When expressing interval(s) on which a function is increasing, decreasing, or constant, we consider only values in the domain of the function. Since the domain of this function is $(-\infty ,\infty )$, we consider all real values of x.

1. As x-values (that is, values in the domain) increase from $x=3$ to $x=5$, the y-values (that is, values in the range) increase from −2 to 2. Thus the function is increasing on the interval (3, 5).

2. As x-values increase from negative infinity to −1, y-values decrease; y-values also decrease as x-values increase from 5 to positive infinity. Thus the function is decreasing on the intervals $(-\infty ,-1)$ and $(5,\infty )$.

3. As x-values increase from −1 to 3, y remains −2. The function is constant on the interval (−1, 3).

We see that the relative maximum value of the function is 2.004. It occurs when $x=-0.082$. We also see the relative minimum: −1.604 at $x=4.082$.

We note that the graph starts rising, or increasing, from the left and stops increasing at the relative maximum. From this point, the graph decreases to the relative minimum and then begins to rise again. The function is increasing on the intervals

and decreasing on the interval

Let’s summarize our results.

1. Suppose 1 hr goes by. At that time, Kiara has traveled 35 mi and Matias has traveled 40 mi. We can use the Pythagorean theorem to find the distance between them. This distance would be the length of the hypotenuse of a right triangle with legs measuring 35 mi and 40 mi. After 2 hr, the triangle’s legs would measure $2\cdot 35$, or 70 mi, and $2\cdot 40$, or 80 mi. Noting that the distances will always be changing, we make a drawing and let $t=$ the time, in hours, that Kiara and Matias have been driving since leaving the hospital.

   

   After t hours, Kiara has traveled 35t miles and Matias 40t miles. We now use the Pythagorean theorem:

   $${[d(t)]}^2={(35t)}^2+{(40t)}^2.$$

   Because distance must be nonnegative, we need consider only the positive square root when solving for d(t):

   $$\begin{array}{rcll}d(t)& =& \sqrt{{(35t)}^2+{(40t)}^2}& \\ & =& \sqrt{1225t^2+1600t^2}& \\ & =& \sqrt{2825t^2}& \\ & \approx & 53.15\left|t\right|& \mathbf{\text{Approximating the root to two decimal places}}\\ & \approx & 53.15t.& \mathbf{\text{Since }}\mathit{t}\mathbf{\ge }\mathbf{0}\mathbf{,}\mathbf{\left|}\mathit{t}\mathbf{\right|}\mathbf{=}\mathit{t}\mathbf{.}\end{array}$$

   Thus, $d(t)=53.15t,t\ge 0$.

2. Since the time traveled, t, must be nonnegative, the domain is the set of nonnegative real numbers $[0,\infty )$.

1. Note that the dividers will form two sides of a rectangle. If, for example, 14 ft of dividers are used for the length of the rectangle, that would leave $30-14$, or 16 ft of dividers for the width. Thus if $x=$ the length, in feet, of the rectangle, then $30-x=$ the width. We represent this information in a drawing, as shown below.

   

   The area, $A(x)$, is given by

   $$\begin{array}{rcll}A(x)& =& x(30-x)& \mathbf{\text{Area}}\mathbf{=}\mathbf{\text{length}}\mathbf{\cdot }\mathbf{\text{width}}\mathbf{\text{.}}\\ & =& 30x-x^2.& \end{array}$$

   The function $A(x)=30x-x^2$ can be used to express the rectangle’s area as a function of the length.

2. Because the rectangle’s length and width must be positive and only 30 ft of dividers are available, we restrict the domain of A to $\left\{x\right|0<x<30\}$—that is, the interval (0, 30).

3. On the graph of the function shown at the top of the page, the maximum value of the area on the interval (0, 30) appears to be 225 when $x=15$. Thus the dimensions that maximize the area are

   • Length $=x=15$ ft and

   • Width $=30-x=30-15=15$ ft.

First, we determine which part of the domain contains the given input. Then we use the corresponding formula to find the output.

Since $-5<-2$, we use the formula $f(x)=x+1:$

Since $-3<-2$, we use the formula $f(x)=x+1$ again:

Since $-2\le 0\le 3$, we use the formula $f(x)=5:$

Since $-2\le 3\le 3$, we use the formula $f(x)=5$ a second time:

Since $4>3$, we use the formula $f(x)=x^2:$

Since $10>3$, we once again use the formula $f(x)=x^2:$

Since the function is defined in two pieces, or parts, we create the graph in two parts.

1. We graph $g(x)=\frac{1}{3}x+3$ only for inputs x less than 3. That is, we use $g(x)=\frac{1}{3}x+3$ only for x-values in the interval $(-\infty ,3)$. Some ordered pairs that are solutions of this piece of the function are shown in Table 1.

   x $(x<3)$	$g(x)=\frac{1}{3}x+3$
   −3	2
   0	3
   2	$3\frac{2}{3}$

2. We graph $g(x)=-x$ only for inputs x greater than or equal to 3. That is, we use $g(x)=-x$ only for x-values in the interval $[3,\infty )$. Some ordered pairs that are solutions of this piece of the function are shown in Table 2.

   x $(x\ge 3)$	$g(x)=-x$
   3	−3
   4	−4
   6	−6

   

We create the graph in three pieces, or parts.

1. We graph $f(x)=4$ only for inputs x less than or equal to 0. That is, we use $f(x)=4$ only for x-values in the interval $(-\infty ,0]$. Some ordered pairs that are solutions of this piece of the function are shown in Table 3.

   x $(x\le 0)$	$f(x)=4$
   −5	4
   −2	4
   0	4

2. We graph $f(x)=4-x^2$ only for inputs x greater than 0 and less than or equal to 2. That is, we use $f(x)=4-x^2$ only for x-values in the interval $(0,2]$. Some ordered pairs that are solutions of this piece of the function are shown in Table 4.

   x $(0<x\le 2)$	$f(x)=4-x^2$
   $\frac{1}{2}$	$3\frac{3}{4}$
   1	3
   2	0

3. We graph $f(x)=2x-6$ only for inputs x greater than 2. That is, we use $f(x)=2x-6$ only for x-values in the interval $(2,\infty )$. Some ordered pairs that are solutions of this piece of the function are shown in Table 5.

   x $(x>2)$	$f(x)=2x-6$
   $2\frac{1}{2}$	−1
   3	0
   5	4

   

When $x\ne -2$, the denominator of $(x^2-4)/(x+2)$ is nonzero, so we can simplify:

Thus,

The graph of this part of the function consists of a line with a “hole” at the point $(-2,-4)$, indicated by the open circle. The hole occurs because the piece of the function represented by $(x^2-4)/(x+2)$ is not defined for $x=-2$. By the definition of the function, we see that $f(-2)=3$, so we plot the point (−2, 3) above the open circle.

The greatest integer function can also be defined as a piecewise function with an infinite number of statements.

We see that the domain of this function is the set of all real numbers, $(-\infty ,\infty )$, and the range is the set of all integers, $\{\dots ,-3,-2,-1,0,1,2,3,\dots \}$.