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

calculus 
bobpursley
distance^2=(x4)^2 + (y)^2
distnace^2=(x4)^2 + x^2
d distance/x * 2 * distance=0=2(x4)+ 2x
or x=x4 or x=2
and y= sqrt2. Point is (2, sqrt2)
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