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Consider the differential equation dy/dx = -1 + (y^2/ x).

Let y = g(x) be the particular solution to the differential equation dy/ dx = -1 + (y^2/ x) with initial condition g(4) = 2. Does g have a relative minimum, a relative maximum, or neither at ? Justify your answer.

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  1. Well where is dy/dx = 0 ? (horizontal either max or min or inflection)
    dy/dx = -1 + (y^2/ x) = 0
    or y^2 = x
    It said start at (2, 4)
    well that very spot is horizontal 4 = 2^2
    now is it a min or a max there?
    what is d^2y/dx^2 ? at (2,4) ?
    dy/dx = -1 + (y^2/ x)
    d/dx(dy/x) = 0 + d/dx(y^2/x)= [x(-2y dy/dx) -y^2] /x^2
    at (2,4)= [ 2*-8* (0 at that point) - 16 ] / 4 = -32/4 oh that is a maximum because headed down

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