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Math Help - Finding open intervals of a function and extremas

  1. #1
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    Finding open intervals of a function and extremas

    Find all open intervals on which the function is increasing or decreasing, and locate all relative extrema. Graph it.
    f(x) = (x2 - 2x + 1 )/(x + 1)

    What I did:
    I know the veritcal asymptote is -1, which is the discontinuity
    I know the horizontal asymptote is 1 since the factor for the top is (x + 1)(x - 1)
    The derivative is: ((x + 1)(2x - 2) + (x2 - 2x + 1)(1))/(x + 1)2 which is simplified into (2x2 + 2x - 2x - 2 + x2 - 2x + 1)/(x + 1)2
    Which is simplified into: (3x2 - 2x - 1)/(x + 1)2
    The numerator is factored into (3x + 1)(x - 1), but do I get my critical numbers from this numerator?
    Because it would be x=(-1/3) and 1, which is different from the answer

    Also does anyone know how to do the 'sign chart?' That way I could get my minimum and maximum.
    An EXAMPLE of a sign chart is:
    x (-∞, -3), -3x, (-3, 0), 0, (0, 3), 3, (3, ∞),
    f(x) positive, undefined, positive, 0, negative, undefined, negative

    Answers are:
    Critical numbers: x = -3, 1
    Discontinuity: x = -1
    Increasing on (-∞, -3) and (1, ∞)
    Decreasing on (-3, -1) and (-1, 1)
    Relative maximum (-3, -8)
    Relative minimum: (1, 0)
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  2. #2
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    Re: Finding open intervals of a function and extremas

    Quote Originally Posted by Chaim View Post
    Find all open intervals on which the function is increasing or decreasing, and locate all relative extrema. Graph it.
    f(x) = (x2 - 2x + 1 )/(x + 1)

    What I did:
    I know the veritcal asymptote is -1, which is the discontinuity
    Yes.

    Quote Originally Posted by Chaim View Post
    I know the horizontal asymptote is 1 since the factor for the top is (x + 1)(x - 1)
    Double check this.
    \lim_{x \to \infty}\frac{x^2 - 2x + 1}{x+1} = \lim_{x \to \infty} \left [ x - 3+ \frac{4}{x+1} \right ]
    which does not fall to zero at infinity. This is actually a slant asymptote.

    Quote Originally Posted by Chaim View Post
    The derivative is: ((x + 1)(2x - 2) + (x2 - 2x + 1)(1))/(x + 1)2 which is simplified into (2x2 + 2x - 2x - 2 + x2 - 2x + 1)/(x + 1)2
    Which is simplified into: (3x2 - 2x - 1)/(x + 1)2
    The numerator is factored into (3x + 1)(x - 1), [B]but do I get my critical numbers from this numerator?
    Actually f'(x) = \frac{x^2 + 2x - 3}{(x + 1)^2} = \frac{(x - 1)(x + 3)}{(x + 1)^2}

    -Dan
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  3. #3
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    Re: Finding open intervals of a function and extremas

    f(x) = \frac{x^2-2x+1}{x+1}

    f'(x) = \frac{(x+1)(2x-2) - (x^2-2x+1)}{(x+1)^2}

    f'(x) = \frac{(2x^2-2) - (x^2-2x+1)}{(x+1)^2}

    f'(x) = \frac{x^2+2x-3}{(x+1)^2}

    try again ...
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  4. #4
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    Re: Finding open intervals of a function and extremas

    Ah thanks! That makes more sense now for the critical numbers and sign chart!
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