Final answer:
To determine if f(x) = x²(log x)³ is O(g(x)) where g(x) = x³(log x)², we compare their growth rates by examining the exponents of x and log x in each function. Since x² grows slower than x³ and the additional power of log x in f(x) does not impact the growth rate significantly, f(x) is indeed O(g(x)).
Step-by-step explanation:
To determine if f(x) = x²(log x)³ is O(g(x)) where g(x) = x³(log x)², we need to check if there exists a constant C and x_0 such that for all x > x_0, f(x) ≤ C*g(x). Using properties of logarithms, specifically that the logarithm of a number raised to an exponent is the product of the exponent and the logarithm of the number, we can rewrite f(x) and g(x) to compare them more easily.
Let's compare the exponents of x in f(x) and g(x). In f(x), we have x², and in g(x), we have x³. Thus, as x increases, g(x) will grow faster than f(x) because the cubic term will dominate the quadratic term.
Additionally, by comparing the powers of log x, we note that f(x) has a cubic term (log x)³ while g(x) has a square term (log x)². The difference here does not change the order of growth significantly because the x terms are more influential as x becomes large. Therefore, f(x) is indeed O(g(x)), meaning f(x) grows at most as quickly as a constant multiple of g(x) as x approaches infinity.