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Math Help - Examining a Curve - # 2

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    Examining a Curve - # 2

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

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

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

    f''(x) = 12x

    Find Critical Points:

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

    4x^{3} - 32 = 0

    4x^{3} = 32

    x^{3} = 8

    x = 2 - 3rd root of 8 is  2

    What is the maximum and minimum?

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

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

    Where is the interval increasing or decreasing?

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

    For x<2

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

    For 2>x

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

    What is the inflection point?

    f''(x) = 12x

    12x = 0

    x = 0

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

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

    Inflection Point (0,1)

    Where is the curve concave up, or concave down?

    For x<0

    f''(x) = 12x

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

    For x<0

    f''(x) = 12x

    f''(-1) = 12(-1) = -12 Positive so concave down
    Last edited by Jason76; October 29th 2013 at 12:52 AM.
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  2. #2
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    Re: Examining a Curve - # 2

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

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

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

    f''(x) = 12x^{2}

    Find Critical Points:

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

    4x^{3} - 32 = 0

    4x^{3} = 32

    x^{3} = 8

    x = 2 - 3rd root of 8 is  2

    What is the maximum and minimum?

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

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

    Where is the interval increasing or decreasing?

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

    For x<2

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

    For 2>x

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

    What is the inflection point?

    f''(x) = 12x^{2}

    12x^{2} = 0

    x^{2} = 0

    x = 0

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

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

    Inflection Point (0,1)

    Where is the curve concave up, or concave down?

    For x<0

    f''(x) = 12x^{2}

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

    For 0 < x

    f''(x) = 12x

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