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Math Help - tangent to graph and parallel to

  1. #1
    Super Member bigwave's Avatar
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    tangent to graph and parallel to

    find an equation of the line that is tangent to the graph of

     f(x) = \frac{1}{\sqrt{x}}

    and parallel to the line of x + 2y - 6 = 0

    I know that

     \frac {d}{dx} \left(\frac{1}{\sqrt{x}}\right) = -\frac{1}{2x^{3/2}}

    and that the parallel line can be rewritten as

    y = -\frac{1}{2}x+3

    but not sure how you use this to get the equation of the line that is tangent to the graph.
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  2. #2
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    Quote Originally Posted by bigwave View Post
    find an equation of the line that is tangent to the graph of

     f(x) = \frac{1}{\sqrt{x}}

    and parallel to the line of x + 2y - 6 = 0

    I know that

     \frac {d}{dx} \left(\frac{1}{\sqrt{x}}\right) = -\frac{1}{2x^{3/2}}

    and that the parallel line can be rewritten as

    y = -\frac{1}{2}x+3

    but not sure how you use this to get the equation of the line that is tangent to the graph.
    The gradient of the line is -\frac{1}{2}. Parallel lines have the same gradient.


    So the gradient of the graph will have to be -\frac{1}{2}.


    Therefore

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


    Solve for x.
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  3. #3
    Super Member bigwave's Avatar
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    Cool

    ok, just from observation x = 1 and so y = 1

    and the slope is -\frac{1}{2} but when I try to build an eq I get

    y=-\frac{1}{2}\left(x-1\right)+1

    which appears to be the tangent we want but a little foggy still how this works. if indeed that is the answer
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  4. #4
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    Quote Originally Posted by bigwave View Post
    ok, just from observation x = 1 and so y = 1

    and the slope is -\frac{1}{2} but when I try to build an eq I get

    y=-\frac{1}{2}\left(x-1\right)+1

    which appears to be the tangent we want but a little foggy still how this works. if indeed that is the answer
    You have m = -\frac{1}{2} and (x, y) = (1, 1).


    So building the equation for the tangent:

    y = mx + c

    1 = -\frac{1}{2}(1) + c

    1 = -\frac{1}{2} + c

    c = \frac{3}{2}.


    So the equation of the tangent is

    y = -\frac{1}{2}x + \frac{3}{2}

    or

    y = -\frac{1}{2}(x - 3).
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