Final answer:
After analyzing the derivative f'(x)=5(x-1)²sin(x²), we determine that out of the given options, x=1.772 is the value where the function f has a local minimum on the interval (1,3).
Step-by-step explanation:
To determine at which value of x the function f has a local minimum on the interval (1,3), we analyze the function's derivative f'(x) = 5(x-1)²sin(x²). A local minimum can occur where the derivative changes from negative to positive, which are potential critical points. We need to find where f'(x) is zero and then determine the sign changes of f'(x) around those points.
Since (x-1)² is always non-negative and sin(x²) fluctuates between -1 and 1, we can deduce that critical points occur when sin(x²) = 0. Analyzing the given options and using trigonometric principles, we can pinpoint where the sine function would be zero. Out of the choices, 1.772 is the value closest to the square root of a multiple of π, which would give us sin(x²) = 0.
However, to ensure that this point is indeed a local minimum and not a local maximum or a point of inflection, we must verify that f'(x) changes from negative to positive at x = 1.772. For x just less than 1.772, sin(x²) is still positive, and for x just more than 1.772, sin(x²) becomes positive again after being zero, indicating that f'(x) changes from negative to positive, confirming a local minimum at x = 1.772 (Choice B).