As we move on the graph of a function from left to right (which corresponds to the increase in the argument $x$) if the graph constantly moves upwards, we say that the function is increasing (Figure 1(a)), and if the graph constantly moves downward, we say that the function is decreasing (Figure 1(b)).

 (a) Graph of an increasing function (b) Graph of a decreasing function

Figure 1

Definition: Let $f(x)$ be defined on an interval $I$.

• We say $f(x)$ is increasing on $I$, if for every pair of points $x_{1},x_{2}$ in $I$ satisfying the condition $x_{1}<x_{2}$, we have $f(x_{1})<f(x_{2})$.
•  We say $f(x)$ is decreasing on $I$, if for every pair of points $x_{1},x_{2}$ in $I$ satisfying the condition $x_{1}<x_{2}$, we have $f(x_{1})>f(x_{2})$.
• Note that $I$ can be finite (or bounded) or infinite (unbounded).
•  An interval on which the function is increasing is called an interval of increase of a function while an interval on which the function is decreasing is called an interval of decrease.
• A function that is either increasing or decreasing on an interval is said to be monotonic on the interval.

For example, $f(x)=x^{2}$ is decreasing on $(-\infty,0]$ and is increasing on $[0,\infty)$ (Figure 2(a)). The function $g(x)=x^{3}$ is increasing on $(-\infty,\infty)$ (Figure 2(b)).

 (a) Function $f(x)=x^{2}$ is decreasing on $(-\infty,0]$ and is increasing on $[0,\infty)$ . (b) Function $g(x)=x^{3}$ is increasing in its entire domain $(-\infty,\infty)$ .
Figure 2
Example 1
Let
$g(x)=\begin{cases} x^{2} & x\leq0\\ 1-x^{2} & x\geq1 \end{cases}$ Determine whether $g(x)$ is monotonic on $(-\infty,\infty)$.
Solution
The best way to determine the behavior of $g$ is to graph it. We have learned how to graph simple functions like $g$ in Section 2.11.The graph of $g$ is shown in the following figure.

From Figure 3, we note that $g$ is decreasing on $(\infty,0]$ and on $[1,\infty)$; the point of the graph always moves downward as we move from left to right. However, $g$ is not a decreasing function on its entire domain $(-\infty,0]\cup[1,\infty)$ because if $x_{1}=0$ and $x_{2}=1$, although $x_{1}<x_{2}$, we have $g(x_{1})=g(x_{2})=0$. Thus $g$ is not monotonic on $(-\infty,0]\cup[1,\infty)$.