Even and odd functions

Let's consider the graph of the following function $f(x)=x^2$:

We observe that any number $x$ and its opposite $-x$ have the same image. In this case we say that the function is a pair.

A function $f$ is a pair if for any $x$ in the domain we have: $$f(x)=f(-x)$$

Note that even functions are symmetrical with regard to the vertical axis.

Let's consider now the function $f(x)=x^3$:

We know that any number $x$ and its inverse $-x$ have inverse images. In this case we say that the function $f$ is odd.

A function $f$ is odd if for any $x$ in the domain we have:$$f(x)=-f(-x)$$

Given following functions, decide which of them are even or odd: $$\displaystyle f(x)=\frac{1}{x} \mbox{ and } g(x)=x^2-2$$

Let's verify if the functions are even: $$f(x)=f(-1) \Longleftrightarrow \displaystyle \frac{1}{x}=\frac{1}{-x} \Longleftrightarrow 1= -1 !! \\ g(x)=g(-x) \Longleftrightarrow x^2-2=(-x)^2-2=x^2-2 \mbox{ OK }$$

Let's verify if the function $f$ is odd ($g$ will not be odd since it is an even function): $$\displaystyle f(x)=-f(-x) \Longleftrightarrow \frac{1}{x}=-\frac{1}{-x}=\frac{1}{x} \mbox{ OK }$$

Therefore the function $f$ is odd, and the function $g$ is a pair.