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Module 2 Lecture Notes

Contents2.1 Even and Odd Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2.2 Increase and Decrease of Functions . . . . . . . . . . . . . . . . . . . . . 5

2.3 Concavity (REQUIRED) . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.4 Using a Graphing Calculator to Determine Local Maximums andLocal Minimums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.5 Piecewise-defined Functions . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.5.2 Evalutating Piecewise-Defined Functions . . . . . . . . . . . . . . . . . . . 13

2.5.3 Graphing Piecewise-Defined Functions . . . . . . . . . . . . . . . . . . . . 15

2.5.4 Finding the Formula (or Definition) of Piecewise-Defined Functions . . . . 21

2.5.5 Applications to Piecewise-Defined Functions . . . . . . . . . . . . . . . . . 23

2.1 Even and Odd Functions

The word symmetry means “the correspondence in size, form, and arrangement of parts on oppositesides of a plane, line, or point.” (Random House Dictionary). The primary two types of symmetrywe will look at classify functions as even symmetry and odd symmetry.

Math 111 Module 2 Lecture Notes

Definition 1: A function f is even if for every x in the domain of f it holds that f(−x) =

f(x). Visually, an even function is symmetric about the y-axis.

A function f is odd if for every x in the domain of f it holds that f(−x) = −f(x). Visually,

an odd function is symmetric about the origin.

Example 1: Two classic examples of even and odd functions are f(x) = x2 and g(x) = x3,respectively, as shown in Figures 2.1 and 2.2 below.

Figure 2.1: Graph of y = f(x)

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

x

y

Figure 2.2: Graph of y = g(x)

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

x

y

Algebraically verify that f is an even function and that g is an odd function.

f(−x) = (−x)2

= x2

As f(−x) = f(x) for all values of x, the functionf is even.

g(−x) = (−x)3

= −x3

− g(x)

As g(−x) = −g(x) for all values of x, the func-tion g is odd.

Example 2: Algebraically determine if the following functions are even or odd. Use the graphsin Figures 2.3-2.6 to graphically verify your answers.

(a) h(x) = x3 − x

h(−x) = (−x)3 − (−x)

= −x3 + x

−h(x) = −(x3 − x)

= −x3 + x

As h(−x) = −h(x) for all values of x, the function g is odd.

Instructor: A.E.Cary of 23


(b) g(t) = 12t4 − 1

g(−t) =1

2(−t)4 − 1

=1

2t4 − 1

= g(t)

As g(−t) = g(t) for all values of t, the function g is even.

(c) f(t) = t3 + 1

g(−t) = (−t)3 + 1

= −t3 + 1

−g(t) = −(t3 + 1)

= −t3 − 1

As g(−t) 6= g(t) and g(−t) 6= −g(t), the function g is neither even nor odd.

(d) f(x) = |x| − 4

f(−x) = |(−x)| − 4

= |x| − 4

= f(x)

As f(−x) = f(x), the function f is even.

Graphical Verification: The function graphed in Figures 2.3 is odd as it is symmetric about theorigin. The functions graphed in Figures 2.4 and 2.6 are even as they are symmetric about thevertical axis. In Figure 2.5, we see that the function is neither symmetric about the vertical axisnor the original and is therefore neither even nor odd.

Figure 2.3: y = h(x)

−4 −2 2 4

−4

−2

2

4

x

y

Figure 2.4: y = g(t)

−4 −2 2 4

−4

−2

2

4

t

y

Figure 2.5: y = f(t)

−4 −2 2 4

−4

−2

2

4

t

y

Figure 2.6: y = f(x)

−4 −2 2 4

−4

−2

2

4

x

y



Example 3: Algebraically determine if the function f defined by f(x) = − 2x3 − x

3x4 + 5x2.

f(−x) = − 2(−x)3 − (−x)

3(−x)4 + 5(−x)2

= −−2x3 + x

3x4 + 5x2

=−1

1· −2x3 + x

3x4 + 5x2

=2x3 − x

3x4 + 5x2

−f(x) = −(− 2x3 − x

3x4 + 5x2

)=

2x3 − x

3x4 + 5x2

As f(−x) = −f(x) for all values of x, the function f is odd.

Example 4: Algebraically determine if the function g defined by g(x) =x2

x4 + 5.

g(−x) =(−x)2

(−x)4 + 5

=x2

x4 + 5

= g(x)

As g(−x) = g(x) for all values of x, the function g is even.

Example 5: Algebraically determine if the function h defined by h(x) = 5x3 + 3x2.

h(−x) = 5(−x)3 + 3(−x)2

= −5x3 + 3x2

−h(x) = −(5x3 + 3x2

)= −5x3 − 3x2

As h(−x) 6= h(x) and h(−x) 6= −h(x), the function h is neither even nor odd.



2.2 Increase and Decrease of Functions

Definition 2: A function f is increasing on an open interval I if for every x1 and x2 in I

with x1 < x2 we have f(x2) > f(x1).

A function f is decreasing on an open interval I if for every x1 and x2 in I with x1 < x2 we

have f(x2) < f(x1).

A function f is constant on an open interval I if for every x1 and x2 in I with x1 < x2 we

have f(x2) = f(x1).

Example 6: Determine the intervals where the function f is increasing, decreasing, or constantin Figure 2.7.


−4 −2 2 4

−4

−2

2

4

x

y

(a) Increasing:

(−5,−3), (1, 2)

(b) Decreasing:

(−3, 1), (4, 5)

(c) Constant:

(2, 4)

(d) Domain of f :

[−5, 5)

(e) Range of f :

(−3, 3]



Definition 3: A function has a local maximum at c if there exists an open interval I con-

taining c so that for all x not equal to c in I, it holds that f(x) < f(c). The output f(c) is

referred to as the local maximum of f .

A function has a local minimum at c if there exists an open interval I containing c so that

for all x not equal to c in I, it holds that f(x) > f(c). The output f(c) is referred to as the

local minimum of f .

Example 7: Use Figure 2.7 to answer the following:

(a) Identify all local maximum values of f and state where they occur.

The function f has a local maximum of 3. This occurs at -3.

(b) Identify all local minimum values of f and state where they occur.

The function f has a local minimum of -1. This occurs at 1.

Definition 4: Let f be a function defined on an interval I.

A function has an absolute maximum at u if it holds that f(x) ≤ f(u) for all x in the

interval I. The output f(u) is referred to as the absolute maximum of f .

A function has an absolute minimum at u if it holds that f(x) ≥ f(u) for all x in the interval

I. The output f(u) is referred to as the absolute minimum of f .



−4 −2 2 4

−4

−2

2

4

x

y (a) Identify all absolute maximum values of f and statewhere they occur.

The function f has an absolute maximum of 4. Thisoccurs at -1.

(b) Identify all absolute minimum values of f and statewhere they occur.

The function f has an absolute minimum of -1. Thisoccurs at -4.




Figure 2.9: y = g(x)

−4 −2 2 4

−4

−2

2

4

x

y

(a) Identify all local maximum values of g and state wherethey occur.

The function g has a local maximum of 3. Thisoccurs at -1.

(b) Identify all local minimum values of g and state wherethey occur.

The function g has a local minimum of -5. Thisoccurs at 1.

(c) Identify all absolute maximum values of g and statewhere they occur.

The function g has an absolute maximum of 5. Thisoccurs at 3.

(d) Identify all absolute minimum values of g and statewhere they occur.

The function g has an absolute minimum of -5. Thisoccurs at 1.



2.3 Concavity (REQUIRED)

So far, we have looked at where a function is increasing and decreasing and where it attainsmaximum and minimum values. We will now study the concept of concavity. This concept involveslooking at the rate at which a function increases or decreases.

Definition 5: The graph of a function f whose rate of change increases (becomes less

negative or more positive as you move left to right) over an interval is concave up on that

interval. Visually, the graph “bends upward.”

The graph of a function f whose rate of change decreases (becomes less positive or more

negative as you move left to right) over an interval is concave down on that interval.

Visually, the graph “bends downward.”

Figure 2.10: Concave UP © Figure 2.11: Concave DOWN §

Example 10: The function defined by f(x) = x2 is concave up on its entire domain. Notice thatit is decreasing on the interval (−∞, 0) and increasing on the interval (0,∞). The functiondefined by f(x) = −x2 is concave down on its entire domain. Notice that it is increasing on theinterval (−∞, 0) and decreasing on the interval (0,∞).

Figure 2.12: Graph of y = x2

−4 −2 2 4

−4

−2

2

4

Figure 2.13: Graph of y = −x2

−4 −2 2 4

−4

−2

2

4



Example 11: The graph of y = h(x) is shown in Figure 2.14. Use this to answer the following.

Figure 2.14: Graph of y = h(x)

−4 −2 2 4 6 8 10

−4

−2

2

4

6

8

10

(a) State the interval(s) where h is positive.

(−2,∞)

(b) State the interval(s) where h is negative.

(−∞,−2)

(c) State the interval(s) where h is increasing.

(−∞, 2), (6,∞)

(d) State the interval(s) where h is decreasing.

(2, 6)

(e) State the interval(s) where h is concave up.

(4,∞)

(f) State the interval(s) where h is concave down.

(−∞, 4)

(g) State any absolute maximum or local minimum values for h and where they occur.

The function h has a local maximum of 8 that occurs at 2. The function h does not havean absolute maximum.

(h) State any local maximum or local minimum values for h and where they occur.

The function h has a local minimum of 4 that occurs at 6. The function h does not havean absolute minimum.



Example 12: When you pour a cup of coffee, the temperature is initially very hot. It coolsquickly at first, but the closer the temperature is to room temperature, the slower it cools. Sketcha graph of a function modeling this scenario. Where is the function be increasing and decreasing?Where is the function be concave up and concave down?

Let t be the number of minutes since the coffee was poured and let T be the temperature of thecoffee in degrees Fahrenheit. The function defined by T = f(t) is be decreasing over its entiredomain. It is be concave up since the rate of change increases (that is, it becomes less negative).

Figure 2.15: Coffee temperature as a function of time

t, minutes after the coffee was poured

T,

Tem

pera

ture

indegre

es

Fahre

nheit

Example 13: When a commercial trend begins (such as iPods, cellphones, Nintendo, Atari), thenumber of people who own such a device grows very slowly at first. As the item becomes morepopular, the number of people who have that particular item grows at a faster and faster rate.There is a point though where most of the people who will ever purchase the item have alreadypurchased it. This causes the number of people purchasing the device to slow down again. Sketcha graph of a function modeling this scenario. Where is the function increasing? Where is thefunction decreasing? Where is the function concave up? Where is it concave down?

A sketch is shown in Figure 2.16. The function is increasing over its entire domain. It is concaveup at first, but then becomes concave down. The approximate point where this change in concavityoccurs (known as an inflection point) is shown in Figure 2.16.

Figure 2.16: Trend Behavior

number of potential purchasers

point where concavity changes

n, units of time

P,number

ofpurchasers



2.4 Using a Graphing Calculator to Determine Local Max-

imums and Local Minimums

Example 14: Graph the function defined by k(x) = 2x4−6x3−6x2 + 22x+ 2 on your calculator.

(a) Determine an appropriate window that shows the important features (such as the x-intercept(s),y-intercept, and any local maximums or minimums).

One appropriate window choice is [−8, 8] for the x-values and [−20, 20] for the y-values.

(b) Use the MAXIMUM and MINIMUM features to find any local maximums and minimumsand where they occur. Round each value accurate to three decimal places. Use these todetermine the intervals of increase and decrease for this function.

There is a local minimum of approximately -18.612 that occurs at approximately -1.147.There is also a local minimum of approximately 3.651 that occurs at approximately 2.397.There is a local maximum of 14 that occurs at 1.

The function is increasing over the approximate intervals (−1.147, 1) and (2.397,∞). Thefunction is decreasing over the approximate intervals (−∞,−1.147) and (1, 2.397).

(c) (Review) Use the ZERO feature and the VALUE feature to determine the x-intercepts andy-intercepts.

The zeros of the function are approximately -1.817 and -0.089. The x-intercepts are ap-proximately (−1.817, 0) and (−0.089). The y-intercept is (0, 2).



2.5 Piecewise-defined Functions

2.5.1 Introduction

In Table 2.1, the 2011 federal income tax rates1 for 2011 are shown.

Table 2.1: Federal Income Tax Percentage Rates for 2014 (Single Filing Status)

Income Amount (m) Percentage of Income Taxed (P (m), in %)

0 ≤ m < 9075 10

9075 ≤ m < 36900 15

36900 ≤ m < 89350 25

89350 ≤ m < 186350 28

186350 ≤ m < 405100 33

405100 ≤ m < 406750 35

m ≥ 406750 39.6

Notice that for each interval, the percentage of income taxed as a function of income is constant.If we graph each piece over its respective interval, we obtain the following:

Figure 2.17: Graph of y = P (m)

50 100 150 200 250 300 350 400 450

5

10

15

20

25

30

35

40

m, income amount in $1,000s

y,percentagetaxed

We see that the function above has different formulas defined on different pieces of its domain.Such a function is known as a piecewise-defined function.

A function that is defined by different formulas on different parts of its domain is apiecewise-defined function.

1http://www.irs.gov/newsroom/article/0,,id=233465,00.html


http://www.irs.gov/newsroom/article/0,,id=233465,00.html


2.5.2 Evalutating Piecewise-Defined Functions

Example 15: Use the piecewise-defined function f defined below to answer the following.

f(x) =

3

x− 4if x ≤ −2

7x− 8 if − 2 < x ≤ 5

−11 if x > 5

(a) f(0)

Since 0 is included in −2 < x ≤ 5, we’ll use the second piece to find f(0):

f(0) = 7(0) + 8

= 8

(b) f(−6)

Since −6 is included in x ≤ −2, we’ll use the first piece to find f(−6):

f(−6) =3

−6− 4

= − 3

10

(c) f(8)

Since 8 is included in x > 5, we’ll use the third piece to find f(8):

f(8) = −11

(d) f(−2)

Since −2 is included in x ≤ −2, we’ll use the first piece to find f(−2):

f(−2) =3

−2− 4

= −1

2

(e) f(5)

Since 5 is included in −2 < x ≤ 5, we’ll use the second piece to find f(5):

f(5) = 7(5)− 8

= 27



Example 16: Use the piecewise-defined function g graphed in Figure 2.37 to answer the following.

(a) State the domain and range of g.

Domain: (−∞, 3)⋃

(3, 7]

Range: (−∞, 3)

(b) Evaluate g(6).

g(6) = 2

(c) Evaluate g(−2).

g(−2) = −4

(d) Solve g(x) = −3.

Solution Set: {1}

(e) Solve g(x) = −5.

Solution Set: {−4, 2}

Figure 2.18

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

x

y

Example 17: Use the function f graphed in Figure 2.19 below to answer the following.

(a) State the domain and range of f .

Domain: (−∞,∞)

Range: (−1,∞)

(b) Find f(0).

f(0) = 2

(c) Find f(−4).

f(−4) = 15

(d) Solve f(x) = 0 using the graph of y = f(x).

Solution Set: {−1}

Figure 2.19: Graph of y = f(x)

−3 −2 −1 1 2 3

−3

−2

−1

1

2

3

x

y



2.5.3 Graphing Piecewise-Defined Functions

Example 18: As a prelude to graphing piecewise functions, let’s graph just a few of the “pieces.”

(a) Graph the linear function defined by f(x) = −x− 2 for values of x where −4 < x ≤ −1.

To start, we need to recognize that this function is linear and written in slope-interceptform. In general, we would use the slope and y-intercept to graph this function. But sincewe’re not graphing it for x = 0, we’ll use more rudimentary means to graph it: We’ll find onepoint and use the slope to find at least one other point. Here, we need to know where theends of this graph occur. At −4, the value of f would be −(−4)− 2 or 2. Since we are onlygraphing this for −4 < x ≤ −1, we will graph an open circle at (−4, 2). At −1, the value off is −(−1)− 2 or −1. Since −1 is included in the piece we are graphing, the point (−1,−1)will be a closed circle. Lastly, we can connect these with a straight line and visually verifythat the slope is correct.

Figure 2.20

−4 −2 2 4

−4

−2

2

4

x

y

(b) Graph the constant function defined by f(x) = 3 for values of x where 2 < x < 4.

This function is a constant function. If we graphed it for all values of x, it would just bea horizontal line with y = 3. With this in mind, we’ll just graph a horizontal line betweenx = 2 and x = 4. Because we see strict inequalities, we’ll plot open circles at (2, 3) and (4, 3).

Figure 2.21

−4 −2 2 4

−4

−2

2

4

x

y



(c) Graph the linear function defined by f(x) = x2 for values of x where x ≥ −1.

This function is a parabola. We know that if it were graphed for values of x, it would passthrough (−3, 9), (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), and (3, 9). Here, we’ll plot that samegraph but only for values of x where x ≥ −1.

Figure 2.22

−4 −2 2 4

−4

−2

2

4

x

y

Example 19: Graph y = h(x) for the piecewise-defined function h given below.

h(x) =

{−x2 + 4 if − 3 < x < 3

−5 if x ≥ 4

Piece 1: This piece is defined byy = −x2 + 4 over −3 < x < 3. We knowthat this is a downward facing parabolawith vertex (0, 4). To find other points,we plug in a few other values for x. We’llplot open circles at both (−3,−5) and (3, 5)since this piece is defined for −3 < x < 3.

x −x2 + 4-3 -5-2 0-1 30 41 32 03 -5

Figure 2.23

−4 −2 2 4

−4

−2

2

4

Piece 2: This piece is defined by y = −5 over x ≥ 4.This is a constant function, so we’ll graph a horizontal linestarting at (4,−5) with a closed circle and continuing rightwith an arrow on the end.The final graph shown is the graph of the entire function.To determine the graph of the entire function, we refer tothe graph of each piece but only graph each piece over theinterval for which it is defined.

Figure 2.24

−4 −2 2 4

−4

−2

2

4



Example 20: Graph y = g(x) for the piecewise-defined function g given below.

g(x) =

−3x− 2 if x < −1

4 if − 1 ≤ x < 232x− 4 if 2 < x ≤ 4

Piece 1: This piece is defined by y = −3x − 2 over x <−1, which is a linear function with a slope of -3. If itwere graphed for all real numbers, the y-intercept would be(0,−2). Instead of using the y-intercept, we’ll find anotherpoint and use the slope to graph the rest of this piece. Thevalue of −3x−2 at x = −1 is −3(−1)−2, which is 1. Thuswe’ll graph (−1, 1) and use the slope of −3 to then graph(−2, 4). Since this piece is defined for all x where x > −1,we’ll put an arrow on the end.

Figure 2.25

−4 −2 2 4

−4

−2

2

4

Piece 2: This piece is defined by y = 4 over −1 ≤ x < 2,which is constant function. So we’ll graph a flat line from(−1, 4) to (2, 4). Since we have 1 ≤ x, we’ll make (−1, 4)a closed circle and since x < 2 we’ll make (2, 4) an opencircle.

Figure 2.26

−4 −2 2 4

−4

−2

2

4

Piece 3: This piece is defined by y = 32x−4 over 2 < x ≤ 4,

which is a linear function with a slope of 32. We’ll just find

the values at the endpoints and visually verify that theslope is correct. Since 3

2(2) − 4 = −1, we’ll plot an open

circle at (2,−1). Since 32(4) − 4 = 2, we’ll plot a closed

circle at (4, 2).The final graph shown is the graph of the entire function.To determine the graph of the entire function, we refer tothe graph of each piece but only graph each piece over theinterval for which it is defined.

Figure 2.27

−4 −2 2 4

−4

−2

2

4



Example 21: Graph y = f(x) for the piecewise-defined function f defined below.

f(x) =

x2 − 1 if x < 0

2 if x = 0√x if x > 0

To determine the graph of the entire function, we will refer to the graph of each piece but onlygraph each piece over the interval for which it is defined.

For the first piece, we need to graph y = x2 − 1but only for values of x where x < 0. We know thatthis function is a parabola with vertex (0,−1). Sincethe function is not defined at zero by this function,we will graph (0,−1) as an open circle. Two otherpoints can be found on this piece ((−2, 3) and (−1, 0)as f(−2) = 3 and f(−1) = 0, respectively). We canthen connect the parabola:

Figure 2.28: Step 1

−3 −2 −1 1 2 3

−3

−2

−1

1

2

3

x

y

For the second piece, we note that this is a constantfunction (which appears as a horizontal line). Sincethe function is only defined by a constant for x = 0,we plot the point (0, 2):

Figure 2.29: Step 2

−3 −2 −1 1 2 3

−3

−2

−1

1

2

3

x

y

For the third piece, we note that this is the squareroot function. Since the function is only defined forx > 0, we plot the point (0, 0) using an open circle.Two other points that can be found for this piece are(1, 1) and (4, 2). We connect this piece in a smoothcurve:

Figure 2.30: Step 3

−3 −2 −1 1 2 3

−3

−2

−1

1

2

3

x

y



Example 22: Graph y = g(t) for the piecewise-defined function g defined below.

g(t) =

t− 2 if − 4 ≤ t < −1

−3 if − 1 ≤ t < 2

5 if t = 2

−2t + 1 if 3 < t ≤ 4

To determine the graph of the entire function, we will refer to the graph of each piece but onlygraph each piece over the interval for which it is defined.

For the first piece, we need to graph y = t − 2 butonly for values of t where −4 ≤ t < −1. We know thatthis function is a linear function with slope 1. The y-intercept would be (0,−2), but we are not graphingthe entire function. The quickest way to graph thispiece is to plot the endpoints (and visually verify thatour slope is correct). The endpoints of this piece are(−4,−6) and (−1,−3). We plot (−4,−6) as a filledcircle and (−1,−3) as an open circle. The last step isto connect these with a straight line:

Figure 2.31: Step 1

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

t

y

For the second and third pieces, we note that theseare constant functions (which appears as a horizontalline). Since the function is defined by -3 for −1 ≤ t <2, we graph a horizontal line over this interval. The leftendpoint is closed; the right endpoint is open. Sincethe function is only defined by the constant function5 for t = 2, we plot the point (2, 5):

Figure 2.32: Step 2

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

t

y

For the fourth piece, we will graph y = −2t + 1 for3 < t ≤ 4. We note that this is a linear function witha slope of −2 and would have a y-intercept of (0, 1)(if the domain were not restricted). Again, the fastestway to plot this piece is to determine the endpointsand to connect them with a straight line. The end-points are (3,−5) and (4,−7). We plot (3,−5) as anopen circle and (4,−7) as a closed circle. Then weconnect them with a straight line:

Figure 2.33: Step 3

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

t

y



Example 23: Graph y = h(x) for the piecewise-defined function h given below.

h(x) =

{(x− 1)2 − 4 if − 2 < x < 2

2 if x ≥ 4

For the first piece, we need to graph y = (x− 1)2− 4but only for values of x where −2 < x < 2. Weknow that this function is a quadratic function and itsgraph is an upward-facing parabola with vertex (1, 4).The fastest way to plot this piece is to make a table ofvalues (including the vertex) and then to connect thesein a smooth curve. Note that since the inequality state−2 < x < 2, we will plot open circles at −2 and at 2.

x y = (x− 1)2 − 4 point open/closed

-2 5 (−2, 5) open

-1 0 (−1, 0) closed

0 -3 (0,−3) closed

1 -4 (1,−4) closed

2 -3 (2,−3) open

Figure 2.34: Step 1

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

t

y

For the second piece we will graph y = 2 for x ≥ 4.We note that this is constant functions (which appearsas a horizontal line). We plot a closed circle at (4, 2)and make sure to put an arrow on the end:

Figure 2.35: Step 2

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

t

y



2.5.4 Finding the Formula (or Definition) of Piecewise-Defined Func-tions

Example 24: Find the formula (or “definition”) for the piecewise-defined function f graphed inFigure 2.36 below.

Figure 2.36

−4 −3 −2 −1 1 2 3 4

−4

−3

−2

−1

1

2

3

4

x

y

To start, we’ll identify how each individual piece is defined and the values of x for which it isdefined.Piece 1: This piece is a linear function. We can see from the graph that its slope is 1 and that they-intercept is (0, 3). Thus it’s defined by y = x + 3. Since there is a closed circle at (−4,−1) andat (1, 4), it’s defined for −4 ≤ x ≤ 1.

Piece 2: This piece is a constant function defined by y = −4. Since there is an open circle at(2,−4) and an arrow on the end, it’s defined for x > 2.

Formula:

f(x) =

{x + 3 if − 4 ≤ x ≤ 1

−4 if x > 2



Example 25: Write the formula for the piecewise-defined function g in Figure 2.37.

To determine the formula of this function, we first need to determine the formula for each piece.

Figure 2.37

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

x

y

Piece 1: The first piece of the domain we willlook at is values of x where x ≤ −2.The functionis linear here so this piece will be written in theform y = mx+b. The slope, m, can be read fromthe graph and is 1

2. The value of b is not obvious

since the y-intercept is not shown. To find thisvalue of b, we can use another point on the linesuch as (−2,−3):

y = mx + b

−3 =1

2(−2) + b

−3 = −1 + b

−2 = b

The formula for this piece is y = 12x− 2.

Piece 2: The second piece of the domain we will look at is values of x where −2 < x ≤ 3. Thispiece is constant, and as such is defined by y = 5.

Piece 3: The third piece of the domain we will look at is values of x where 3 < x < 7.The functionis linear here, so we will need to write this piece in the form y = mx + b. The slope, m, can beread from the graph and is −2. The value of b is not obvious since the y-intercept is not shown.To find this value of b, we can use another point on the line such as (4,−1):

y = mx + b

−1 = −2(4) + b

−1 = −8 + b

7 = b

The formula for this piece is y = −2x + 7.

The Whole Formula: We must find the formula of each piece, but the function is a single functionand must be defined as such:

g(x) =

12x− 2 if x ≤ −2

5 if − 2 < x < 2

−2x + 7 if 3 < x < 7



2.5.5 Applications to Piecewise-Defined Functions

Example 26: When calculating your electricity bill, PGE uses the follows rates: It costs 5.124cents per kWh for the first 250 kWh used in a month. After the first 250 kWh, it costs 6.899 centsfor each additional kWh used. Let C(x) represent the amount due for a PGE residential electricitybill (in dollars) when x kWh of energy are used in a given month.

(a) Write the formula piecewise-defined function for C.

First consider if 250 kWh or less are used. The rate is 5.124 cents per kWh, so we modelthis piece by y = 0.05124x, where y is the amount due in dollars. Next consider whathappens if you use more than 250 kWh. The first 250 kWh will be charged at a rate of 5.124cents per kWh. The total for the first 250 kWh will then be 0.05124$/kWh×250, or $12.81.Any kWh of energy used over 250 will be charged at a rate of 6.899 kWh. The expressionx − 250 represents the number of kWh use over 250. The formula for the second piece isthen y = 0.06899(x− 250) + 12.81.

The formula for C(x) is then:

C(x) =

{0.05124x, if x ≤ 250

0.06889(x− 250) + 12.81, if x > 250

(b) Use that formula to determine the amount due (before taxes and other fees) when you use325 kWh of electricity in a month.

C(325) = 0.06899(325− 250) + 12.81

= 0.06899(75) + 12.81

≈ 17.98

The amount due if 325 kWh are used is $17.98.


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