To show that the function $f(n) = 3n^3 + 7n + 10$ belongs to $\theta(n^3)$, we need to demonstrate that there exist positive constants $c_1$, $c_2$, and $n_0$ such that for all $n \geq n_0$:

$c_1 n^3 \leq f(n) \leq c_2 n^3$

Step 1: Upper Bound

We first show that $f(n) \leq c_2 n^3$ for some constant $c_2$ and sufficiently large $n$.

$f(n) = 3n^3 + 7n + 10$

For large $n$, the term $3n^3$ will dominate, but we must include all terms for correctness:

$f(n) \leq 3n^3 + 7n^3 + 10n^3 = 3n^3 + \frac{7}{n^2} n^3 + \frac{10}{n^3} n^3$

Now for large enough $n$, the terms $\frac{7}{n^2} n^3$ and $\frac{10}{n^3} n^3$ become negligible compared to $3n^3$. However, let’s combine all terms under a single constant:

$f(n) \leq (3 + \epsilon) n^3 \quad \text{for some small } \epsilon > 0$

Therefore, for a sufficiently large $n$, we can choose $c_2 = 4$ (since 3 + any small $\epsilon$ will be less than 4):

$f(n) \leq 4n^3$

Step 2: Lower Bound

We now show that $f(n) \geq c_1 n^3$ for some constant $c_1$ and sufficiently large $n$.

Notice:

$f(n) = 3n^3 + 7n + 10 \geq 3n^3$

This holds for any $n \geq 1$, so we can take $c_1 = 3$.

Step 3: Conclusion

We have shown that:

$3n^3 \leq f(n) \leq 4n^3$

Thus, $f(n) = 3n^3 + 7n + 10$ satisfies the definition of $\theta(n^3)$, meaning:

$f(n) \in \theta(n^3)$

This concludes the proof.

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Would you like to see more details or have any questions? Here are 5 related questions:

1. What are other examples of functions that belong to $\theta(n^3)$?
2. How does the concept of $\theta$-notation differ from $O$-notation?
3. Can you explain how to determine $\theta$-notation for a more complex function?
4. How would you show that a function belongs to $\theta(n^2)$ instead?
5. What is the significance of asymptotic notations in algorithm analysis?

Tip: When proving that a function belongs to a particular asymptotic notation, always remember to analyze both the upper and lower bounds carefully.