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Math Help - Examining a Curve - # 3

  1. #1
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    Examining a Curve - # 3

    All of this is wrong, so please look over.

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

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

     f''(x) = 24x + 18

    Find Critical Numbers:

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

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

     6(2x + 3x - 9) = 0

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

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

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

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

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

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

    Find Max Min

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

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

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

    Where does the interval increase or decrease?


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

    For x < -3

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

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

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

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

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

    For \dfrac{3}{4} < x

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

    What is the infection point?

     f''(x) = 24x + 18

     24x + 18 = 0

     24x = -18

     x = -\dfrac{4}{3}

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

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

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

     f(-\dfrac{4}{3}) = -\dfrac{256}{67}) + 16 + \dfrac{216}{3}) + 8
    Last edited by Jason76; October 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

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

     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?

     f''(x) = 24x + 18

     24x + 18 = 0

     24x = -18
    Good...

    Quote Originally Posted by Jason76 View Post

     x = -\dfrac{4}{3}
    The fraction is upside down! Try again.
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  3. #3
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    Re: Examining a Curve - # 3

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

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

     f''(x) = 24x + 18

    Find Critical Numbers:

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

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

     6(2x + 3x - 9) = 0

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

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

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

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

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

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

    Find Max Min

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

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

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

    Where does the interval increase or decrease?

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

    For x < -3

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

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

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

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

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

    For x >\dfrac{3}{2}

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

    What is the infection point?

     f''(x) = 24x + 18

     24x + 18 = 0

     24x = -18

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

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

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

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

     f(-\dfrac{4}{3}) = -\dfrac{256}{67}) + 16 + \dfrac{216}{3}) + 8
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    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

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

     f(\dfrac{3}{2}) = 4(\dfrac{3}{2})^{3} + 9(\dfrac{3}{2})^{2} - 54(\dfrac{3}{2}) + 8 =43
    I get:
     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
     f(-3) = 4(-3)^{3} + 9(-3)^{2} - 54(-3) + 8 = 287
    No:
     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?

     f''(x) = 24x + 18

     24x + 18 = 0

     24x = -18

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