Answer is: option4

\( \frac{5}{2} \)

Solution:

Compute \( f'(x) \) using the product rule:

Let \( u = (x-1)^3 \) and \( v = (3-x) \).

Then:

\( u' = 3(x-1)^2 \), \( v' = -1 \)

Using the product rule:

\( f'(x) = u'v + uv' \)

\( f'(x) = 3(x-1)^2 (3-x) + (x-1)^3 (-1) \)

Factor out \( (x-1)^2 \):

\( f'(x) = (x-1)^2 [3(3-x) - (x-1)] \)

Simplify inside the brackets:

\( 3(3-x) - (x-1) = 9 - 3x - x + 1 = 10 - 4x \)

Thus,

\( f'(x) = (x-1)^2 (10-4x) \)

Setting \( f'(x) = 0 \):

\( (x-1)^2 (10-4x) = 0 \)

This gives two possible solutions:

1. \( (x-1)^2 = 0 \Rightarrow x = 1 \)
2. \( 10-4x = 0 \Rightarrow x = \frac{10}{4} = \frac{5}{2} \)

To determine which critical point gives the absolute maximum, we check \( f(x) \) at these points.

Evaluate \( f(x) \) at Critical Points:

\( f(1) = (1-1)^3 (3-1) = 0 \)

\( f(5/2) = \left(\frac{5}{2} - 1\right)^3 \left(3 - \frac{5}{2}\right) \)

\( = \left(\frac{3}{2}\right)^3 \left(\frac{1}{2}\right) \)

\( = \frac{27}{8} \times \frac{1}{2} = \frac{27}{16} \)

Since \( f(1) = 0 \) and \( f(5/2) = \frac{27}{16} \), the absolute maximum occurs at: \( \frac{5}{2} \)