Find the point on the graph of √(x+1) that lies closest to point ( 4,0).

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Find the point on the graph of √(x+1) that lies closest to point ( 4,0).

OCW Scholar - Single Variable Calculus
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Overall concept: Substitute points into the distance formula find a minimum using derivatives. \[d=\sqrt{(x _{2}-x _{1})^{2}+(y _{2}-y _{1})^{2}}\]where\[(x _{1}, y _{1})=(4,0)\]\[(x _{2},y _{2})=(x, \sqrt{x+1})\]Simplify. Take the derivative. Find critical points by setting the derivative to zero. Determine if there is a relative minimum and what it is. Plug that back into the original function to find the y value of the point.

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