Answer
$2^{x}+17$ is $O(3^{x})$
Work Step by Step
Solution:
Given: $O\left(3^{x}\right),$ thus $g(x)=3^{x}$ in the definition of $Big-O$ Notation.
$f(x)=2^{x}+17$
When $x>1$, we have the property note$3x>2x$ (note: not true for powers smaller than $1$)
Since $32=9$ and $32=27$, we obtain the property $3x>17$ when $x>2$.
For convenience sake, we will then choose $k=2$ and thus use $x>2$. (note: you could choose a Different value of $k$, which will lead to different value for $C$).
$|f(x)|=\left|2^{x}+17\right|$
$\quad \leq\left|2^{x}\right|+|17|$
$\quad=2^{x}+17$
$<3^{x}+17$
$<3^{x}+3^{x}$
$=2 \cdot 3^{x}$
$=2\left|3^{x}\right|$
thus we need to choose $C$ to be at least $2$. Let us then take $C=2$.
Hence by the definition of Big-O notation, $f(x)=O(3^{x})$ with $k=2$ and $C=2$.
