Final answer:
The sign of the function f(x)=(2x-1)(3x^5)(x^2-1) on the interval -5/3 < x < 1/2 is negative because within this range, specifically between -1 and 1/2, the factors when multiplied result in a negative value.
Step-by-step explanation:
The function given is f(x)=(2x-1)(3x^5)(x^2-1), and we want to determine the sign of f(x) on the interval -5/3 < x < 1/2. The zeros provided are at x=-5/3, x=-1, and x=1/2. To find the sign of f(x) on the interval, we should examine the behavior of each factor separately between the zeros. The factor (2x-1) will be negative between its zero at x=1/2 and positive elsewhere. The factor (3x^5) will be negative for x>-5/3 since it's an odd power and x is negative in that interval. Finally, (x^2-1) will be positive except for its zero at x=-1 because the square of a real number is always non-negative.
When we multiply these together between -5/3 and -1, we will have a negative times a negative times a positive, which gives a positive result. Between -1 and 1/2, it will be a negative times a positive times a positive, giving a negative result. Since the question asks for the interval -5/3 < x < 1/2, and x=-1 is within this interval, the sign of f(x) will ultimately be negative between -1 and 1/2, which is part of the interval of interest.
Therefore, on the interval -5/3 < x < 1/2, the sign of f(x) is Negative.