Even Functions
A function $y=f(x)$ defined on an interval $(-a,a)$ is called even if upon changing the sign of any value of $x$ that belongs to this interval, the value of the function does not change, namely:
\[
f(-x)=f(x)\qquad\text{for every }x\text{ in the interval}(-a,a).
\]
Figure 1: Graph of an even function is symmetric about the $y$-axis |
If a function is even, its graph is symmetric about the $y$-axis. That is, if we draw a horizontal straight line from a point on the graph to the $y$-axis and continue in an equal distance on the other side of the $y$-axis, we will come to another point on the graph(see Figure 1). Equivalently we can say that the graph of $y=f(x)$ is symmetric about the $y$-axis if we fold the coordinate plane along the $y$-axis, the graph which lies in the left half of the plane completely coincides with that in the right half. Examples of even functions are $y=x^{2}$, $y=1/x^{2}$, and $y=\cos x$ (see Figure 2).
(a) Graph of $y=x^{2} $ | (b) Graph of $y=1/x^{2} $ | (c) Graph of $y=\cos x $ |
Figure 2
Odd Functions
A function $y=f(x)$ defined on an interval $(-a,a)$ is called odd if upon changing the sign of any value of $x$ belonging to this interval, only the sign (and not the absolute value) of the function changes, namely
\[
f(-x)=-f(x)\qquad\text{for every }x\text{ in the interval}(-a,a).
\]
If a function is odd, its graph is symmetric about the origin. That is, if we draw a straight line from any point on the graph to the origin and continue in an equal distance on the other side of the origin, we will come to another point on the graph (see Figure 3). Another way of saying that is: the graph of $y=f(x)$ is symmetric about the origin if for any point $(x,y)$ on the graph, $(-x,-y)$ also lies on the graph. Examples of odd functions are $y=x^{3},y=1/x$, and $y=\sin x$ (see Figure 4).
Figure 3: Graph of an odd function is symmetric about the origin |
(a) graph of $ y=x^{3} $ | (b) Graph of $y=1/x $ | (c) Graph of $y=\sin x $ |
Figure 4
- If an odd function $f$ is defined at $x=0$, then its value must be zero there:
\[
f(0)=0.
\] [Suppose $f(0)=b$. Because the point $(0,b)$ lies on the graph of $f$ and $f$ is odd, $(0,-b)$ must also lie on the graph of $f$.Therefore, we must have $b=0$, otherwise $f$ would assign two different values $\pm b$ to $x=0$ which is impossible.]