# How do you analyse the time complexity of a piecewise function?

Is there a method to apply when trying to finding out the time complexity of a piecewise function?

\begin{align*} F(x) &= \begin{cases} 2^x & x < 8, \\ x^2 & x \geq 8. \end{cases} \\ \end{align*}

This case seems straightforward as the function is $\Theta(x^2)$ for a large enough value of $x$, but what should be done in a more complex case... for example a function that gives a result for an odd value of $x$ and another one for an even value of $x$?

There seems to be a misconception here: every function $f$ has a "nice" asymptotic form, in the sense that $f = \Theta(g)$ for some function $g$ given by a reasonable formula. Some examples of reasonable formulas include $c^nn^k\log^\ell n$ for various values of $c,k,\ell$. However, this isn't always the case. Consider for example the function $$f(n) = \begin{cases} n & \text{if n is odd}, \\ n^2 & \text{if n is even}. \end{cases}$$ There is no "nice" function $g$ such that $f = \Theta(g)$. It is apparent that $f = \Omega(n)$ and that $f = O(n^2)$, but there is no value of $k$ for which $f = \Theta(n^k)$.
This discussion is related to another misconception, that we can always compare two functions asymptotically. That is, that for any two functions $f,g$, either $f = o(g)$, or $f = \Theta(g)$, or $f = \omega(g)$. While these three possibilities are indeed mutually exclusive, there is a fourth possibility, namely, that $f$ and $g$ are incomparable. As an example, if we take the function $f$ considered above and try to compare it to $g(n) = n^{3/2}$, then we find that none of $f = o(g)$, $f = \Theta(g)$, $f = \omega(g)$ holds. However, the trichotomy does hold when $f$ and $g$ are "nice" functions (defined using only arithmetic operations, logarithm, and exponential).
Finally, there is one special case which is easy to handle: if $$f(n) = \begin{cases} g(n) & \text{if } n < C, \\ h(n) & \text{if } n \geq C, \end{cases}$$ Then $f(n) = \Theta(h(n))$. This is the case in your example.