I am trying to determine whether my piecewise function is even or odd or neither. If it wasn't a piecewise I would use the trick of subbing in a negative x but when there are two parts to it I don't believe that would work. Is the best way just to observe a sketch of the function?
Cheers
You can still plug in $-x$ and compare. As a simple example, consider the function $$ f(x)=\begin{cases} 2x, \quad &x<5 \\ 15-x, \quad &x\ge 5 \end{cases} \tag1$$ Replace $x$ with $-x$ throughout: $$ f(-x)=\begin{cases} -2x, \quad &-x<5 \\ 15+x, \quad &-x\ge 5 \end{cases} \tag2$$ Rewrite the conditons on the right in a more digestible form: $$ f(-x)=\begin{cases} -2x, \quad &x>-5 \\ 15+x, \quad &x\le -5 \end{cases} \tag3$$ So, $f(-x)$ has a transition at $-5$ instead of $5$. This already suggests that $f$ is neither even nor odd. But to be absolutely sure, calculate $$ f(x)+f(-x) = \begin{cases} 15+3x, \quad &x\le -5 \\ 0 & -5<x<5\\ 15-3x, \quad &x\ge 5 \end{cases} \tag4$$ Since $f(x)+f(-x)$ is not identically zero, $f$ is not odd.
Similarly, $$ f(x)-f(-x) = \begin{cases} x-15, \quad &x\le -5 \\ 4x & -5<x<5\\ 15+x, \quad &x\ge 5 \end{cases} \tag5$$ Since $f(x)-f(-x)$ is not identically zero, $f$ is not even.