Identify any extrema of the function $$ g(x, y)=(x-3)^{2}+(y-2)^{2} $$ by recognizing its given form or its form after completing the square.

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1. The function is given by
   $$
   g(x, y) = (x-3)^2 + (y-2)^2.
   $$
2. This function represents the squared distance from a point $$(x, y)$$ to the fixed point $$(3, 2)$$. Since a squared distance is always non-negative, we have
   $$
   g(x, y) \geq 0 \quad \text{for all } (x, y).
   $$
3. The minimum value of $$g(x, y)$$ occurs when both squared terms are equal to zero, namely when
   $$
   (x-3)^2 = 0 \quad \text{and} \quad (y-2)^2 = 0.
   $$
   This gives
   $$
   x = 3 \quad \text{and} \quad y = 2.
   $$
4. Substituting $$x = 3$$ and $$y = 2$$ into the function, we have
   $$
   g(3, 2) = (3-3)^2 + (2-2)^2 = 0.
   $$
5. Because the function $$g(x, y)$$ grows without bound as $$(x, y)$$ moves away from $$(3, 2)$$, there is no global maximum.

Thus, the function has a global minimum at $$(3, 2)$$ with a value of $$0$$ and no global maximum.
The function $$ g(x, y) = (x-3)^2 + (y-2)^2 $$ has a global minimum at the point $$(3, 2)$$ with a value of $$0$$. There is no global maximum for this function.

Identify any extrema of the function by recognizing its given form or its form
after completing the square.

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Identify any extrema of the function by recognizing its given form or its form after completing the square.