Final answer:
The range of the function V(x, y) = 3x² + y² is [0, ∞), as the squared terms are always non-negative and there is no upper bound on the values of V.
Step-by-step explanation:
The range of a function in mathematics typically refers to the set of all possible output values the function can produce. For the given function V(x, y) = 3x² + y², the range is the set of all possible values of V based on the input values x and y. Since 3x² and y² are both squares of real numbers, they are always non-negative. Consequently, the smallest value V(x, y) can take is 0, when both x and y are 0.
There is no upper limit to the value of V, as increasing x or y will increase the value of V(x, y). Therefore, the range of V(x, y) is [0, ∞).