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Thread: Examining a Curve - # 2

  1. #1
    MHF Contributor Jason76's Avatar
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    Examining a Curve - # 2

    All of this is wrong. Please look thru for mistakes.

    $\displaystyle f(x) = x^{4} - 32x + 1$

    $\displaystyle f'(x) = 4x^{3} - 32$

    $\displaystyle f''(x) = 12x$

    Find Critical Points:

    $\displaystyle f'(x) = 4x^{3} - 32$

    $\displaystyle 4x^{3} - 32 = 0$

    $\displaystyle 4x^{3} = 32$

    $\displaystyle x^{3} = 8$

    $\displaystyle x = 2$ - $\displaystyle 3rd$ root of $\displaystyle 8$ is $\displaystyle 2$

    What is the maximum and minimum?

    $\displaystyle f(x) = x^{4} - 32x + 1$

    $\displaystyle f(2) = (2)^{2} - 32(2) + 1 = -59$

    Where is the interval increasing or decreasing?

    $\displaystyle f'(x) = 4x^{3} - 32$

    For $\displaystyle x<2$

    $\displaystyle f'(1) = 4(1)^{3} - 32 = -21$ Negative so decreasing

    For $\displaystyle 2>x$

    $\displaystyle f'(3) = 4(x)^{3} - 32 = 4$ Positive so increasing

    What is the inflection point?

    $\displaystyle f''(x) = 12x$

    $\displaystyle 12x = 0$

    $\displaystyle x = 0$

    $\displaystyle f(x) = x^{4} - 32x + 1$

    $\displaystyle f(0) = (0)^{4} - 32(0) + 1 = 1$

    Inflection Point $\displaystyle (0,1)$

    Where is the curve concave up, or concave down?

    For $\displaystyle x<0$

    $\displaystyle f''(x) = 12x$

    $\displaystyle f''(-1) = 12(-1) = -12$ Negative so concave up

    For $\displaystyle x<0$

    $\displaystyle f''(x) = 12x$

    $\displaystyle f''(-1) = 12(-1) = -12$ Positive so concave down
    Last edited by Jason76; Oct 29th 2013 at 12:52 AM.
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  2. #2
    MHF Contributor Jason76's Avatar
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    Re: Examining a Curve - # 2

    2nd derivative is corrected. Any hints on other errors?

    $\displaystyle f(x) = x^{4} - 32x + 1$

    $\displaystyle f'(x) = 4x^{3} - 32$

    $\displaystyle f''(x) = 12x^{2}$

    Find Critical Points:

    $\displaystyle f'(x) = 4x^{3} - 32$

    $\displaystyle 4x^{3} - 32 = 0$

    $\displaystyle 4x^{3} = 32$

    $\displaystyle x^{3} = 8$

    $\displaystyle x = 2$ - $\displaystyle 3rd$ root of $\displaystyle 8$ is $\displaystyle 2$

    What is the maximum and minimum?

    $\displaystyle f(x) = x^{4} - 32x + 1$

    $\displaystyle f(2) = (2)^{2} - 32(2) + 1 = -59$

    Where is the interval increasing or decreasing?

    $\displaystyle f'(x) = 4x^{3} - 32$

    For $\displaystyle x<2$

    $\displaystyle f'(1) = 4(1)^{3} - 32 = -21$ Negative so decreasing

    For $\displaystyle 2>x$

    $\displaystyle f'(3) = 4(x)^{3} - 32 = 4$ Positive so increasing

    What is the inflection point?

    $\displaystyle f''(x) = 12x^{2}$

    $\displaystyle 12x^{2} = 0$

    $\displaystyle x^{2} = 0$

    $\displaystyle x = 0$

    $\displaystyle f(x) = x^{4} - 32x + 1$

    $\displaystyle f(0) = (0)^{4} - 32(0) + 1 = 1$

    Inflection Point $\displaystyle (0,1)$

    Where is the curve concave up, or concave down?

    For $\displaystyle x<0$

    $\displaystyle f''(x) = 12x^{2}$

    $\displaystyle f''(-1) = 12(-1)^{2} = 12$ Negative so concave up

    For $\displaystyle 0 < x$

    $\displaystyle f''(x) = 12x$

    $\displaystyle f''(1) = 12(1)^{2} = 12$ Positive so concave up
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