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Thread: Examining a Curve - # 3

  1. #1
    MHF Contributor Jason76's Avatar
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    Examining a Curve - # 3

    All of this is wrong, so please look over.

    $\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8 $

    $\displaystyle f'(x) = 12x^{2} + 18x - 54$

    $\displaystyle f''(x) = 24x + 18$

    Find Critical Numbers:

    $\displaystyle f'(x) = 12x^{2} + 18x - 54$

    $\displaystyle 12x^{2} + 18x - 54 = 0$

    $\displaystyle 6(2x + 3x - 9) = 0$

    $\displaystyle \dfrac{-3 \pm \sqrt{3^{2} - 4(2)(-9)}}{2(2)}$

    $\displaystyle \dfrac{-3 \pm \sqrt{81}}{4}$

    $\displaystyle \dfrac{-3 \pm 9}{4}$

    $\displaystyle \dfrac{-3 + 9}{4} = \dfrac{6}{4} = \dfrac{3}{2} $

    $\displaystyle \dfrac{-3 - 9}{4} = \dfrac{-12}{4} = -3 $

    Critical Numbers: $\displaystyle \dfrac{3}{2}, -3$

    Find Max Min

    $\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8 $

    $\displaystyle f(\dfrac{3}{2}) = 4(\dfrac{3}{2})^{3} + 9(\dfrac{3}{2})^{2} - 54(\dfrac{3}{2}) + 8 = -\dfrac{370}{8}$

    $\displaystyle f(-3) = 4(-3)^{3} + 9(-3)^{2} - 54(-3) + 8 = 215 $

    Where does the interval increase or decrease?


    $\displaystyle f'(x) = 12x^{2} + 18x - 54$

    For $\displaystyle x < -3$

    $\displaystyle f'(-4) = 12(-4)^{2} + 18(-4) - 54 = $

    $\displaystyle f'(-4) = 192 -72 - 54 = 66 $ Positive so increasing

    For $\displaystyle -3 < x <\dfrac{3}{2}$

    $\displaystyle f'(-1) = 12(-1)^{2} + 18(-1) - 54 = $

    $\displaystyle f'(-1) = 12 - 18 - 54 = -60$ Negative so decreasing

    For $\displaystyle \dfrac{3}{4} < x$

    $\displaystyle f'(1) = 12(1)^{2} + 18(1) - 54 = - 24 $ Negative so decreasing

    What is the infection point?

    $\displaystyle f''(x) = 24x + 18$

    $\displaystyle 24x + 18 = 0$

    $\displaystyle 24x = -18$

    $\displaystyle x = -\dfrac{4}{3}$

    $\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8 $

    $\displaystyle f(-\dfrac{4}{3}) = 4(-\dfrac{4}{3})^{3} + 9(-\dfrac{4}{3})^{2} - 54(-\dfrac{4}{3}) + 8 $

    $\displaystyle f(-\dfrac{4}{3}) = 4(-\dfrac{64}{67}) + 9(\dfrac{16}{9}) - 54(-\dfrac{4}{3}) + 8 $

    $\displaystyle f(-\dfrac{4}{3}) = -\dfrac{256}{67}) + 16 + \dfrac{216}{3}) + 8 $
    Last edited by Jason76; Oct 29th 2013 at 02:00 AM.
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Examining a Curve - # 3

    Quote Originally Posted by Jason76 View Post
    All of this is wrong, so please look over.
    It's not all wrong at all - just a few arithmetic errors:

    Quote Originally Posted by Jason76 View Post

    $\displaystyle f(\dfrac{3}{2}) = 4(\dfrac{3}{2})^{3} + 9(\dfrac{3}{2})^{2} - 54(\dfrac{3}{2}) + 8 = -\dfrac{370}{8}$

    $\displaystyle f(-3) = 4(-3)^{3} + 9(-3)^{2} - 54(-3) + 8 = 215 $
    Both of these are correct as set up but you made a mistakes with the actual calculation. Try them again.

    Quote Originally Posted by Jason76 View Post
    What is the infection point?

    $\displaystyle f''(x) = 24x + 18$

    $\displaystyle 24x + 18 = 0$

    $\displaystyle 24x = -18$
    Good...

    Quote Originally Posted by Jason76 View Post

    $\displaystyle x = -\dfrac{4}{3}$
    The fraction is upside down! Try again.
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  3. #3
    MHF Contributor Jason76's Avatar
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    Re: Examining a Curve - # 3

    $\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8 $

    $\displaystyle f'(x) = 12x^{2} + 18x - 54$

    $\displaystyle f''(x) = 24x + 18$

    Find Critical Numbers:

    $\displaystyle f'(x) = 12x^{2} + 18x - 54$

    $\displaystyle 12x^{2} + 18x - 54 = 0$

    $\displaystyle 6(2x + 3x - 9) = 0$

    $\displaystyle \dfrac{-3 \pm \sqrt{3^{2} - 4(2)(-9)}}{2(2)}$

    $\displaystyle \dfrac{-3 \pm \sqrt{81}}{4}$

    $\displaystyle \dfrac{-3 \pm 9}{4}$

    $\displaystyle \dfrac{-3 + 9}{4} = \dfrac{6}{4} = \dfrac{3}{2} $

    $\displaystyle \dfrac{-3 - 9}{4} = \dfrac{-12}{4} = -3 $

    Critical Numbers: $\displaystyle \dfrac{3}{2}, -3$

    Find Max Min

    $\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8 $

    $\displaystyle f(\dfrac{3}{2}) = 4(\dfrac{3}{2})^{3} + 9(\dfrac{3}{2})^{2} - 54(\dfrac{3}{2}) + 8 =43$

    $\displaystyle f(-3) = 4(-3)^{3} + 9(-3)^{2} - 54(-3) + 8 = 287 $

    Where does the interval increase or decrease?

    $\displaystyle f'(x) = 12x^{2} + 18x - 54$

    For $\displaystyle x < -3$

    $\displaystyle f'(-4) = 12(-4)^{2} + 18(-4) - 54 = $

    $\displaystyle f'(-4) = 192 -72 - 54 = 66 $ Positive so increasing

    For $\displaystyle -3 < x <\dfrac{3}{2}$

    $\displaystyle f'(-1) = 12(-1)^{2} + 18(-1) - 54 = $

    $\displaystyle f'(-1) = 12 - 18 - 54 = -60$ Negative so decreasing

    For $\displaystyle x >\dfrac{3}{2}$

    $\displaystyle f'(1) = 12(3)^{2} + 18(3) - 54 = 108 $ Positive so increasing

    What is the infection point?

    $\displaystyle f''(x) = 24x + 18$

    $\displaystyle 24x + 18 = 0$

    $\displaystyle 24x = -18$

    $\displaystyle x = -\dfrac{4}{3}$ This looks right to me cause $\displaystyle 6$ goes into $\displaystyle 24$, $\displaystyle 6$ times and $\displaystyle 6$ goes into $\displaystyle 18$, $\displaystyle 3$ times.

    $\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8 $

    $\displaystyle f(-\dfrac{4}{3}) = 4(-\dfrac{4}{3})^{3} + 9(-\dfrac{4}{3})^{2} - 54(-\dfrac{4}{3}) + 8 $

    $\displaystyle f(-\dfrac{4}{3}) = 4(-\dfrac{64}{67}) + 9(\dfrac{16}{9}) - 54(-\dfrac{4}{3}) + 8 $

    $\displaystyle f(-\dfrac{4}{3}) = -\dfrac{256}{67}) + 16 + \dfrac{216}{3}) + 8 $
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  4. #4
    MHF Contributor ebaines's Avatar
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    Re: Examining a Curve - # 3

    Still teh same errors!

    Quote Originally Posted by Jason76 View Post
    Find Max Min

    $\displaystyle f(x) = 4x^{3} + 9x^{2} - 54x + 8 $

    $\displaystyle f(\dfrac{3}{2}) = 4(\dfrac{3}{2})^{3} + 9(\dfrac{3}{2})^{2} - 54(\dfrac{3}{2}) + 8 =43$
    I get:
    $\displaystyle f(\dfrac{3}{2}) = 4(\dfrac{3}{2})^{3} + 9(\dfrac{3}{2})^{2} - 54(\dfrac{3}{2}) + 8 =13.5 + 20.25 -81+8 = -39.25$

    Quote Originally Posted by Jason76 View Post
    $\displaystyle f(-3) = 4(-3)^{3} + 9(-3)^{2} - 54(-3) + 8 = 287 $
    No:
    $\displaystyle f(-3) = 4(-3)^{3} + 9(-3)^{2} - 54(-3) + 8 = -108+81+162+8=143 $

    Quote Originally Posted by Jason76 View Post
    What is the infection point?

    $\displaystyle f''(x) = 24x + 18$

    $\displaystyle 24x + 18 = 0$

    $\displaystyle 24x = -18$

    $\displaystyle x = -\dfrac{4}{3}$ This looks right to me cause $\displaystyle 6$ goes into $\displaystyle 24$, $\displaystyle 6$ times and $\displaystyle 6$ goes into $\displaystyle 18$, $\displaystyle 3$ times.
    If $\displaystyle 24x = -18 $ then divide both sides by 24: $\displaystyle x = \frac {-18}{24} = -\frac 3 4$

    By the way - I suggest that to help double-check your work you should plot the function and see whether your answers make sense.
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