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    Re: Examining a Curve

    $\displaystyle f(x) = -8x^{3} + 12x^{2} + 8$

    $\displaystyle f'(x) = -24x^{2} + 24x$

    $\displaystyle f''(x) = -48x + 24$

    Find Critical Numbers:

    $\displaystyle f'(x) = -24x^{2} + 24x$

    $\displaystyle -24x^{2} + 24x = 0$

    $\displaystyle 24x(-x + 1) = 0$ a different approach

    $\displaystyle 24x = 0$

    $\displaystyle x = 0$

    $\displaystyle (-x + 1) = 0$

    $\displaystyle -x = -1$

    $\displaystyle x = 1$ and $\displaystyle x = 0$ (Critical Number)


    Find Min Max

    $\displaystyle f(x) = -8x^{3} + 12x^{2} + 8$

    $\displaystyle f(1) = -8(1)^{3} + 12(1)^{2} + 8 = 12$ - Max

    $\displaystyle f(0) = -8(0)^{3} + 12(0)^{2} + 8 = 8$ - Min
    Last edited by Jason76; Oct 29th 2013 at 12:17 PM.
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