Let's analyze the asymptotic growth rates of the given functions: ### Functions: . \( f_1(n) 10^n \) 2. \( f_2(n) = n \log n \) 3. \( f_3(n) = n^{\sqrt{n}} \) --- ### Step-by-step comparison: #### 1. \( f_2(n) = n \log n \) - Grows faster than polynomial functions like \( n \) but slower than exponential functions like \( 10^n \). - It is known to grow slower than \( 10^n \) and \( n^{\sqrt{n}} \). #### 2. \( f_3(n) = n^{\sqrt{n}} \) - This is a super-polynomial function. Since \( \sqrt{n} \) increases without bound, \( n^{\sqrt{n}} \) grows faster than any polynomial but slower than exponential functions with fixed bases, but in this case, it is extremely large. - For large \( n \), \( n^{\sqrt{n}} = e^{\sqrt{n} \log n} \). Since \( \sqrt{n} \log n \) grows faster than \( n \) or \( \log n \), this function grows faster than polynomial and exponential functions with fixed bases. #### 3. \( f_1(n) = 10^n \) - Pure exponential growth, but with base 10. --- ### Ordering: - \( n \log n \) (slowest) - \( n^{\sqrt{n}} \) (faster than polynomial but slower than \( 10^n \)) - \( 10^n \) (fastest) ### Final order: \[ f_2(n) < f_3(n) < f_1(n) \] ### Corresponding option: **f\(_2\), f\(_3\), f\(_1\)** --- ### Correct answer: **f\(_2\), f\(_3\), f\(_1\)** --- **Note:** The accepted answer in the question is **f₁, f₂, f₃**, but based on the asymptotic analysis, the correct order should be **f₂, f₃, f₁**. There might be an inconsistency in the provided options or question, but mathematically, this is the correct sequence. --- **Summary:** - \( n \log n \) grows slowest. - \( n^{\sqrt{n}} \) grows faster than \( n \log n \). - \( 10^n \) grows the fastest. **Therefore, the increasing order is:** \[ \boxed{ f_2, \quad f_3, \quad f_1 } \]