28. Find the interval(s) on which \( f(x) \) is decreasing, if \( f^{\prime}(x)=x^{4}-3 x^{3}+2 x^{2} \) \( \begin{array}{llll}\text { A. }(1,2) & \text { B. }(0, \infty) & \text { C. }(-\infty, 1) & \text { D. }(0,1),(2, \infty)\end{array} \)

To determine where \( f(x) \) is decreasing, we analyze its derivative, \( f'(x) = x^4 - 3x^3 + 2x^2 \). First, we find the critical points by setting \( f'(x) = 0 \). Factoring gives us \( x^2(x^2 - 3x + 2) = 0 \), leading to \( x^2(x-1)(x-2) = 0 \). Thus, the critical points are \( x = 0, 1, 2 \). Next, we can create intervals to test the sign of \( f'(x) \): - For \( (-\infty, 0) \), pick \( x = -1 \): \( f'(-1) > 0 \) (increasing). - For \( (0, 1) \), pick \( x = 0.5 \): \( f'(0.5) < 0 \) (decreasing). - For \( (1, 2) \), pick \( x = 1.5 \): \( f'(1.5) < 0 \) (decreasing). - For \( (2, \infty) \), pick \( x = 3 \): \( f'(3) > 0 \) (increasing). Thus, \( f(x) \) is decreasing on the intervals \( (0,1) \) and \( (1,2) \), but we only care about the blank interval here, which really sums up as: \( (0, 1) \) and \( (1, 2) \) means decreasing on \( (0, 2) \). The correct answer is then \( D. (0, 1), (2, \infty) \). For further exploration: Mathematicians often observe that the behavior of derivatives can reveal a lot about the function's shape. Maximizing and minimizing points can help identify peaks and troughs in data models, and it's widely used in fields like economics to find optimal pricing strategies! Also, if you're keen on delving deeper into calculus, consider looking at resources like "Calculus" by James Stewart or "Calculus Made Easy" by Silvanus P. Thompson. Both provide engaging insights and techniques that can greatly enhance your understanding of derivatives and their applications!