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Precalculus Study Guide

Chapter 10: Piecewise-Defined Functions

Introduction

A piecewise-defined function is an amalgam of two or more rules, each operative on its own domain. In general, these domains are contiguous, but that is not an essential element of the definition.

Some piecewise functions have special names that are used the way the name of the sine or exponential functions might be used. Two, in particular, are $|x|$ and $[x]$ , the first being the absolute-value function, and the second, the greatest-integer function. These are met in the next two sections below. The general manipulation of piecewise-defined functions is thus the object of this chapter.

Chapter Glossary

The following terms in Chapter 10 are linked to the Maple Math Dictionary.

absolute value

additive inverse

denominator

domain

exponential

extension

greatest integer function

inequality

interval

negative

nonnegative

numerator

positive

real line

real number

sine

singularity

subset

The Piecewise-Defined Function $|x|$

As was seen in Chapter 1, the absolute value function, written as

$f (x) = |x|$

is always a nonnegative quantity. For example, the absolute value of $- 3$ is the positive number 3.

Viewing the change from $- 3$ to 3 as "dropping the minus sign" obscures the abstract definition of this function.

It's much better to see the transition of $- 3$ to 3 as attained by multiplying the negative number $- 3$ by $- 1$ to form +3, the additive inverse of $- 3$ . Hence, the action of the absolute value function should be seen as

$|- 3| = - (- 3)$ = 3

that is, don't drop a minus sign, but make the negative number positive by introducing a second minus sign. This is worth repeating.

Make a negative number positive by negating the negative number

If this is completely understood, then the abstract definition of the absolute value function, namely,

$|x| = {\begin{array}{c} - x & x < 0 \\ 0 & x = 0 \\ x & x > 0 \end{array}$

is easier to understand. If $x$ is already know to be a negative number, it is made positive by putting a minus sign in front of it. If $x$ is zero, its absolute value remains zero. If $x$ is positive, then its absolute value is itself, so $x$ remains unchanged.

The abstract definition of the absolute value function just given is an example of a piecewise-defined function. In general, a piecewise-defined function has more than one rule, and each rule is applied for a specific set of values in the domain of the function.

A Maple implementation of the absolute-value function as a piecewise-defined function is given in Table 10.0.1.

$convert (|x|, piecewise)$

Table 10.0.1 Maple's implementation of $|x|$ as a piecewise function

Notice how the value at $x = 0$ is included in the case of positive $x$ .

Table 10.0.2 shows the Maple piecewise command used to reproduce the absolute-value function.

$piecewise (x < 0, - x, x)$

Table 10.0.2 The absolute-value function via Maple's piecewise command

Note the use of "otherwise" to represent the complement of the case $x < 0$ . Hence, it represents the case $x \geq 0$ .

The Greatest-Integer Function $[x]$

If $x$ is a real number, the symbol $[x]$ typically denotes the greatest integer that is less than or equal to $x$ . This defines the greatest-integer function

$[x]$ = the greatest integer $n$ that satisfies the condition $n \leq x$

In effect, this is a piecewise-defined function with an infinite number of rules. A piecewise representation of the greatest-integer function would look something like the following.

$[x] = {\begin{array}{c} ⋮ & ⋮ \\ - 2 & - 2 \leq x < - 1 \\ - 1 & - 1 \leq x < 0 \\ 0 & 0 \leq x < 1 \\ 1 & 1 \leq x < 2 \\ 2 & 2 \leq x < 3 \\ ⋮ & ⋮ \end{array}$

Those who think visually and geometrically would see $[x]$ as $x$ itself whenever $x$ is an integer, and as the integer on the number line that is just to the left of $x$ whenever $x$ is not an integer.

Maple's floor function is an implementation of the greatest-integer function. Some specific values are listed in Table 10.0.3.

$[- 2.4] = floor (- 2.4) &semi; [- 1.5] = floor (- 1.5) &semi; [- 0.7] = floor (- .7) &semi; [0.8] = floor (.8) &semi; [1.25] = floor (1.25) &semi; [2.78] = floor (2.78)$

Table 10.0.3 Some values of $[x]$ as given by Maple's floor command

A portion of the graph of $f (x) = [x]$ is shown in Figure 10.0.1.

Figure 10.0.1 Portion of the graph of $[x] = floor (x)$

Typical Problems

10.1. Graph the function $f (x) = \{\begin{array}{c} 2 x + 1 & x < - 1 \\ 1 & x \leq 1 \\ x^{2} & x < 2 \\ 3 - x & otherwise \end{array}$ on the interval $[- 2, 4]$ , and obtain the value $f (2)$ .

10.2. Compare the functions $f (x) = \frac{x^{2} - 1}{x - 1}, g (x) = x + 1$ , and $h (x) = {\begin{array}{c} \frac{x^{2} - 1}{x - 1} & x \neq 1 \\ 2 & x = 1 \end{array}$ .

10.3. Graph $f (x) = x + [x]$ on the interval $[- 3, 3]$ .

10.4. Graph $f (x) = x - [x]$ on the interval $[- 3, 3]$ .

10.5. Represent $f (x) = |3 x - 5| + 2$ as a piecewise-defined function, and obtain its graph on the interval $[- 1, 4]$ .

10.6. Represent $f (x) = |1 - 2 x - 3 x^{2}|$ as a piecewise-defined function, and obtain its graph on the interval $[- 2, 2]$ .

Maple Initializations

Before accessing the Maple version of the solutions to the typical problems stated above, initialize Maple by pressing the button provided on the right.

Solutions

Problem 10.1

10.1 - Mathematical Solution

The graph of the function

$f (x) = \{\begin{array}{c} 2 x + 1 & x < - 1 \\ 1 & x \leq 1 \\ x^{2} & x < 2 \\ 3 - x & otherwise \end{array}$

on the interval $[- 2, 4]$ appears in Figure 10.1.1.

At $x = 2$ , the value of $f (x)$ is $f (2) = 1$ .

Figure 10.1.1 Graph of piecewise-defined $f (x)$

10.1 - Maplet Solution

A graph of the function

$f (x) = \{\begin{array}{c} 2 x + 1 & x < - 1 \\ 1 & x \leq 1 \\ x^{2} & x < 2 \\ 3 - x & otherwise \end{array}$

on the interval $[- 2, 4]$ , and the value $f (2)$ , can be found with the Piecewise Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 10.1.2.

A piecewise function with up to four different rules can be entered rule-by-rule as a function of type Format 2.

Figure 10.1.2 Thumbnail image of the Piecewise Function Tutor

The Graph Format 2 button is used to generate the required graph. The Plot Options button should be used to limit the domain to the given interval $[- 2, 4]$ .

The value $f (2) = 1$ is found by entering $x = 2$ in the Evaluation section, and clicking the button labeled Evaluate $f (x)$ Format 2.

To launch the Piecewise Function Tutor, click the following link: Piecewise Function Tutor

10.1 - Interactive Solution

Enter the piecewise function

A piecewise function can be created interactively by means of $\{\begin{array}{c} - x & x < a \\ x & x \geq a \end{array}$ , the piecewise template in the Expression palette. Move through the fields of the template using the Tab key.

Use Control + Shift + r to generate additional rows.

•	Complete the equation $f (x) = \dots$ with the piecewise function.

•	Context Panel: Assign Function

Graph $f (x)$

•	Type $f (x)$ and press the Enter key.

•	Context Panel: Plots≻Plot Builder $- 2 \leq x \leq 4$ Options: Find Discontinuities

Obtain $f (2)$

•	Type $f (2)$

•	Context Panel: Evaluate and Display Inline

10.1 - Programmatic Solution

If the function

$f (x) = \{\begin{array}{c} 2 x + 1 & x < - 1 \\ 1 & x \leq 1 \\ x^{2} & x < 2 \\ 3 - x & otherwise \end{array}$

is entered into Maple as the expression

>	$f ≔ piecewise (x < - 1, 2 x + 1, x \leq 1, 1, x < 2, x^{2}, 3 - x)$

then $f (2)$ is obtained with

>	$eval (f, x = 2)$

and the graph of $f (x)$ is obtained as Figure 10.1.2 with

>	$plot (f, x = - 2 .. 4, discont = true)$

Figure 10.1.2 Graph of piecewise-defined $f (x)$

If the given function is entered into Maple as a Maple function, then appropriate syntax would be

>	$F ≔ x \to piecewise (x < - 1, 2 x + 1, x \leq 1, 1, x < 2, x^{2}, 3 - x)$

Because this mode of entry does not display the "echelon" form of the piecewise function, the following alternate strategy can be used.

>	$F ≔ x \to piecewise (x < - 1, 2 x + 1, x \leq 1, 1, x < 2, x^{2}, 3 - x) &colon;' F' (x) = F (x)$

in which case the value $f (2)$ is obtained with

$F (2)$

Problem 10.2

10.2 - Mathematical Solution

Let us compare the functions $f (x) = \frac{x^{2} - 1}{x - 1}, g (x) = x + 1$ , and $h (x) = \{\begin{array}{c} \frac{x^{2} - 1}{x - 1} & x \neq 1 \\ 2 & x = 1 \end{array}$ .

It is tempting to write

$f (x) = \frac{x^{2} - 1}{x - 1}$ = $\frac{(x + 1) (x - 1)}{x - 1} = x + 1$ = $g (x)$

but this is naive and incorrect. Although more cumbersome, it would be clearer to write

$f (x) = \frac{x^{2} - 1}{x - 1}$ = $\frac{(x + 1) (x - 1)}{x - 1}$ = ${\begin{array}{c} x + 1 & x \neq 1 \\ undefined & x = 1 \end{array}$

By convention, the domain of a function is the largest subset of the real numbers for which its rule is valid.

The function $f (x)$ is not defined at $x = 1$ , but the function $g (x)$ is defined there, and has the value 2. Thus, $f (x) \neq g (x)$ , that is, the functions $f (x)$ and $g (x)$ are not the same. Their domains are not the same. The domain of $g (x)$ contains one more point than the domain of $f (x)$ . In that sense, $g (x)$ is an extension of $f (x)$ ; that is, $g (x)$ agrees with $f (x)$ at all points in their common domains, but $g (x)$ is defined for at least one point where $f (x)$ isn't.

The function $h (x)$ agrees everywhere with $g (x)$ . Their domains are the same, namely, the set of all real numbers, and their values agree at every real number. Hence, $h (x) = g (x)$ .

10.2 - Maplet Solution

The three functions

$f (x) = \frac{x^{2} - 1}{x - 1}$

$g (x) = x + 1$

$h (x) = \{\begin{array}{c} \frac{x^{2} - 1}{x - 1} & x \neq 1 \\ 2 & x = 1 \end{array}$

can be compared with the Removable Singularity Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 10.2.1.

Figure 10.2.1 Thumbnail image of the Removable Singularity Tutor

The numerator and denominator of the function $f (x)$ are entered separately, and the button labeled Set of singularities provides the set $A = \{1\}$ containing the singularities of $f (x)$ .

The button labeled Piecewise form for $f (x)$ expresses $f (x)$ in the piecewise form

$f (x) = {\begin{array}{c} \frac{x^{2} - 1}{x - 1} & x \neq 1 \\ undefined & x = 1 \end{array}$

The button labeled Incorrect simplification of $f (x)$ produces the function

$g (x) = x + 1$

a function that is not equivalent to $f (x)$ . This function is an extension of $f (x)$ because it agrees with $f (x)$ at all points in the domain of $f (x)$ , but its domain contains (at least) one more point than the domain of $f (x)$ .

The button labeled Piecewise form of $g (x)$ produces the function

$h (x) = \{\begin{array}{c} \frac{x^{2} - 1}{x - 1} & x \neq 1 \\ 2 & x = 1 \end{array}$

which is equivalent to $g (x)$ . Thus, the singularity of $f (x)$ at $x = 1$ can be removed by defining a new function that gives a value for $f (1)$ . This new function can be written either in piecewise form as $h (x)$ or as $g (x)$ .

To launch the Removable Singularity Tutor, click the following link: Removable Singularity Tutor

10.2 - Interactive Solution

Type $\frac{x^{2} - 1}{x - 1}$ , the rule for $f (x)$ Context Panel: Simplify≻Simplify
Maple simplifies $f (x)$ to $x + 1$ because Maple functions and expressions don't carry domain information. Unfortunately, this means Maple is incorrect when making this simplification.
Type $\{\begin{array}{c} \frac{x^{2} - 1}{x - 1} & x \neq 1 \\ 2 & x = 1 \end{array}$ , the rule for $h (x)$ Context Panel: Simplify≻Simplify
Maple simplifies $h (x)$ to $x + 1$ . Although this is a correct simplification, Maple really doesn't "understand" why this is correct and the simplification of $f (x)$ is not.

10.2 - Programmatic Solution

It is tempting to write

$f (x) = \frac{x^{2} - 1}{x - 1}$ = $\frac{(x + 1) (x - 1)}{x - 1} = x + 1$ = $g (x)$

but this is naive and incorrect. Although more cumbersome, it would be better to write

$f (x) = \frac{x^{2} - 1}{x - 1}$ = $\frac{(x + 1) (x - 1)}{x - 1}$ = ${\begin{array}{c} x + 1 & x \neq 1 \\ undefined & x = 1 \end{array}$

In fact, as difficult as it is for Maple to produce, Figure 10.2.2 shows $f (x)$ to be linear, except for the the gap or "hole" at $x = 1$ .

>	$plot (\frac{x^{2} - x}{x - 1}, x = 0 .. 2, sample = [seq (\frac{k}{50}, k = 0 .. 100)], adaptive = false, thickness = 2)$

Figure 10.2.2 Graph of $f (x) = \frac{x^{2} - 1}{x - 1}$

By convention, the domain of a function is the largest subset of the real numbers for which its rule is valid.

The function $h (x)$ agrees everywhere with $g (x)$ . Their domains are the same, namely, the set of all real numbers, and their values agree at every real number. Hence, $h (x) = g (x)$ .

Problem 10.3

10.3 - Mathematical Solution

The graph of

$f (x) = x + [x]$

on the interval $[- 3, 3]$ appears in Figure 10.3.1.

Recall that $[x]$ is the "greatest integer" function that returns the greatest integer that is less than or equal to $x$ .

Figure 10.3.1 Graph of $f (x) = x + [x]$

10.3 - Maplet Solution

Provided that $[x]$ , the greatest-integer function, is rendered as $floor (x)$ in Maple, a graph of the function

$f (x) = x + [x]$

on the interval $[- 3, 3]$ can be obtained with the Piecewise Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 10.3.2.

Enter $f (x) = x + floor (x)$ as a function in Format 1.

Because the domain of $[x]$ is the whole real line, expressing it as a piecewise-defined function would require an infinite number of rules. Hence, the Piecewise Function Tutor does not write this function in "echelon" form.

Figure 10.3.2 Thumbnail image of the Piecewise Function Tutor

The required graph is obtained by clicking on the button labeled Graph Format 1.

To launch the Piecewise Function Tutor, click the following link: Piecewise Function Tutor

10.3 - Interactive Solution

•	Type $x + Floor (x)$

•	Context Panel: Plots≻Plot Builder $- 3 \leq x \leq 3$ Options: Find Discontinuities

Alternatively

•

Type $x$

•	Context Panel: Integer Functions≻Floor

•	Control-drag $Floor (x)$ and append to it $+ x$

•	Press the Enter key.

•	Context Panel: Plots≻Plot Builder

10.3 - Programmatic Solution

The graph of $f (x) = x + [x]$ appears in Figure 10.3.3.

>	$plot (x + floor (x), x = - 3 .. 3, discont = true, scaling = constrained)$

Figure 10.3.3 Graph of $f (x) = x + [x]$

Problem 10.4

10.4 - Mathematical Solution

Figure 10.4.1 contains a graph of

$f (x) = x - [x]$

drawn on the interval $[- 3, 3]$ .

Figure 10.4.1 Graph of $f (x) = x - [x]$

10.4 - Maplet Solution

Provided that $[x]$ , the greatest-integer function, is rendered as $floor (x)$ in Maple, a graph of the function

$f (x) = x - [x]$

on the interval $[- 3, 3]$ can be obtained with the Piecewise Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 10.4.2.

Enter $f (x) = x - floor (x)$ as a function in Format 1.

Figure 10.4.2 Thumbnail image of the Piecewise Function Tutor

The required graph is obtained by clicking on the button labeled Graph Format 1.

To launch the Piecewise Function Tutor, click the following link: Piecewise Function Tutor

10.4 - Interactive Solution

•	Type $x - Floor (x)$

•	Context Panel: Plots≻Plot Builder $- 3 \leq x \leq 3$ Options: Find Discontinuities

Alternatively

•

Type $x$

•	Context Panel: Integer Functions≻Floor

•	Type $x -$ and Control-drag $Floor (x)$

•	Press the Enter key.

•	Context Panel: Plots≻Plot Builder

10.4 - Programmatic Solution

The graph of $f (x) = x - [x]$ appears in Figure 10.4.3.

>	$plot (x - floor (x), x = - 3 .. 3, discont = true, scaling = constrained, tickmarks = [7, 2])$

Figure 10.4.3 Graph of $f (x) = x - [x]$

Problem 10.5

10.5 - Mathematical Solution

Converting the function $f (x) = |3 x - 5| + 2$ from the absolute value notation to piecewise notation requires knowing where $3 x - 5$ changes from negative to positive. Thus, one could solve the equation

$3 x - 5 = 0$

for $x = 5 / 3$ . Alternatively, one could solve the inequalities

$3 x - 5 < 0$

and

$3 x - 5$ $> 0$

for

$x < 5 / 3$

and

$x > 5 / 3$

respectively.

Thus, when $x < 5 / 3$ , the expression $3 x - 5$ is negative, so $f (x)$ becomes $- (3 x - 5) + 2 = 7 - 3 x$ .

When $x > 5 / 3$ , the expression $3 x - 5$ is positive, so $f (x)$ becomes $(3 x - 5) + 2 = 3 x - 3$ .

When $x = 5 / 3$ , both expressions reduce to 2, so it does not matter where the equality condition is included in the piecewise definition.

Thus, the piecewise form of $f (x)$ would be

$f (x) = \{\begin{array}{c} 7 - 3 x & x < \frac{5}{3} \\ 3 x - 3 & x \geq \frac{5}{3} \end{array}$

Figure 10.5.1 provides a graph of this function.

Figure 10.5.1 Graph of $f (x) = |3 x - 5| + 2$

10.5 - Maplet Solution

To represent the function

$f (x) = |3 x - 5| + 2$

as a piecewise-defined function, and to obtain its graph on the interval $[- 1, 4]$ , use the Piecewise Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 10.5.2.

As required for Format 1 functions, enter $|3 x - 5|$ as abs(3*x-5).

Then, click the button labeled Convert to piecewise to obtain the "echelon" representation of this function.

Figure 10.5.2 Thumbnail image of the Piecewise Function Tutor

Clicking the button labeled Graph Format 1 will give the required graph, provided the Plot Options button has been used to set the plot range to the required domain.

To launch the Piecewise Function Tutor, click the following link: Piecewise Function Tutor

10.5 - Interactive Solution

•	Type $\|3 x - 5\| + 2$

•	Context Panel: Apply a Command Apply the command≻convert Append the arguments≻piecewise

•	Context Panel: Plots≻Plot Builder $- 1 \leq x \leq 4$ Options≻ $0 \leq y \leq 11$

Solution from First Principles

•	Type $3 x - 5 < 0$

•	Context Panel: Solve≻Solve

•	Type $3 x - 5 > 0$

•	Context Panel: Solve≻Solve

•	Type $- (3 x - 5) + 2$

•	Context Panel: Evaluate and Display Inline

•	Type $3 x - 5 + 2$

•	Context Panel: Evaluate and Display Inline

•	Use $\{\begin{array}{c} - x & x < a \\ x & x \geq a \end{array}$ , the piecewise function template from the Expression palette to obtain $f (x)$ in piecewise form.

10.5 - Programmatic Solution

The function $f (x) = |3 x - 5| + 2$ , given in Maple by

>	$f ≔ abs (3 x - 5) + 2$

can be represented as the piecewise function

>	$convert (f, piecewise)$

The graph of $f (x)$ is found in Figure 10.5.3.

>	$plot (f, x = - 1 .. 4, y = 0 .. 11)$

Figure 10.5.3 Graph of $f (x) = |3 x - 5| + 2$

Converting from the absolute value notation to the piecewise notation requires knowing where $3 x - 5$ changes from negative to positive. Thus, one could solve the equation

>	$q ≔ 3 x - 5 = 0$

and obtain

>	$solve (q, x)$

On the other hand, one could solve the inequalities

>	$q1 ≔ 3 x - 5 < 0 &semi; q2 ≔ 3 x - 5 > 0$

obtaining

>	$solve (q1, \{x\}) &semi; solve (q2, \{x\})$

Thus, when $x < 5 / 3$ , the expression $3 x - 5$ is negative, so $f (x)$ becomes $- (3 x - 5) + 2 = 7 - 3 x$ .

When $x > 5 / 3$ , the expression $3 x - 5$ is positive, so $f (x)$ becomes $(3 x - 5) + 2 = 3 x - 3$ .

When $x = 5 / 3$ , both expressions reduce to 2, so it does not matter where the equality condition is included in the piecewise definition.

Problem 10.6

10.6 - Mathematical Solution

Converting the function $f (x) = |1 - 2 x - 3 x^{2}|$ from absolute value notation to piecewise notation requires knowing where $1 - 2 x - 3 x^{2}$ changes from negative to positive. Thus, one could solve the equation

$1 - 2 x - 3 x^{2} = 0$

by the quadratic formula, obtaining $x = - 1$ and $x = 1 / 3$ .

Alternatively, one could solve the inequalities

$1 - 2 x - 3 x^{2} < 0$

and

$1 - 2 x - 3 x^{2} > 0$

obtaining

$\{x < - 1\} \cup$ { $x > 1 / 3$ }

and

$- 1 < x < 1 / 3$

respectively.

Thus, when $x < - 1$ , or when $x > 1 / 3$ , the expression $1 - 2 x - 3 x^{2}$ is negative, so $f (x)$ becomes

$- (1 - 2 x - 3 x^{2}) = 3 x^{2} + 2 x - 1$

When $- 1 < x$ $< 1 / 3$ the expression $1 - 2 x - 3 x^{2}$ is positive, so $f (x)$ becomes $1 - 2 x - 3 x^{2}$ .

When $x = 1 / 3$ or when $x = - 1$ , both expressions reduce to 0, so it does not matter where the equality condition is included in the piecewise definition.

This function is graphed in Figure 10.6.1.

Figure 10.6.1 Graph of $f (x) = |1 - 2 x - 3 x^{2}|$

10.6 - Maplet Solution

To represent the function

$f (x) = |1 - 2 x - 3 x^{2}|$

as a piecewise-defined function, and to obtain its graph on the interval $[- 2, 2]$ , use the Piecewise Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 10.6.2.

As required for Format 1 functions, enter $f (x)$ as

abs(1-2*x-3*x^2)

Then, click the button labeled Convert to piecewise to obtain the following "echelon" representation of this function.

Figure 10.6.2 Thumbnail image of the Piecewise Function Tutor

$f (x) = \{\begin{array}{c} - 1 + 2 x + 3 x^{2} & x \leq - 1 \\ 1 - 2 x - 3 x^{2} & x < 1 / 3 \\ - 1 + 2 x + 3 x^{2} & x \geq 1 / 3 \end{array}$

Clicking the button labeled Graph Format 1 will give the required graph, provided the Plot Options button has been used to set the plot range to the required domain.

To launch the Piecewise Function Tutor, click the following link: Piecewise Function Tutor

10.6 - Interactive Solution

•	Type $1 - 2 x - 3 x^{2}$

•	Context Panel: Assign to a Name≻ $g$

•	Type $\|g\|$

•	Context Panel: Apply a Command Apply the command≻convert Append the arguments≻piecewise

•	Context Panel: Plots≻Plot Builder $- 2 \leq x \leq 2$ Options≻ $0 \leq y \leq 15$

Solution from First Principles

•	Type $g < 0$

•	Context Panel: Solve≻Solve

•	Type $g > 0$

•	Context Panel: Solve≻Solve

•	Use $\{\begin{array}{c} - x & x < a \\ x & x \geq a \end{array}$ , the piecewise function template from the Expression palette, to obtain $f (x)$ in piecewise form. Add additional "rows" with Ctrl + Shift + r

10.6 - Programmatic Solution

The function $f (x) = |1 - 2 x - 3 x^{2}|$ , given in Maple by

>	$f ≔ abs (1 - 2 x - 3 x^{2})$

can be represented as the piecewise function

>	$convert (f, piecewise)$

The graph of $f (x)$ is found in Figure 10.6.3.

>	$plot (f, x = - 2 .. 2, y = 0 .. 15)$

Figure 10.6.3 Graph of $f (x) = |1 - 2 x - 3 x^{2}|$

Converting from the absolute value notation to the piecewise notation requires knowing where $1 - 2 x - 3 x^{2}$ changes from negative to positive. Thus, one could solve the equation

>	$q ≔ 1 - 2 x - 3 x^{2} = 0$

and obtain

>	$solve (q, x)$

On the other hand, one could solve the inequalities

>	$q1 ≔ 1 - 2 x - 3 x^{2} < 0 &semi; q2 ≔ 1 - 2 x - 3 x^{2} > 0$

obtaining

>	$solve (q1, \{x\}) &semi; solve (q2, \{x\})$

Thus, when $x < - 1$ , or when $x > 1 / 3$ , the expression $1 - 2 x - 3 x^{2}$ is negative, so $f (x)$ becomes

$- (1 - 2 x - 3 x^{2}) = 3 x^{2} + 2 x - 1$

When $- 1 < x$ $< 1 / 3$ the expression $1 - 2 x - 3 x^{2}$ is positive, so $f (x)$ becomes $1 - 2 x - 3 x^{2}$ .

When $x = 1 / 3$ or when $x = - 1$ , both expressions reduce to 0, so it does not matter where the equality condition is included in the piecewise definition.

Exercises - Chapter 10

10.1. For $x$ in the interval $[- 1, 1]$ , graph the piecewise-defined function $f (x) = {\begin{array}{c} x^{2} - 1 & x < 0 \\ 1 - x^{2} & otherwise \end{array}$ .

10.2. For $x$ in the interval $[- π, π]$ , graph the piecewise-defined function $f (x) = {\begin{array}{c} - \sin (x) & x < 0 \\ \sin (x) & otherwise \end{array}$ .

10.3. For $x$ in the interval $[- 1, 1]$ , graph the piecewise-defined function $f (x) = {\begin{array}{c} 2 + x & x < - 1 \\ 1 & x < 1 \\ 2 - x & otherwise \end{array}$ .

10.4. For $x$ in the interval $[- 3, 3]$ , graph the piecewise-defined function $f (x) = {\begin{array}{c} (1 + x) (x + 3) & x < - 1 \\ 0 & x < 1 \\ (- 1 + x) (3 - x) & otherwise \end{array}$ .

10.5. For $x$ in the interval $[- 1, 1]$ , graph the piecewise-defined function $f (x) = {\begin{array}{c} {&ExponentialE;}^{x} & x < 0 \\ {&ExponentialE;}^{- x} & 0 \leq x \end{array}$ .

10.6. For $x$ in the interval $[- 2, 2]$ , graph the piecewise-defined function $f (x) = {\begin{array}{c} - 2 - x & x < - 1 \\ - 1 & x < 0 \\ 1 & x < 1 \\ 2 - x & otherwise \end{array}$ .

10.7. For $x$ in the interval $[- 3, 3]$ , graph the piecewise-defined function $f (x) = {\begin{array}{c} - 3 - x & x < - 1 \\ - 1 + x & x < 0 \\ 1 + x & x < 1 \\ 3 - x & otherwise \end{array}$ .

In Exercises 10.8 - 10.10, compare the functions $f (x), g (x)$ , and $h (x)$ .

10.8. $f (x) = \frac{x^{2} - x - 2}{x - 2}, g (x) = x + 1$ , $h (x) = {\begin{array}{c} \frac{x^{2} - x - 2}{x - 2} & x \neq 2 \\ 3 & x = 2 \end{array}$

10.9. $f (x) = \frac{3 x^{2} + 8 x - 3}{x + 3}, g (x) = x + 3$ , $h (x) = {\begin{array}{c} \frac{3 x^{2} + 8 x - 3}{x + 3} & x \neq - 3 \\ - 10 & x = - 3 \end{array}$

10.10. $f (x) = \frac{x - 9}{\sqrt{x} - 3}, g (x) = \sqrt{x} + 3$ , $h (x) = {\begin{array}{c} \frac{x - 9}{\sqrt{x} - 3} & x \neq 9 \\ 6 & x = 9 \end{array}$

10.11. Graph $f (x) = 3 x + 2 [x]$ on the interval $[- 3, 3]$ .

10.12. Graph $f (x) = 3 x - 2 [x]$ on the interval $[- 3, 3]$ .

10.13. Represent $f (x) = 3 |1 - 2 x| + 4$ as a piecewise-defined function, and obtain its graph on the interval $[- 2, 3]$ .

10.14. Represent $f (x) = 1 - 2 |3 x + 4|$ as a piecewise-defined function, and obtain its graph on the interval $[- 3, 1]$ .

10.15. Represent $f (x) = |x^{2} - 3 x + 2|$ as a piecewise-defined function, and obtain its graph on the interval $[0, 3]$ .

10.16. Represent $f (x) = |12 + x - x^{2}|$ as a piecewise-defined function, and obtain its graph on the interval $[- 4, 5]$ .

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