We want to find the point on the curve
that is closest to (2,1).
In problems of minimizing or maximizing a distance, there is a trick that is helpful for simplifying computations. It is based on the observation that the distance and the square of the distance have their max and min at the same point. That is why we do not use the square root. Makes things easier anytime you can eliminate a radical.
^{2}+(y-1)^{2}})
^{2}+(-2x^{2}+3x+4)^{2}})
.........[1]
So, from the observation mentioned above, we can use:
^{2}+(-2x^{2}+3x+4)^{2})
Differentiate and get
Set to 0 and solve for x.
You will get 3 solutions. You can use the 2nd derivative test to check the nature of these critical point.
Better yet, just plug your solutions into [1] to see which gives the shortest distance. You can also graph it and see.
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