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Math Help - Derivative/Tangent Line Question

  1. #1
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    Derivative/Tangent Line Question

    1. Find an equation of the tangent line to the graph of f at the indicated point.

    f(x) = 1/x+1
    (0,1)

    2. Find an equation of the line that is tangent to the graph of f and parallel to the given line.

    f(x)=1/sqrtx, x + 2y-6=-

    Any help would be greatly appreciated. Thanks!
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  2. #2
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    Quote Originally Posted by alreadyinuse View Post
    1. Find an equation of the tangent line to the graph of f at the indicated point.

    f(x) = 1/x+1
    (0,1)

    2. Find an equation of the line that is tangent to the graph of f and parallel to the given line.

    f(x)=1/sqrtx, x + 2y-6=-

    Any help would be greatly appreciated. Thanks!
    The slope of the tangent line is the value of the first derivative.

    f(x) = 1/(x+1) = (x+1)^(-1)
    f'(x) = -1(x+1)^(-2)
    So at x=0,
    m = f'(0) = -1(0+1)^(-2) = -1 /(1)^2 = -1
    Using the point-slope form of the line, and the point (0,1):
    (y -y1) = m(x -x1)
    (y -1) = -1(x -0)
    y -1 = -x
    y = -x +1 ---------------the tangent line.

    --------------------
    If two lines are parallel, then they have the same slope.

    f(x) = 1 /sqrt(x) = (x)^(-1/2)
    f'(x) = (-1/2)x^(-3/2)
    f'(x) = -1 /[2sqrt(x^3)] -------(i)

    Slope of the line x + 2y-6=-
    What?
    Suppose the line is x + 2y-6 = 0
    Transform it into the form y = mx +b,
    x +2y -6 = 0
    2y = -x +6
    y = -(1/2)x +3
    So, m = -1/2

    Then, m = f'(x) = -1/2
    We need the point of tangency so that we can use the point-slope form.

    -1/2 = -1/[2sqrt(x^3)]
    Divide both sides by (-1/2),
    1 = 1/sqrt(x^3)
    sqrt(x^3) = 1
    Square both sides,
    x^3 = 1
    x = 1 ------at point of tangency.

    What is the y there?
    y = f(x) = 1/sqrt(x)
    y = 1/sqrt(1)
    y = 1
    So, point of tangency is (1,1)
    Then, with m = -1/2, and (1,1):
    (y -1) = (-1/2)(x -1)
    y = (-1/2)x +1/2 +1
    y = -(1/2)x +3/2 ---------------the tangent line.
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