Final answer:
The correct increasing order of asymptotic complexity for the given functions is f3, f2, f1, f4 (option b). The order is determined by comparing growth rates: logarithmic (f3) is slower than polynomial (f2), which is slower than exponential (f1), which in turn is slower than the variable base exponential function (f4).
Step-by-step explanation:
The question asks for the increasing order of asymptotic complexity of the functions f1(n) = 2ⁿ, f2(n) = n³/², f3(n) = nLogn, and f4(n) = nⁱᵗⁿ. To determine the increasing order, we need to compare the growth rates of these functions as n approaches infinity.
Firstly, we consider f3(n) = nLogn and f2(n) = n³/². It is known that logarithmic growth is slower than polynomial growth, so f3(n) grows more slowly than f2(n). Next, we compare f1(n) = 2ⁿ with the previously mentioned functions. An exponential function grows faster than polynomial and polynomial-logarithmic functions, so f1(n) will grow faster than both f2(n) and f3(n). Lastly, for f4(n) = nⁱᵗⁿ, we note that any function where the base is the variable and the power is a function of that variable will grow faster than exponential with a constant base. This means f4(n) will grow even faster than f1(n). Therefore, the correct answer is (b) f3, f2, f1, f4, which represents these functions in order of increasing complexity from the slowest to the fastest growing function.