How to come up with an interval for this Big Oh Problem?

Question

This is from Discrete Mathematics and its Applications

I'am trying to use the interval method like what was shown in this example

Here is my work so far:

I noticed that when $x > 5$, $2^x > 17$, so when $x > 5$, $2^x + 17 \leq 2^x + 2^x \leq 2*2^x$. This shows that $2^x + 17$ is $O(2^x)$, because there exist constants $C = 2$ and $x_0 = 5$ such that when $x > 5$, $2^x + 17 < 2 * 2^x$.

The next step was to show that $O(2^x)$ is in $O(3^x)$. To do this, I took a function $2^x$ that is in $O(2^x)$, and first observed that when $x > 1$, $2^x \leq 3^x$, so $2^x$ is in $O(3^x)$ because there exist constants $C = 3$ and $x_0 = 1$ such that when $x > 1$, $2^x < 3^x$.

Did i do everything to show that $2^x + 17$ is $O(3^x)$? Is there a more efficient way to do this, to go straight to $O(3^x)$ and not $O(2^x)$ to $O(3^x)$ like I did ?

Answer 1

For all $x>1$ we get $$17+2^x \le 2^5+2^x \le 2^5\cdot 2^{x} \le 2^5 \cdot 3^{x}=O(3^x).$$ The notation of big O is useful for sufficiently large values of $x$, to give an idea of the order of magnitude of some function.

Answer 2

Notice that $2 \cdot 2^x < 2 \cdot 3^x$ if $(\frac{3}{2})^x >1$, hence $\forall x>0$