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Find an equation of the line that is tangent to the graph f and parallel to the given line
Function: f(x)= 1/ square root of x
Line: x+2y-6=0

  • calculus - ,

    slope of f(x) is -1/(2x√x)

    slope of line: -1/2

    so, you want -1/(2x√x) = -1/2

    so, you now have a slope=1 and a point (1,1).

    (y-1) = (-1/2)(x-1)
    y = -x/2 + 3/2

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