Final answer:
The minimum value of the function t^2sin(3t) is -[infinity] because while t^2 is non-negative, the sine function oscillates between -1 and 1, which means the function can become increasingly negative without bound. The minimum value of the function t^2sin(3t) is -[infinity], which corresponds to the option (d).
Step-by-step explanation:
The student's question pertains to finding the minimum value of the function t^2sin(3t). To find the minimum value of this function, we need to understand the behavior of both parts: t^2 and sin(3t). The function t^2 is always non-negative since any real number squared is non-negative. The sine function, however, oscillates between -1 and 1. This means that the product of these two functions t^2sin(3t) will have its minimum value when sin(3t) is at its minimum, which is -1, and t is not zero to avoid the trivial solution.
However, no matter how large t becomes, the sine function can only bring the value to -t^2 at best since its range is restricted to [-1,1]. Therefore, as t increases, the overall value of -t^2 can become arbitrarily large in the negative sense. Hence, the function does not have a bounded minimum; no matter how large the negative value is, there will always be a larger negative value the function can obtain. This leads us to conclude that the minimum value of the function t^2sin(3t) is -[infinity], which corresponds to the option (d).