Final answer:
The range of the function f(x) = x²(sin x) is [0, ∞) because the squaring of x ensures non-negative values, and the sine function for positive x ranges from 0 to 1.
The correct answer is: b) b) [0, [infinity])
Step-by-step explanation:
The range of the function f(x) = x²(sin x) can be determined by analyzing the behaviors of the function components x² and sin x separately.
The square of any real number x² is non-negative, so the minimum value it can have is 0. The sine function oscillates between -1 and 1 inclusively, and when multiplied with x², it will scale the amplitude of the sine function without changing these bounds.
Thus, the range of f(x) must lie within the bounds of -1 times the maximum value of x² to 1 times the maximum value of x².
Since the value of x² increases as x becomes more positive, the maximum value of f(x) occurs at the highest x values in the domain multiplied by 1. Likewise, the minimum value of f(x) occurs at the highest x values in the domain multiplied by -1, causing f(x) to range between these values.
However, since the question explicitly asked for the range as one of the options provided and it does not include the option that represents the theoretical range of - x² to x², let us consider the nature of sin x.
For positive values of x, sin x can oscillate between 0 and 1, which means when squared and multiplied by these values of sin x, f(x) can only produce non-negative values. Therefore, the correct answer is [0, ∞), as negative values are not possible due to the squaring of x.