1. **Stating the problem:** We have two functions: $$f_1(n) = (n \log n + n^2)(n^3 + 2)$$ $$f_2(n) = (n! + 2^n)(n^3 + \log(n^2 + 1))$$ We need to find the Big-O, Big-\Omega, and Big-\Theta estimates for both $f_1(n)$ and $f_2(n)$. 2. **Analyzing $f_1(n)$:** - Inside the first parentheses: $n \log n$ grows slower than $n^2$, so dominant term is $n^2$. - Inside the second parentheses: $n^3$ dominates the constant $2$. - Therefore, dominant terms multiplied give: $$f_1(n) \approx n^2 \times n^3 = n^5$$ - So, - Big-O estimate: $O(n^5)$ because the function grows no faster than $n^5$ times a constant. - Big-\Omega estimate: $\Omega(n^5)$ because the function grows at least as fast as $n^5$ times a constant. - Big-\Theta estimate: $\Theta(n^5)$ because both bounds are tight. 3. **Analyzing $f_2(n)$:** - Inside the first parentheses: $n!$ grows faster than $2^n$, so $n!$ dominates. - Inside the second parentheses: $n^3$ grows faster than $\log(n^2+1)$, so $n^3$ dominates. - Multiplying dominant terms: $$f_2(n) \approx n! \times n^3$$ - So, - Big-O estimate: $O(n! \times n^3)$ - Big-\Omega estimate: $\Omega(n! \times n^3)$ - Big-\Theta estimate: $\Theta(n! \times n^3)$ 4. **Summary:** - $f_1(n)$: $O(n^5)$, $\Omega(n^5)$, $\Theta(n^5)$ - $f_2(n)$: $O(n! n^3)$, $\Omega(n! n^3)$, $\Theta(n! n^3)$