Final answer:
The functions that are uniformly continuous on the specified set are: a) f(x) = x¹⁷sin(x) - eˣcos(3x) on [0, π], b) f(x) = x³ on [0,1], d) f(x) = x³ on ℝ, and e) f(x) = 1/x³ on (0,1].
Step-by-step explanation:
To determine which of the given functions are uniformly continuous on the specified set, we can apply the theorem which states that a function is uniformly continuous on a closed and bounded interval if and only if the function is continuous on the interval.
(a) The function f(x) = x¹⁷sin(x) - eˣcos(3x) is continuous on the closed interval [0, π], so it is uniformly continuous on this interval.
(b) The function f(x) = x³ is continuous on the closed interval [0, 1], so it is uniformly continuous on this interval.
(c) The function f(x) = x³ is not continuous on the open interval (0, 1), so it is not uniformly continuous on this interval.
(d) The function f(x) = x³ is continuous on the entire real line ℝ, so it is uniformly continuous on this interval.
(e) The function f(x) = 1/x³ is continuous on the closed interval (0, 1], so it is uniformly continuous on this interval.
(f) The function f(x) = sin(1/x²) is not continuous at x = 0 on the open interval (0, 1], so it is not uniformly continuous on this interval.
(g) The function f(x) = x²sin(1/x) is not continuous at x = 0 on the open interval (0, 1], so it is not uniformly continuous on this interval.