Final answer:
The function f(x) = x^(1/5) + 1 has no critical points as the derivative is always positive. For f(x) = x^(1/3) + 1, the only critical number is x = 0, and the function is increasing on the intervals (-∞, 0) and (0, ∞).
Step-by-step explanation:
The student is asking about two separate functions, with their corresponding tasks on finding critical numbers and intervals of increase or decrease. For the function f(x) = x^(1/5) + 1, to find critical numbers, one would need to find the values of x where the first derivative f'(x) is equal to zero or undefined. However, this particular function has no critical points since the derivative is always positive, indicating the function is always increasing. On the other hand, for the function f(x) = x^(1/3) + 1, the derivative is f'(x) = (1/3)x^(-2/3) which is undefined when x = 0. Thus, x = 0 is the only critical number. The function f(x) is increasing on the interval (-∞, 0) and on the interval (0, ∞), because the derivative is positive on these intervals.