Final answer:
The absolute maximum of the function occurs at x=4 and the absolute minimum occurs at x=1, which corresponds to the answer choice B.
Step-by-step explanation:
To find the absolute maximum and absolute minimum of the given function f(x) = \frac{1}{3}x^3 - \frac{3}{2}x^2 on the interval [1, 4], we need to take several steps.
- First, find the function's derivative: f'(x) = x^2 - 3x.
- Set the derivative equal to zero to find the critical points: f'(x) = x(x - 3) = 0 gives us x = 0 and x = 3 as critical points. However, x = 0 is not in our interval [1, 4].
- Evaluate the function at the critical point within the interval and the endpoints: f(1), f(3), and f(4).
- Compare these values to determine the absolute maximum and minimum values.
By calculating, we find that f(1) = \frac{1}{3}(1)^3 - \frac{3}{2}(1)^2 = -\frac{7}{6}, f(3) = \frac{1}{3}(3)^3 - \frac{3}{2}(3)^2 = 0, and f(4) = \frac{1}{3}(4)^3 - \frac{3}{2}(4)^2 = \frac{32}{3} - 24 = \frac{8}{3}. Comparing these values reveals that the absolute minimum occurs at x=1 and the absolute maximum occurs at x=4.
Therefore, the correct answer is B) Absolute maximum at x=4, Absolute minimum at x=1