**Showing f(n)+g(n)=O(n2)f(n) + g(n) = O(n^2)f(n)+g(n)=O(n2)**

Given:

f(n)=3n2−n+4f(n) = 3n^2 - n + 4f(n)=3n2−n+4

g(n)=nlog⁡n+5g(n) = n \log n + 5g(n)=nlogn+5

We need to prove that f(n)+g(n)f(n) + g(n)f(n)+g(n) is O(n2)O(n^2)O(n2).

To prove that \( f(n) + g(n) = O(n^2) \), we need to show that there exist positive constants \( c \) and \( n_0 \) such that for all \( n \geq n_0 \):

\[
f(n) + g(n) \leq c \cdot n^2
\]

Given:

\[
f(n) = 3n^2 - n + 4
\]
\[
g(n) = n \log n + 5
\]

### Step 1: Analyze \( f(n) \)
The function \( f(n) = 3n^2 - n + 4 \) is a quadratic function. We know that for sufficiently large \( n \), the \( 3n^2 \) term will dominate the linear term \( -n \) and the constant term \( 4 \). Therefore:

\[
f(n) \approx 3n^2
\]

Clearly, \( f(n) \leq 4n^2 \) for large \( n \). Thus:

\[
f(n) = O(n^2)
\]

### Step 2: Analyze \( g(n) \)
The function \( g(n) = n \log n + 5 \) is composed of a term \( n \log n \) which grows faster than any linear function of \( n \) but slower than any quadratic function of \( n \). We can express \( g(n) \) as:

\[
g(n) \leq n \log n + 5 \leq n^2 \text{ for large } n
\]

This is because \( n \log n \) grows slower than \( n^2 \). Therefore:

\[
g(n) = O(n^2)
\]

### Step 3: Combine \( f(n) \) and \( g(n) \)
Now, consider the sum \( f(n) + g(n) \):

\[
f(n) + g(n) = (3n^2 - n + 4) + (n \log n + 5) = 3n^2 + n \log n - n + 9
\]

For large \( n \), the dominant term here is \( 3n^2 \). Therefore:

\[
f(n) + g(n) \leq 3n^2 + n \log n + 9 \leq 4n^2 \text{ for large } n
\]

Thus, \( f(n) + g(n) = O(n^2) \).

### Conclusion
We have shown that both \( f(n) \) and \( g(n) \) are \( O(n^2) \), and therefore their sum \( f(n) + g(n) \) is also \( O(n^2) \).

Would you like to go over any part of this proof in more detail or do you have any questions?

Here are 5 related questions you might find interesting:
1. How would the analysis change if \( f(n) \) was a cubic function?
2. Can \( f(n) + g(n) \) be \( O(n \log n) \)?
3. What if \( g(n) = n^2 \log n \)? How would that affect the proof?
4. How can we prove that a function is \( \Omega(n^2) \)?
5. What is the difference between \( O(n^2) \) and \( \Theta(n^2) \)?

**Tip:** When proving a function is \( O(g(n)) \), focus on identifying the dominant term for large \( n \).

Explore how to prove that the sum of f(n) and g(n), where f(n) = 3n^2 - n + 4 and g(n) = n log n + 5, is O(n^2). This detailed mathematical analysis covers quadratic and logarithmic functions, suitable for advanced college-level students and mathematicians.

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