Determine the range of values for which the function $$ y = \sqrt{x - 2} + \sqrt{8 - x} $$ is defined.

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To find the domain of the function $$ y = \sqrt{x - 2} + \sqrt{8 - x} $$, we need to ensure that both square root terms are defined. This requires the expressions inside the square roots to be non-negative.

For $$ \sqrt{x - 2} $$, the condition is:
$$
x - 2 \geq 0 \quad \implies \quad x \geq 2.
$$

For $$ \sqrt{8 - x} $$, the condition is:
$$
8 - x \geq 0 \quad \implies \quad x \leq 8.
$$

Combining these two inequalities, we get:
$$
2 \leq x \leq 8.
$$

Thus, the domain of the function is $$ [2, 8].

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