Piecewise Function Help and Determining whether Odd, Even of Neither I have two piecewise functions and I am trying to work out whether they are Odd, Even or neither. I understand the rules behind finding out whether a normal function is odd, even or neither by using $F(x)= F(-x)$ and 
$F(-x)= -F(x)$ etc. With these, however, because some are just straight integers with a domain, I do not know how to apply the rules to test out what they are.
The first function is:
$$
f(x)=\begin{cases} 4, \quad &0<x<2 \\ -4, \quad &-2<x< 0 \end{cases}
$$
This also has a period of $4$. 
The second function is:
$$
f(x)=\begin{cases} \sin x, \quad &0<x<\pi \\ -1, \quad &\pi<x<2\pi \end{cases}
$$
This has a period of $2\pi$. 
I understand how to plot the functions, and how to read them. As for working out whether they are odd, even or neither I am completely stuck. Any help would be great, thank you.
 A: Hint Try plotting the two functions.
[First function] What can you say about $f(1)$ and $f(-1)$? Can you generalize this to $f(x)$ for $x > 0$ and $f(x)$ for $x < 0$ using the given definitions?
[Second function] What is $f(-\pi/2)$ and what is $f(\pi/2)$? Can it be odd or even?
A: A sketch of the graph of the function with period $4$ that satisfies 
$$f(x) = 
\begin{cases}
4 && \text{if $0 < x < 2$}\\
-4 && \text{if $-2 < x < 0$}
\end{cases}
$$
is shown below.

Note that the defining property of even functions that $f(-x) = f(x)$ for each $x$ in its domain implies that a point $(a, b)$ is on the graph of an even function if and only if a point $(-a, b)$ is also on the graph of the function.  Hence, the graph of an even function is symmetric with respect to the $y$-axis. 
The defining property of odd functions that $f(-x) = -f(x)$ for each $x$ in the domain of $f$ implies a point $(a, b)$ is on the graph of an odd function if and only if the point $(-a, -b)$ is also on the graph.  Hence, the graph of an odd function is symmetric with respect to the origin.
The portion of the graph that we can see is symmetric with respect to the origin, which suggests the function may be odd.   
Observe that 
$$f(x) = 
\begin{cases}
4 && \text{if $4n < x < 4n + 2$, $n \in \mathbb{Z}$}\\
-4 && \text{if $4n - 2 < x < 4n$, $n \in \mathbb{Z}$}
\end{cases}
$$ 
To prove that $f$ is odd, we must show that for each $x$ in its domain, $f(-x) = -f(x)$.  Suppose $x$ satisfies $4n < x < 4n + 2$ for some $n \in \mathbb{Z}$.  Then $f(x) = 4$ and 
$$-4n > -x > -4n - 2 \implies 4(-n) > -x > 4(-n) - 2$$
Since $n \in \mathbb{Z}$, $-n \in \mathbb{Z}$.  Hence, $f(-x) = -4 = -f(x)$.  
I will leave it to you to verify that $f(-x) = -f(x)$ if $4n - 2 < x < 4n$ for some $n \in \mathbb{Z}$.  
A sketch of the graph with period $2\pi$ that satisfies
$$f(x) = 
\begin{cases}
\sin x && \text{if $0 < x < \pi$}\\
-1 && \text{if $\pi < x < 2\pi$}
\end{cases}
$$ 
is shown below.

The graph does not exhibit symmetry with respect to either the $y$-axis or the origin, which suggests that the function is neither even nor odd.  We can confirm this by observing that 
$$f\left(\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \neq -1 = f\left(-\frac{\pi}{6}\right)$$
so the function is not even, and 
$$f\left(-\frac{\pi}{6}\right) = -1 \neq -\frac{1}{2} = -\sin\left(\frac{\pi}{6}\right)$$
so the function is not odd.  
A: When it comes to periodic functions, only the central repeating part needs to be considered for function parity.
For Function 1, begin with $0<x<2$; $f(x)=4$ for any choice. Now consider its reflection, $-2<x<0$, for which the function is $-4$ for any choice. Since $f(-x)=-f(x)$ everywhere, the function is odd.
Function 2 is neither, and I only need one value to show it. Move the second piece back one period to $-\pi<x<0$, and take $x=\pi/6$. Here $f(x)=1/2$ but $f(-x)=-1$, and the latter is neither the same as nor the negation of the former.
