Is there a function $f(x)$ and a constant $k$ so that $f(x-k) \neq O(f(x))$? Is there a function $f(x)$ and a constant $k$ so that $f(x-k) \neq O(f(x))$?
If it exist, what is the function, and if it doesn't, what is the proof to that?
My intuition is that big O refers to the worst case, meaning worst x, so no constant should have a significant effect on the time complexity, but I'm not sure at all, and I'm struggling to write it formally.
As Christian Blatter commented, let $f(x)=e^{-x^2}$. Then
$$
\begin{align}
f(x-k)
&amp;=e^{-(x-k)^2}\\
&amp;=e^{2kx-k^2}e^{-x^2}\\
&amp;=e^{2kx-k^2}f(x)
\end{align}
$$
That is, for $k\gt0$, $\frac{f(x-k)}{f(x)}$ grows without bound as $x\to\infty$.
This means that $f(x-k)\not\in O(f(x))$.
I got an idea, would love some verification:
I will prove that no such function exist and that for any function $f(x-k)=O(f(x))$.

To prove that, I will use proof by contradiction assuming that there is a function that for $f(x-k)\neq O(f(x))$ does hold.
There for $\lim \frac{f(x-k)}{f(x)}=\infty$
$\lim f(x)=\lim f(x-k)$
If so, then $\lim \frac{f(x-k)}{f(x)}=\frac{\lim f(x-k)}{\lim f(x)}=1$, in contradiction to our assumption that $f(x-k)\neq O(f(x))\Leftrightarrow \lim \frac{f(x-k)}{f(x)}=\infty$.
There for indeed $f(x-k)=O(f(x))\blacksquare$