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Math Help - Two problems dealing with linear approximation and inflection points

  1. #1
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    Two problems dealing with linear approximation and inflection points

    Okay here are two problems that have my friends and I stumped as we can't figure out a correct answer.

    Problem #1:
    At the point where x=3 on the curve y = (ax+1)/(x-2) , where a is a constant the slop of the normal is 1/5. Find the value of a.

    Its multiple choice and the answers they give us are 1/2, 2, 0, 3, 1/3.
    However, I keep getting -5/3 for a.


    Problem #2:
    The function y=x^4 - 4x^3 + 14 has a horizontal inflection at (A, B). The value of A+B is what?

    Yet again it is another multiple choice question. The answers given are -10, 0, -39, 14, or none of these. I have a question though. In class we talk about points of inflection but in this question it talks about horizontal inflection. Is there a difference? We haven't touched base at all in class if it is different.
    We found two points of inflection and they were (-2, 2) and (0,14) but we don't know which one is the correct answer if any. Its either 0 or 14 or maybe not if we screwed up in our calculations.

    Any help would be appreciated.
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  2. #2
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    Quote Originally Posted by forkball42 View Post
    Problem #1:
    At the point where x=3 on the curve y = (ax+1)/(x-2) , where a is a constant the slop of the normal is 1/5. Find the value of a.

    Its multiple choice and the answers they give us are 1/2, 2, 0, 3, 1/3.
    However, I keep getting -5/3 for a.
    The slope of the tangent to the curve can be found by
    y(x) = \frac{ax + 1}{x - 2}

    y^{\prime}(x) = \frac{a(x - 2) - (ax + 1)(1)}{(x - 2)^2}

    y^{\prime}(x) = -\frac{2a + 1}{(x - 2)^2}

    So at x = 3, the tangent to the curve has the value
    y^{\prime}(3) = -\frac{2a + 1}{(3 - 2)^2} = -(2a + 1)

    Now, the normal to the curve at this point is a line with a slope perpendicular to the tangent line. So the slope of the normal line is
    -\frac{1}{-(2a + 1)} = \frac{1}{2a + 1} = \frac{1}{5}

    So solving this I get that a = 2.

    -Dan
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  3. #3
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    For #1 the slope is given by y'=\frac{-(2a+1)}{(x-2)^{2}}

    The normal has slope 1/5, therefore, the slope of the tangent at x=3 must have slope -5.

    So, when x=3, \frac{-(2a+1)}{(3-2)^{2}}=-5

    Solving for a we see a=2
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by forkball42 View Post
    Problem #2:
    The function y=x^4 - 4x^3 + 14 has a horizontal inflection at (A, B). The value of A+B is what?

    Yet again it is another multiple choice question. The answers given are -10, 0, -39, 14, or none of these. I have a question though. In class we talk about points of inflection but in this question it talks about horizontal inflection. Is there a difference? We haven't touched base at all in class if it is different.
    We found two points of inflection and they were (-2, 2) and (0,14) but we don't know which one is the correct answer if any. Its either 0 or 14 or maybe not if we screwed up in our calculations.
    This is my best guess after doing a search: A "horizontal inflection point" likely what you would simply call an inflection point. Now, recall that a critical point exists when the first derivative is either 0 or undefined. If the critical point is one where the first derivative is undefined and the second derivative at this point is also undefined, we have what you might call a "vertical inflection point."

    In any event, all the inflection points here have a horizontal "tangent" so I would imagine they fall under the category of horizontal inflection points.

    y(x) = x^4 - 4x^3 + 14

    y^{\prime}(x) = 4x^3 - 12x^2

    So we have critical points at
    y^{\prime}(x) = 4x^3 - 12x^2 = 0 \implies x = 0, 3

    The second derivative is
    y^{\prime \prime}(x) = 12x^2 - 24x

    y^{\prime \prime}(0) = 0

    y^{\prime \prime}(3) = 36

    So I get that there is only one inflection point for the function: at x = 0. The y value at x = 0 is y = 14. So the point (A, B) = (0, 14) which implies that A + B = 0 + 14 = 14.

    -Dan
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