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calculus

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5. Find the point (x, y) on the graph of y=sqr(x) nearest the point (4, 0).

  • calculus -

    distance^2=(x-4)^2 + (y)^2

    distnace^2=(x-4)^2 + x^2
    d distance/x * 2 * distance=0=2(x-4)+ 2x

    or -x=x-4 or x=2
    and y= sqrt2. Point is (2, sqrt2)

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