Subsection 3.6.4 Symmetries
Before we proceed to some examples, we should examine some simple symmetries possessed by some functions. We’ll look at three symmetries — evenness, oddness and periodicity. If a function possesses one of these symmetries then it can be exploited to reduce the amount of work required to sketch the graph of the function.
Let us start with even and odd functions.
Definition 3.6.6.
A function \(f(x)\) is said to be even if \(f(-x)=f(x)\) for all \(x\text{.}\)
Definition 3.6.7.
A function \(f(x)\) is said to be odd if \(f(-x)=-f(x)\) for all \(x\text{.}\)
Example 3.6.8. An even function and an odd funtion.
Let \(f(x) = x^2\) and \(g(x)=x^3\text{.}\) Then
\begin{align*}
f(-x) &= (-x)^2 = x^2 = f(x)\\
g(-x) &= (-x)^3 = -x^3 = -g(x)
\end{align*}
Hence \(f(x)\) is even and \(g(x)\) is odd.
Notice any polynomial involving only even powers of \(x\) will be even
\begin{align*}
f(x) &= 7x^6+2x^4-3x^2+5 & \text{remember that } 5=5x^0\\
f(-x) &= 7(-x)^6+2(-x)^4-3(-x)^2+5\\
&= 7x^6+2x^4-3x^2+5 = f(x)
\end{align*}
Similarly any polynomial involving only odd powers of \(x\) will be odd
\begin{align*}
g(x) &= 2x^5-8x^3-3x\\
g(-x) &= 2(-x)^5-8(-x)^3-3(-x)\\
&= -2x^5+8x^3+3x = -g(x)
\end{align*}
Not all even and odd functions are polynomials. For example
\begin{align*}
|x| && \cos x && \text{ and } (e^x + e^{-x})
\end{align*}
are all even, while
\begin{align*}
\sin x && \tan x && \text{ and } (e^x-e^{-x})
\end{align*}
are all odd. Indeed, given any function \(f(x)\text{,}\) the function
\begin{align*}
g(x) &= f(x)+f(-x) & \text{ will be even, and}\\
h(x) &= f(x)-f(-x) & \text{ will be odd.}
\end{align*}
Now let us see how we can make use of these symmetries to make graph sketching easier. Let \(f(x)\) be an even function. Then
\begin{gather*}
\text{the point } (x_0,y_0)\text{ lies on the graph of }y=f(x)
\end{gather*}
if and only if \(y_0= f(x_0) = f(-x_0)\) which is the case if and only if
\begin{gather*}
\text{the point }(-x_0,y_0)\text{ lies on the graph of }y=f(x).
\end{gather*}
Notice that the points \((x_0,y_0)\) and \((-x_0,y_0)\) are just reflections of each other across the \(y\)-axis. Consequently, to draw the graph \(y=f(x)\text{,}\) it suffices to draw the part of the graph with \(x\ge 0\) and then reflect it in the \(y\)–axis. Here is an example. The part with \(x\ge 0\) is on the left and the full graph is on the right.
Very similarly, when \(f(x)\) is an odd function then
\begin{gather*}
(x_0,y_0)\text{ lies on the graph of }y=f(x)
\end{gather*}
if and only if
\begin{gather*}
(-x_0,-y_0)\text{ lies on the graph of }y=f(x)
\end{gather*}
Now the symmetry is a little harder to interpret pictorially. To get from \((x_0,y_0)\) to \((-x_0,-y_0)\) one can first reflect \((x_0,y_0)\) in the \(y\)–axis to get to \((-x_0,y_0)\) and then reflect the result in the \(x\)–axis to get to \((-x_0,-y_0)\text{.}\) Consequently, to draw the graph \(y=f(x)\text{,}\) it suffices to draw the part of the graph with \(x\ge 0\) and then reflect it first in the \(y\)–axis and then in the \(x\)–axis. Here is an example. First, here is the part of the graph with \(x\ge 0\text{.}\)
Next, as an intermediate step (usually done in our heads rather than on paper), we add in the reflection in the \(y\)–axis.
Finally to get the full graph, we reflect the dashed line in the \(x\)–axis
and then remove the dashed line.
Let’s do a more substantial example of an even function
Example 3.6.9. An even rational function.
Consider the function
\begin{align*}
g(x) &= \frac{x^2-9}{x^2+3}
\end{align*}
The function is even since
\begin{align*}
g(-x) &= \frac{(-x)^2-9}{(-x)^2+3} = \frac{x^2-9}{x^2+3} = g(x)
\end{align*}
Thus it suffices to study the function for
\(x\geq0\) because we can then use the even symmetry to understand what happens for
\(x \lt 0\text{.}\)
The function is defined on all real numbers since its denominator \(x^2+3\) is never zero. Hence it has no vertical asymptotes.
The \(y\)-intercept is \(g(0) = \frac{-9}{3} = -3\text{.}\) And \(x\)-intercepts are given by the solution of \(x^2-9=0\text{,}\) namely \(x=\pm 3\text{.}\) Note that we only need to establish \(x=3\) as an intercept. Then since \(g\) is even, we know that \(x=-3\) is also an intercept.
To find the horizontal asymptotes we compute the limit as
\(x\to+\infty\)
\begin{align*}
\lim_{x\to \infty} g(x)
&= \lim_{x\to \infty} \frac{x^2-9}{x^2+3}\\
&= \lim_{x\to \infty} \frac{x^2(1-9/x^2)}{x^2(1+3/x^2)}\\
&= \lim_{x\to \infty} \frac{1-9/x^2}{1+3/x^2} = 1
\end{align*}
Thus
\(y=1\) is a horizontal asymptote. Indeed, this is also the asymptote as
\(x\to-\infty\) since by the even symmetry
\begin{align*}
\lim_{x\to -\infty} g(x)
&=\lim_{x\to \infty} g(-x)
= \lim_{x\to \infty} g(x).
\end{align*}
-
We can already produce a quite reasonable sketch just by putting in the horizontal asymptote and the intercepts and drawing a smooth curve between them.
Note that we have drawn the function as never crossing the asymptote \(y=1\text{,}\) however we have not yet proved that. We could by trying to solve \(g(x)=1\text{.}\)
\begin{align*}
\frac{x^2-9}{x^2+3} &= 1\\
x^2-9 &= x^2+3\\
-9=3 & \text{ so no solutions.}
\end{align*}
Alternatively we could analyse the first derivative to see how the function approaches the asymptote.
Now we turn to the first derivative:
\begin{align*}
g'(x) &= \frac{(x^2+3)(2x) - (x^2-9)(2x)}{(x^2+3)^2}\\
&= \frac{24x}{(x^2+3)^2}
\end{align*}
There are no singular points since the denominator is nowhere zero. The only critical point is at
\(x=0\text{.}\) Thus we must find the sign of
\(g'(x)\) on the intervals
\begin{align*}
(-\infty,0) && (0,\infty)
\end{align*}
When \(x \gt 0\text{,}\) \(24x \gt 0\) and \((x^2+3) \gt 0\text{,}\) so \(g'(x) \gt 0\) and the function is increasing. By even symmetry we know that when \(x \lt 0\) the function must be decreasing. Hence the critical point \(x=0\) is a local minimum of the function.
Notice that since the function is increasing for \(x \gt 0\) and the function must approach the horizontal asymptote \(y=1\) from below. Thus the sketch above is quite accurate.
Now consider the second derivative:
\begin{align*}
g''(x) &= \diff{}{x} \frac{24x}{(x^2+3)^2}\\
&= \frac{(x^2+3)^2 \cdot 24 - 24x\cdot 2 (x^2+3)\cdot2x}{(x^2+3)^4}\\
\end{align*}
cancel a factor of \((x^2+3)\)
\begin{align*}
&= \frac{(x^2+3) \cdot 24 - 96x^2}{(x^2+3)^3}\\
&= \frac{72(1-x^2)}{(x^2+3)^3}
\end{align*}
It is clear that
\(g''(x) = 0\) when
\(x=\pm 1\text{.}\) Note that, again, we can infer the zero at
\(x=-1\) from the zero at
\(x=1\) by the even symmetry. Thus we need to examine the sign of
\(g''(x)\) the intervals
\begin{align*}
(-\infty,-1)&&(-1,1)&&(1,\infty)
\end{align*}
When \(|x| \lt 1\) we have \((1-x^2) \gt 0\) so that \(g''(x) \gt 0\) and the function is concave up. When \(|x| \gt 1\) we have \((1-x^2) \lt 0\) so that \(g''(x) \lt 0\) and the function is concave down. Thus the points \(x=\pm 1\) are inflection points. Their coordinates are \((\pm1, g(\pm1)) =(\pm 1,-2)\text{.}\)
-
Putting this together gives the following sketch:
Another symmetry we should consider is periodicity.
Definition 3.6.10.
A function \(f(x)\) is said to be periodic, with period \(P \gt 0\text{,}\) if \(f(x+P)=f(x)\) for all \(x\text{.}\)
Note that if \(f(x+P)=f(x)\) for all \(x\text{,}\) then replacing \(x\) by \(x+P\text{,}\) we have
\begin{gather*}
f(x+2P)=f(x+P+P)=f(x+P)=f(x).
\end{gather*}
More generally \(f(x+kP)=f(x)\) for all integers \(k\text{.}\) Thus if \(f\) has period \(P\text{,}\) then it also has period \(nP\) for all natural numbers \(n\text{.}\) The smallest period is called the fundamental period.
Example 3.6.11. \(\sin x\) is periodic.
The classic example of a periodic function is \(f(x)=\sin x\text{,}\) which has period \(2\pi\) since \(f(x+2\pi)=\sin(x+2\pi)=\sin x=f(x)\text{.}\)
If \(f(x)\) has period \(P\) then
\begin{gather*}
(x_0,y_0)\text{ lies on the graph of }y=f(x)\\
\end{gather*}
if and only if \(y_0=f(x_0)=f(x_0+P)\) which is the case if and only if
\begin{gather*}
(x_0+P,y_0)\text{ lies on the graph of }y=f(x)
\end{gather*}
and, more generally,
\begin{gather*}
(x_0,y_0)\text{ lies on the graph of }y=f(x)\\
\end{gather*}
if and only if
\begin{gather*}
(x_0+nP,y_0)\text{ lies on the graph of }y=f(x)
\end{gather*}
for all integers \(n\text{.}\)
Note that the point \((x_0+P,y_0)\) can be obtained by translating \((x_0,y_0)\) horizontally by \(P\text{.}\) Similarly the point \((x_0+nP,y_0)\) can be found by repeatedly translating \((x_0,y_0)\) horizontally by \(P\text{.}\)
Consequently, to draw the graph \(y=f(x)\text{,}\) it suffices to draw one period of the graph, say the part with \(0\le x\le P\text{,}\) and then translate it repeatedly. Here is an example. Here is a sketch of one period
and here is the full sketch.
