3.2 math homework answers

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3.2: Relations and Functions

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• Page ID 56726

Learning Objectives

By the end of this section, you will be able to:

• Find the domain and range of a relation
• Determine if a relation is a function
• Find the value of a function

Before you get started, take this readiness quiz.

• Evaluate \(3x−5\) when \(x=−2\). If you missed this problem, review [link] .
• Evaluate \(2x^2−x−3\) when \(x=a\). If you missed this problem, review [link] .
• Simplify: \(7x−1−4x+5\). If you missed this problem, review [link] .

Find the Domain and Range of a Relation

As we go about our daily lives, we have many data items or quantities that are paired to our names. Our social security number, student ID number, email address, phone number and our birthday are matched to our name. There is a relationship between our name and each of those items.

When your professor gets her class roster, the names of all the students in the class are listed in one column and then the student ID number is likely to be in the next column. If we think of the correspondence as a set of ordered pairs, where the first element is a student name and the second element is that student’s ID number, we call this a relation .

\[(\text{Student name}, \text{ Student ID #})\nonumber \]

The set of all the names of the students in the class is called the domain of the relation and the set of all student ID numbers paired with these students is the range of the relation.

There are many similar situations where one variable is paired or matched with another. The set of ordered pairs that records this matching is a relation.

Definition: Relation

A relation is any set of ordered pairs, \((x,y)\). All the x -values in the ordered pairs together make up the domain . All the y -values in the ordered pairs together make up the range .

Example \(\PageIndex{1}\)

For the relation \({(1,1),(2,4),(3,9),(4,16),(5,25)}\):

• Find the domain of the relation.
• Find the range of the relation.

\[\begin{array} {ll} {} &{ {\{(1,1), (2,4), (3,9), (4,16), (5,25) }\} } \\ {ⓐ\text{ The domain is the set of all x-values of the relation.}} &{ {\{1,2,3,4,5}\} } \\ {ⓑ\text{ The range is the set of all y-values of the relation.}} &{ {\{1,4,9,16,25}\} } \\ \nonumber \end{array}\]

Example \(\PageIndex{2}\)

For the relation \({\{(1,1),(2,8),(3,27),(4,64),(5,125)}\}\):

\({\{1,2,3,4,5}\}\)

\({\{1,8,27,64,125}\}\)

Example \(\PageIndex{3}\)

For the relation \({\{(1,3),(2,6),(3,9),(4,12),(5,15)}\}\):

\({\{3,6,9,12,15}\}\)

A mapping is sometimes used to show a relation. The arrows show the pairing of the elements of the domain with the elements of the range.

Example \(\PageIndex{4}\)

Use the mapping of the relation shown to

• list the ordered pairs of the relation,
• find the domain of the relation, and
• find the range of the relation.

ⓐ The arrow shows the matching of the person to their birthday. We create ordered pairs with the person’s name as the x -value and their birthday as the y -value.

{(Alison, April 25), (Penelope, May 23), (June, August 2), (Gregory, September 15), (Geoffrey, January 12), (Lauren, May 10), (Stephen, July 24), (Alice, February 3), (Liz, August 2), (Danny, July 24)}

ⓑ The domain is the set of all x -values of the relation.

{Alison, Penelope, June, Gregory, Geoffrey, Lauren, Stephen, Alice, Liz, Danny}

ⓒ The range is the set of all y -values of the relation.

{January 12, February 3, April 25, May 10, May 23, July 24, August 2, September 15}

Example \(\PageIndex{5}\)

• list the ordered pairs of the relation
• find the domain of the relation

ⓐ (Khanh Nguyen, kn68413), (Abigail Brown, ab56781), (Sumantha Mishal, sm32479), (Jose Hern and ez, jh47983)

ⓑ {Khanh Nguyen, Abigail Brown, Sumantha Mishal, Jose Hern and ez}

ⓒ {kn68413, ab56781, sm32479, jh47983}

Example \(\PageIndex{6}\)

ⓐ (Maria, November 6), (Arm and o, January 18), (Cynthia, December 8), (Kelly, March 15), (Rachel, November 6)

ⓑ {Maria, Arm and o, Cynthia, Kelly, Rachel}

ⓒ{November 6, January 18, December 8, March 15}

A graph is yet another way that a relation can be represented. The set of ordered pairs of all the points plotted is the relation. The set of all x -coordinates is the domain of the relation and the set of all y -coordinates is the range. Generally we write the numbers in ascending order for both the domain and range.

Example \(\PageIndex{7}\)

Use the graph of the relation to

ⓐ The ordered pairs of the relation are: \[{\{(1,5),(−3,−1),(4,−2),(0,3),(2,−2),(−3,4)}\}.\nonumber\]

ⓑ The domain is the set of all x -values of the relation: \(\quad {\{−3,0,1,2,4}\}\).

Notice that while \(−3\) repeats, it is only listed once.

ⓒ The range is the set of all y -values of the relation: \(\quad {\{−2,−1,3,4,5}\}\).

Notice that while \(−2\) repeats, it is only listed once.

Example \(\PageIndex{8}\)

a.    \((−3,3),(−2,2),(−1,0),\), \((0,−1),(2,−2),(4,−4)\)

b.    \({\{−3,−2,−1,0,2,4}\}\)

c.    \({\{3,2,0,−1,−2,−4}\}\)

Example \(\PageIndex{9}\)

ⓐ \((−3,0),(−3,5),(−3,−6),\) \((−1,−2),(1,2),(4,−4)\) ⓑ \({\{−3,−1,1,4}\}\) ⓒ \({\{−6,0,5,−2,2,−4}\}\)

Determine if a Relation is a Function

A special type of relation, called a function , occurs extensively in mathematics. A function is a relation that assigns to each element in its domain exactly one element in the range. For each ordered pair in the relation, each x -value is matched with only one y -value.

Definition: Function

A function is a relation that assigns to each element in its domain exactly one element in the range.

The birthday example from Example helps us understand this definition. Every person has a birthday but no one has two birthdays. It is okay for two people to share a birthday. It is okay that Danny and Stephen share July 24 th as their birthday and that June and Liz share August 2 nd . Since each person has exactly one birthday, the relation in Example is a function.

The relation shown by the graph in Example includes the ordered pairs \((−3,−1)\) and \((−3,4)\). Is that okay in a function? No, as this is like one person having two different birthdays.

Example \(\PageIndex{10}\)

Use the set of ordered pairs to (i) determine whether the relation is a function (ii) find the domain of the relation (iii) find the range of the relation.

• \({\{(−3,27),(−2,8),(−1,1),(0,0),(1,1),(2,8),(3,27)}\}\)
• \({\{(9,−3),(4,−2),(1,−1),(0,0),(1,1),(4,2),(9,3)}\}\)

ⓐ \({\{(−3,27),(−2,8),(−1,1),(0,0),(1,1),(2,8),(3,27)}\}\)

(i) Each x -value is matched with only one y -value. So this relation is a function.

(ii) The domain is the set of all x -values in the relation. The domain is: \({\{−3,−2,−1,0,1,2,3}\}\).

(iii) The range is the set of all y -values in the relation. Notice we do not list range values twice. The range is: \({\{27,8,1,0}\}\).

ⓑ \({\{(9,−3),(4,−2),(1,−1),(0,0),(1,1),(4,2),(9,3)}\}\)

(i) The x -value 9 is matched with two y -values, both 3 and \(−3\). So this relation is not a function.

(ii) The domain is the set of all x -values in the relation. Notice we do not list domain values twice. The domain is: \({\{0,1,2,4,9}\}\).

(iii) The range is the set of all y -values in the relation. The range is: \({\{−3,−2,−1,0,1,2,3}\}\).

Example \(\PageIndex{11}\)

Use the set of ordered pairs to (i) determine whether the relation is a function (ii) find the domain of the relation (iii) find the range of the function.

• \({\{(−3,−6),(−2,−4),(−1,−2),(0,0),(1,2),(2,4),(3,6)}\}\)
• \({\{(8,−4),(4,−2),(2,−1),(0,0),(2,1),(4,2),(8,4)}\}\)

ⓐ Yes; \({\{−3,−2,−1,0,1,2,3}\}\); \({\{−6,−4,−2,0,2,4,6}\}\) ⓑ No; \({\{0,2,4,8}\}\); \({\{−4,−2,−1,0,1,2,4}\}\)

Example \(\PageIndex{12}\)

• \({\{(27,−3),(8,−2),(1,−1),(0,0),(1,1),(8,2),(27,3)}\}\)
• \({\{(7,−3),(−5,−4),(8,−0),(0,0),(−6,4),(−2,2),(−1,3)}\}\)

ⓐ No; \({\{0,1,8,27}\}\); \({\{−3,−2,−1,0,2,2,3}\}\) ⓑ Yes; \({\{7,−5,8,0,−6,−2,−1}\}\); \({\{−3,−4,0,4,2,3}\}\)

Example \(\PageIndex{13}\)

Use the mapping to

• determine whether the relation is a function

ⓐ Both Lydia and Marty have two phone numbers. So each x -value is not matched with only one y -value. So this relation is not a function.

ⓑ The domain is the set of all x -values in the relation. The domain is: {Lydia, Eugene, Janet, Rick, Marty}

ⓒ The range is the set of all y -values in the relation. The range is:

\({\{321-549-3327, 427-658-2314, 321-964-7324, 684-358-7961, 684-369-7231, 798-367-8541}\}\)

Example \(\PageIndex{14}\)

Use the mapping to ⓐ determine whether the relation is a function ⓑ find the domain of the relation ⓒ find the range of the relation.

ⓐ no ⓑ {NBC, HGTV, HBO} ⓒ {Ellen Degeneres Show, Law and Order, Tonight Show, Property Brothers, House Hunters, Love it or List it, Game of Thrones, True Detective, Sesame Street}

Example \(\PageIndex{15}\)

ⓐ No ⓑ {Neal, Krystal, Kelvin, George, Christa, Mike} ⓒ {123-567-4839 work, 231-378-5941 cell, 743-469-9731 cell, 567-534-2970 work, 684-369-7231 cell, 798-367-8541 cell, 639-847-6971 cell}

In algebra, more often than not, functions will be represented by an equation. It is easiest to see if the equation is a function when it is solved for y . If each value of x results in only one value of y , then the equation defines a function.

Example \(\PageIndex{16}\)

Determine whether each equation is a function.

• \(y=x^2+1\)
• \(x+y^2=3\)

ⓐ \(2x+y=7\)

For each value of x , we multiply it by \(−2\) and then add 7 to get the y -value

We have that when \(x=3\), then \(y=1\). It would work similarly for any value of x . Since each value of x , corresponds to only one value of y the equation defines a function.

ⓑ \(y=x^2+1\)

For each value of x , we square it and then add 1 to get the y -value.

We have that when \(x=2\), then \(y=5\). It would work similarly for any value of x . Since each value of x , corresponds to only one value of y the equation defines a function.

We have shown that when \(x=2\), then \(y=1\) and \(y=−1\). It would work similarly for any value of x . Since each value of x does not corresponds to only one value of y the equation does not define a function.

Example \(\PageIndex{17}\)

• \(4x+y=−3\)
• \(x+y^2=1\)
• \(y−x^2=2\)

ⓐ yes ⓑ no ⓒ yes

Example \(\PageIndex{18}\)

• \(x+y^2=4\)
• \(y=x^2−7\)
• \(y=5x−4\)

ⓐ no ⓑ yes ⓒ yes

Find the Value of a Function

It is very convenient to name a function and most often we name it f , g , h , F , G , or H . In any function, for each x -value from the domain we get a corresponding y -value in the range. For the function \(f\), we write this range value \(y\) as \(f(x)\). This is called function notation and is read \(f\) of \(x\) or the value of \(f\) at \(x\). In this case the parentheses does not indicate multiplication.

Definition: Function Notation

For the function \(y=f(x)\)

\[\begin{array} {l} {f\text{ is the name of the function}} \\{x \text{ is the domain value}} \\ {f(x) \text{ is the range value } y \text{ corresponding to the value } x} \\ \nonumber \end{array}\]

We read \(f(x)\) as \(f\) of \(x\) or the value of \(f\) at \(x\).

We call x the independent variable as it can be any value in the domain. We call y the dependent variable as its value depends on x .

INDEPENDENT AND DEPENDENT VARIABLES

For the function \(y=f(x)\),

\[\begin{array} {l} {x \text{ is the independent variable as it can be any value in the domain}} \\ {y \text{ the dependent variable as its value depends on } x} \\ \nonumber \end{array}\]

Much as when you first encountered the variable x , function notation may be rather unsettling. It seems strange because it is new. You will feel more comfortable with the notation as you use it.

Let’s look at the equation \(y=4x−5\). To find the value of y when \(x=2\), we know to substitute \(x=2\) into the equation and then simplify.

The value of the function at \(x=2\) is 3.

We do the same thing using function notation, the equation \(y=4x−5\) can be written as \(f(x)=4x−5\). To find the value when \(x=2\), we write:

This process of finding the value of \(f(x)\) for a given value of x is called evaluating the function.

Example \(\PageIndex{19}\)

For the function \(f(x)=2x^2+3x−1\), evaluate the function.

• \(f(−2)\)

Example \(\PageIndex{20}\)

For the function \(f(x)=3x^2−2x+1\), evaluate the function.

• \(f(−1)\)

ⓐ \(f(3)=22\) ⓑ \(f(−1)=6\) ⓒ \(f(t)=3t^2−2t−1\)

Example \(\PageIndex{21}\)

For the function \(f(x)=2x^2+4x−3\), evaluate the function.

• \(f(−3)\)

ⓐ \((2)=13\) ⓑ \(f(−3)=3\) ⓒ \(f(h)=2h2+4h−3\)

In the last example, we found \(f(x)\) for a constant value of x . In the next example, we are asked to find \(g(x)\) with values of x that are variables. We still follow the same procedure and substitute the variables in for the x .

Example \(\PageIndex{22}\)

For the function \(g(x)=3x−5\), evaluate the function.

• \(g(x)+g(2)\)

Notice the difference between part ⓑ and ⓒ. We get \(g(x+2)=3x+1\) and \(g(x)+g(2)=3x−4\). So we see that \(g(x+2)\neq g(x)+g(2)\).

Example \(\PageIndex{23}\)

For the function \(g(x)=4x−7\), evaluate the function.

• \(g(x−3)\)
• \(g(x)−g(3)\)

ⓐ \(4m^2−7\) ⓑ \(4x−19\) ⓒ \(x−12\)

Example \(\PageIndex{24}\)

For the function \(h(x)=2x+1\), evaluate the function.

• \(h(x)+h(1)\)

ⓐ \(2k^2+1\) ⓑ \(2x+3\) ⓒ \(2x+4\)

Many everyday situations can be modeled using functions.

Example \(\PageIndex{25}\)

The number of unread emails in Sylvia’s account is 75. This number grows by 10 unread emails a day. The function \(N(t)=75+10t\) represents the relation between the number of emails, N , and the time, t , measured in days.

• Determine the independent and dependent variable.
• Find \(N(5)\). Explain what this result means.

ⓐ The number of unread emails is a function of the number of days. The number of unread emails, N , depends on the number of days, t . Therefore, the variable N , is the dependent variable and the variable tt is the independent variable.

ⓑ Find \(N(5)\). Explain what this result means.

Since 5 is the number of days, \(N(5)\), is the number of unread emails after 5 days. After 5 days, there are 125 unread emails in the account.

Example \(\PageIndex{26}\)

The number of unread emails in Bryan’s account is 100. This number grows by 15 unread emails a day. The function \(N(t)=100+15t\) represents the relation between the number of emails, N , and the time, t , measured in days.

• Find \(N(7)]\). Explain what this result means.

ⓐ t IND; N DEP ⓑ 205; the number of unread emails in Bryan’s account on the seventh day.

Example \(\PageIndex{27}\)

The number of unread emails in Anthony’s account is 110. This number grows by 25 unread emails a day. The function \(N(t)=110+25t\) represents the relation between the number of emails, N , and the time, t , measured in days.

• Find \(N(14)\). Explain what this result means.

ⓐ t IND; N DEP ⓑ 460; the number of unread emails in Anthony’s account on the fourteenth day

Access this online resource for additional instruction and practice with relations and functions.

• Introduction to Functions

Key Concepts

• f is the name of the function
• x is the domain value
• \(f(x)\) is the range value y corresponding to the value x We read \(f(x)\) as f of x or the value of f at x .
• x is the independent variable as it can be any value in the domain
• y is the dependent variable as its value depends on x

3.1 Functions and Function Notation

• ⓑ yes (Note: If two players had been tied for, say, 4th place, then the name would not have been a function of rank.)

w = f ( d ) w = f ( d )

g ( 5 ) = 1 g ( 5 ) = 1

m = 8 m = 8

y = f ( x ) = x 3 2 y = f ( x ) = x 3 2

g ( 1 ) = 8 g ( 1 ) = 8

x = 0 x = 0 or x = 2 x = 2

• ⓐ yes, because each bank account has a single balance at any given time;
• ⓑ no, because several bank account numbers may have the same balance;
• ⓒ no, because the same output may correspond to more than one input.
• ⓐ Yes, letter grade is a function of percent grade;
• ⓑ No, it is not one-to-one. There are 100 different percent numbers we could get but only about five possible letter grades, so there cannot be only one percent number that corresponds to each letter grade.

No, because it does not pass the horizontal line test.

3.2 Domain and Range

{ − 5 , 0 , 5 , 10 , 15 } { − 5 , 0 , 5 , 10 , 15 }

( − ∞ , ∞ ) ( − ∞ , ∞ )

( − ∞ , 1 2 ) ∪ ( 1 2 , ∞ ) ( − ∞ , 1 2 ) ∪ ( 1 2 , ∞ )

[ − 5 2 , ∞ ) [ − 5 2 , ∞ )

• ⓐ values that are less than or equal to –2, or values that are greater than or equal to –1 and less than 3
• ⓑ { x | x ≤ − 2 or − 1 ≤ x < 3 } { x | x ≤ − 2 or − 1 ≤ x < 3 }
• ⓒ ( − ∞ , − 2 ] ∪ [ − 1 , 3 ) ( − ∞ , − 2 ] ∪ [ − 1 , 3 )

domain =[1950,2002] range = [47,000,000,89,000,000]

domain: ( − ∞ , 2 ] ; ( − ∞ , 2 ] ; range: ( − ∞ , 0 ] ( − ∞ , 0 ]

3.3 Rates of Change and Behavior of Graphs

$ 2.84 − $ 2.31 5 years = $ 0.53 5 years = $ 0.106 $ 2.84 − $ 2.31 5 years = $ 0.53 5 years = $ 0.106 per year.

a + 7 a + 7

The local maximum appears to occur at ( − 1 , 28 ) , ( − 1 , 28 ) , and the local minimum occurs at ( 5 , − 80 ) . ( 5 , − 80 ) . The function is increasing on ( − ∞ , − 1 ) ∪ ( 5 , ∞ ) ( − ∞ , − 1 ) ∪ ( 5 , ∞ ) and decreasing on ( − 1 , 5 ) . ( − 1 , 5 ) .

3.4 Composition of Functions

( f g ) ( x ) = f ( x ) g ( x ) = ( x − 1 ) ( x 2 − 1 ) = x 3 − x 2 − x + 1 ( f − g ) ( x ) = f ( x ) − g ( x ) = ( x − 1 ) − ( x 2 − 1 ) = x − x 2 ( f g ) ( x ) = f ( x ) g ( x ) = ( x − 1 ) ( x 2 − 1 ) = x 3 − x 2 − x + 1 ( f − g ) ( x ) = f ( x ) − g ( x ) = ( x − 1 ) − ( x 2 − 1 ) = x − x 2

No, the functions are not the same.

A gravitational force is still a force, so a ( G ( r ) ) a ( G ( r ) ) makes sense as the acceleration of a planet at a distance r from the Sun (due to gravity), but G ( a ( F ) ) G ( a ( F ) ) does not make sense.

f ( g ( 1 ) ) = f ( 3 ) = 3 f ( g ( 1 ) ) = f ( 3 ) = 3 and g ( f ( 4 ) ) = g ( 1 ) = 3 g ( f ( 4 ) ) = g ( 1 ) = 3

g ( f ( 2 ) ) = g ( 5 ) = 3 g ( f ( 2 ) ) = g ( 5 ) = 3

[ − 4 , 0 ) ∪ ( 0 , ∞ ) [ − 4 , 0 ) ∪ ( 0 , ∞ )

Possible answer:

g ( x ) = 4 + x 2 h ( x ) = 4 3 − x f = h ∘ g g ( x ) = 4 + x 2 h ( x ) = 4 3 − x f = h ∘ g

3.5 Transformation of Functions

The graphs of f ( x ) f ( x ) and g ( x ) g ( x ) are shown below. The transformation is a horizontal shift. The function is shifted to the left by 2 units.

g ( x ) = 1 x - 1 + 1 g ( x ) = 1 x - 1 + 1

g ( x ) = − f ( x ) g ( x ) = − f ( x )

h ( x ) = f ( − x ) h ( x ) = f ( − x )

Notice: g ( x ) = f ( − x ) g ( x ) = f ( − x ) looks the same as f ( x ) f ( x ) .

g ( x ) = 3 x - 2 g ( x ) = 3 x - 2

g ( x ) = f ( 1 3 x ) g ( x ) = f ( 1 3 x ) so using the square root function we get g ( x ) = 1 3 x g ( x ) = 1 3 x

3.6 Absolute Value Functions

using the variable p p for passing, | p − 80 | ≤ 20 | p − 80 | ≤ 20

f ( x ) = − | x + 2 | + 3 f ( x ) = − | x + 2 | + 3

x = − 1 x = − 1 or x = 2 x = 2

3.7 Inverse Functions

h ( 2 ) = 6 h ( 2 ) = 6

The domain of function f − 1 f − 1 is ( − ∞ , − 2 ) ( − ∞ , − 2 ) and the range of function f − 1 f − 1 is ( 1 , ∞ ) . ( 1 , ∞ ) .

• ⓐ f ( 60 ) = 50. f ( 60 ) = 50. In 60 minutes, 50 miles are traveled.
• ⓑ f − 1 ( 60 ) = 70. f − 1 ( 60 ) = 70. To travel 60 miles, it will take 70 minutes.

x = 3 y + 5 x = 3 y + 5

f − 1 ( x ) = ( 2 − x ) 2 ; domain of f : [ 0 , ∞ ) ; domain of f − 1 : ( − ∞ , 2 ] f − 1 ( x ) = ( 2 − x ) 2 ; domain of f : [ 0 , ∞ ) ; domain of f − 1 : ( − ∞ , 2 ]

3.1 Section Exercises

A relation is a set of ordered pairs. A function is a special kind of relation in which no two ordered pairs have the same first coordinate.

When a vertical line intersects the graph of a relation more than once, that indicates that for that input there is more than one output. At any particular input value, there can be only one output if the relation is to be a function.

When a horizontal line intersects the graph of a function more than once, that indicates that for that output there is more than one input. A function is one-to-one if each output corresponds to only one input.

not a function

f ( − 3 ) = − 11 ; f ( − 3 ) = − 11 ; f ( 2 ) = − 1 ; f ( 2 ) = − 1 ; f ( − a ) = − 2 a − 5 ; f ( − a ) = − 2 a − 5 ; − f ( a ) = − 2 a + 5 ; − f ( a ) = − 2 a + 5 ; f ( a + h ) = 2 a + 2 h − 5 f ( a + h ) = 2 a + 2 h − 5

f ( − 3 ) = 5 + 5 ; f ( − 3 ) = 5 + 5 ; f ( 2 ) = 5 ; f ( 2 ) = 5 ; f ( − a ) = 2 + a + 5 ; f ( − a ) = 2 + a + 5 ; − f ( a ) = − 2 − a − 5 ; − f ( a ) = − 2 − a − 5 ; f ( a + h ) = 2 − a − h + 5 f ( a + h ) = 2 − a − h + 5

f ( − 3 ) = 2 ; f ( − 3 ) = 2 ; f ( 2 ) = 1 − 3 = − 2 ; f ( 2 ) = 1 − 3 = − 2 ; f ( − a ) = | − a − 1 | − | − a + 1 | ; f ( − a ) = | − a − 1 | − | − a + 1 | ; − f ( a ) = − | a − 1 | + | a + 1 | ; − f ( a ) = − | a − 1 | + | a + 1 | ; f ( a + h ) = | a + h − 1 | − | a + h + 1 | f ( a + h ) = | a + h − 1 | − | a + h + 1 |

g ( x ) − g ( a ) x − a = x + a + 2 , x ≠ a g ( x ) − g ( a ) x − a = x + a + 2 , x ≠ a

a. f ( − 2 ) = 14 ; f ( − 2 ) = 14 ; b. x = 3 x = 3

a. f ( 5 ) = 10 ; f ( 5 ) = 10 ; b. x = − 1 x = − 1 or x = 4 x = 4

• ⓐ f ( t ) = 6 − 2 3 t ; f ( t ) = 6 − 2 3 t ;
• ⓑ f ( − 3 ) = 8 ; f ( − 3 ) = 8 ;
• ⓒ t = 6 t = 6
• ⓐ f ( 0 ) = 1 ; f ( 0 ) = 1 ;
• ⓑ f ( x ) = − 3 , x = − 2 f ( x ) = − 3 , x = − 2 or x = 2 x = 2

not a function so it is also not a one-to-one function

one-to- one function

function, but not one-to-one

f ( x ) = 1 , x = 2 f ( x ) = 1 , x = 2

f ( − 2 ) = 14 ; f ( − 1 ) = 11 ; f ( 0 ) = 8 ; f ( 1 ) = 5 ; f ( 2 ) = 2 f ( − 2 ) = 14 ; f ( − 1 ) = 11 ; f ( 0 ) = 8 ; f ( 1 ) = 5 ; f ( 2 ) = 2

f ( − 2 ) = 4 ;    f ( − 1 ) = 4.414 ; f ( 0 ) = 4.732 ; f ( 1 ) = 5 ; f ( 2 ) = 5.236 f ( − 2 ) = 4 ;    f ( − 1 ) = 4.414 ; f ( 0 ) = 4.732 ; f ( 1 ) = 5 ; f ( 2 ) = 5.236

f ( − 2 ) = 1 9 ; f ( − 1 ) = 1 3 ; f ( 0 ) = 1 ; f ( 1 ) = 3 ; f ( 2 ) = 9 f ( − 2 ) = 1 9 ; f ( − 1 ) = 1 3 ; f ( 0 ) = 1 ; f ( 1 ) = 3 ; f ( 2 ) = 9

[ 0 , 100 ] [ 0 , 100 ]

[ − 0.001 , 0 .001 ] [ − 0.001 , 0 .001 ]

[ − 1 , 000 , 000 , 1,000,000 ] [ − 1 , 000 , 000 , 1,000,000 ]

[ 0 , 10 ] [ 0 , 10 ]

[ −0.1 , 0.1 ] [ −0.1 , 0.1 ]

[ − 100 , 100 ] [ − 100 , 100 ]

• ⓐ g ( 5000 ) = 50 ; g ( 5000 ) = 50 ;
• ⓑ The number of cubic yards of dirt required for a garden of 100 square feet is 1.
• ⓐ The height of a rocket above ground after 1 second is 200 ft.
• ⓑ The height of a rocket above ground after 2 seconds is 350 ft.

3.2 Section Exercises

The domain of a function depends upon what values of the independent variable make the function undefined or imaginary.

There is no restriction on x x for f ( x ) = x 3 f ( x ) = x 3 because you can take the cube root of any real number. So the domain is all real numbers, ( − ∞ , ∞ ) . ( − ∞ , ∞ ) . When dealing with the set of real numbers, you cannot take the square root of negative numbers. So x x -values are restricted for f ( x ) = x f ( x ) = x to nonnegative numbers and the domain is [ 0 , ∞ ) . [ 0 , ∞ ) .

Graph each formula of the piecewise function over its corresponding domain. Use the same scale for the x x -axis and y y -axis for each graph. Indicate inclusive endpoints with a solid circle and exclusive endpoints with an open circle. Use an arrow to indicate − ∞ − ∞ or ∞ . ∞ . Combine the graphs to find the graph of the piecewise function.

( − ∞ , 3 ] ( − ∞ , 3 ]

( − ∞ , − 1 2 ) ∪ ( − 1 2 , ∞ ) ( − ∞ , − 1 2 ) ∪ ( − 1 2 , ∞ )

( − ∞ , − 11 ) ∪ ( − 11 , 2 ) ∪ ( 2 , ∞ ) ( − ∞ , − 11 ) ∪ ( − 11 , 2 ) ∪ ( 2 , ∞ )

( − ∞ , − 3 ) ∪ ( − 3 , 5 ) ∪ ( 5 , ∞ ) ( − ∞ , − 3 ) ∪ ( − 3 , 5 ) ∪ ( 5 , ∞ )

( − ∞ , 5 ) ( − ∞ , 5 )

[ 6 , ∞ ) [ 6 , ∞ )

( − ∞ , − 9 ) ∪ ( − 9 , 9 ) ∪ ( 9 , ∞ ) ( − ∞ , − 9 ) ∪ ( − 9 , 9 ) ∪ ( 9 , ∞ )

domain: ( 2 , 8 ] , ( 2 , 8 ] , range [ 6 , 8 ) [ 6 , 8 )

domain: [ − 4 , 4], [ − 4 , 4], range: [ 0 , 2] [ 0 , 2]

domain: [ − 5 , 3 ) , [ − 5 , 3 ) , range: [ 0 , 2 ] [ 0 , 2 ]

domain: ( − ∞ , 1 ] , ( − ∞ , 1 ] , range: [ 0 , ∞ ) [ 0 , ∞ )

domain: [ − 6 , − 1 6 ] ∪ [ 1 6 , 6 ] ; [ − 6 , − 1 6 ] ∪ [ 1 6 , 6 ] ; range: [ − 6 , − 1 6 ] ∪ [ 1 6 , 6 ] [ − 6 , − 1 6 ] ∪ [ 1 6 , 6 ]

domain: [ − 3 , ∞ ) ; [ − 3 , ∞ ) ; range: [ 0 , ∞ ) [ 0 , ∞ )

domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

f ( − 3 ) = 1 ; f ( − 2 ) = 0 ; f ( − 1 ) = 0 ; f ( 0 ) = 0 f ( − 3 ) = 1 ; f ( − 2 ) = 0 ; f ( − 1 ) = 0 ; f ( 0 ) = 0

f ( − 1 ) = − 4 ; f ( 0 ) = 6 ; f ( 2 ) = 20 ; f ( 4 ) = 34 f ( − 1 ) = − 4 ; f ( 0 ) = 6 ; f ( 2 ) = 20 ; f ( 4 ) = 34

f ( − 1 ) = − 5 ; f ( 0 ) = 3 ; f ( 2 ) = 3 ; f ( 4 ) = 16 f ( − 1 ) = − 5 ; f ( 0 ) = 3 ; f ( 2 ) = 3 ; f ( 4 ) = 16

domain: ( − ∞ , 1 ) ∪ ( 1 , ∞ ) ( − ∞ , 1 ) ∪ ( 1 , ∞ )

window: [ − 0.5 , − 0.1 ] ; [ − 0.5 , − 0.1 ] ; range: [ 4 , 100 ] [ 4 , 100 ]

window: [ 0.1 , 0.5 ] ; [ 0.1 , 0.5 ] ; range: [ 4 , 100 ] [ 4 , 100 ]

[ 0 , 8 ] [ 0 , 8 ]

Many answers. One function is f ( x ) = 1 x − 2 . f ( x ) = 1 x − 2 .

• ⓐ The fixed cost is $500.
• ⓑ The cost of making 25 items is $750.
• ⓒ The domain is [0, 100] and the range is [500, 1500].

3.3 Section Exercises

Yes, the average rate of change of all linear functions is constant.

The absolute maximum and minimum relate to the entire graph, whereas the local extrema relate only to a specific region around an open interval.

4 ( b + 1 ) 4 ( b + 1 )

4 x + 2 h 4 x + 2 h

− 1 13 ( 13 + h ) − 1 13 ( 13 + h )

3 h 2 + 9 h + 9 3 h 2 + 9 h + 9

4 x + 2 h − 3 4 x + 2 h − 3

increasing on ( − ∞ , − 2.5 ) ∪ ( 1 , ∞ ) , ( − ∞ , − 2.5 ) ∪ ( 1 , ∞ ) , decreasing on ( − 2.5 , 1 ) ( − 2.5 , 1 )

increasing on ( − ∞ , 1 ) ∪ ( 3 , 4 ) , ( − ∞ , 1 ) ∪ ( 3 , 4 ) , decreasing on ( 1 , 3 ) ∪ ( 4 , ∞ ) ( 1 , 3 ) ∪ ( 4 , ∞ )

local maximum: ( − 3 , 60 ) , ( − 3 , 60 ) , local minimum: ( 3 , − 60 ) ( 3 , − 60 )

absolute maximum at approximately ( 7 , 150 ) , ( 7 , 150 ) , absolute minimum at approximately ( −7.5 , −220 ) ( −7.5 , −220 )

Local minimum at ( 3 , − 22 ) , ( 3 , − 22 ) , decreasing on ( − ∞ , 3 ) , ( − ∞ , 3 ) , increasing on ( 3 , ∞ ) ( 3 , ∞ )

Local minimum at ( − 2 , − 2 ) , ( − 2 , − 2 ) , decreasing on ( − 3 , − 2 ) , ( − 3 , − 2 ) , increasing on ( − 2 , ∞ ) ( − 2 , ∞ )

Local maximum at ( − 0.5 , 6 ) , ( − 0.5 , 6 ) , local minima at ( − 3.25 , − 47 ) ( − 3.25 , − 47 ) and ( 2.1 , − 32 ) , ( 2.1 , − 32 ) , decreasing on ( − ∞ , − 3.25 ) ( − ∞ , − 3.25 ) and ( − 0.5 , 2.1 ) , ( − 0.5 , 2.1 ) , increasing on ( − 3.25 , − 0.5 ) ( − 3.25 , − 0.5 ) and ( 2.1 , ∞ ) ( 2.1 , ∞ )

b = 5 b = 5

2.7 gallons per minute

approximately –0.6 milligrams per day

3.4 Section Exercises

Find the numbers that make the function in the denominator g g equal to zero, and check for any other domain restrictions on f f and g , g , such as an even-indexed root or zeros in the denominator.

Yes. Sample answer: Let f ( x ) = x + 1 and  g ( x ) = x − 1. f ( x ) = x + 1 and  g ( x ) = x − 1. Then f ( g ( x ) ) = f ( x − 1 ) = ( x − 1 ) + 1 = x f ( g ( x ) ) = f ( x − 1 ) = ( x − 1 ) + 1 = x and g ( f ( x ) ) = g ( x + 1 ) = ( x + 1 ) − 1 = x . g ( f ( x ) ) = g ( x + 1 ) = ( x + 1 ) − 1 = x . So f ∘ g = g ∘ f . f ∘ g = g ∘ f .

( f + g ) ( x ) = 2 x + 6 , ( f + g ) ( x ) = 2 x + 6 , domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

( f − g ) ( x ) = 2 x 2 + 2 x − 6 , ( f − g ) ( x ) = 2 x 2 + 2 x − 6 , domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

( f g ) ( x ) = − x 4 − 2 x 3 + 6 x 2 + 12 x , ( f g ) ( x ) = − x 4 − 2 x 3 + 6 x 2 + 12 x , domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

( f g ) ( x ) = x 2 + 2 x 6 − x 2 , ( f g ) ( x ) = x 2 + 2 x 6 − x 2 , domain: ( − ∞ , − 6 ) ∪ ( − 6 , 6 ) ∪ ( 6 , ∞ ) ( − ∞ , − 6 ) ∪ ( − 6 , 6 ) ∪ ( 6 , ∞ )

( f + g ) ( x ) = 4 x 3 + 8 x 2 + 1 2 x , ( f + g ) ( x ) = 4 x 3 + 8 x 2 + 1 2 x , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f − g ) ( x ) = 4 x 3 + 8 x 2 − 1 2 x , ( f − g ) ( x ) = 4 x 3 + 8 x 2 − 1 2 x , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f g ) ( x ) = x + 2 , ( f g ) ( x ) = x + 2 , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f g ) ( x ) = 4 x 3 + 8 x 2 , ( f g ) ( x ) = 4 x 3 + 8 x 2 , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f + g ) ( x ) = 3 x 2 + x − 5 , ( f + g ) ( x ) = 3 x 2 + x − 5 , domain: [ 5 , ∞ ) [ 5 , ∞ )

( f − g ) ( x ) = 3 x 2 − x − 5 , ( f − g ) ( x ) = 3 x 2 − x − 5 , domain: [ 5 , ∞ ) [ 5 , ∞ )

( f g ) ( x ) = 3 x 2 x − 5 , ( f g ) ( x ) = 3 x 2 x − 5 , domain: [ 5 , ∞ ) [ 5 , ∞ )

( f g ) ( x ) = 3 x 2 x − 5 , ( f g ) ( x ) = 3 x 2 x − 5 , domain: ( 5 , ∞ ) ( 5 , ∞ )

• ⓑ f ( g ( x ) ) = 2 ( 3 x − 5 ) 2 + 1 f ( g ( x ) ) = 2 ( 3 x − 5 ) 2 + 1
• ⓒ f ( g ( x ) ) = 6 x 2 − 2 f ( g ( x ) ) = 6 x 2 − 2
• ⓓ ( g ∘ g ) ( x ) = 3 ( 3 x − 5 ) − 5 = 9 x − 20 ( g ∘ g ) ( x ) = 3 ( 3 x − 5 ) − 5 = 9 x − 20
• ⓔ ( f ∘ f ) ( − 2 ) = 163 ( f ∘ f ) ( − 2 ) = 163

f ( g ( x ) ) = x 2 + 3 + 2 , g ( f ( x ) ) = x + 4 x + 7 f ( g ( x ) ) = x 2 + 3 + 2 , g ( f ( x ) ) = x + 4 x + 7

f ( g ( x ) ) = x + 1 x 3 3 = x + 1 3 x , g ( f ( x ) ) = x 3 + 1 x f ( g ( x ) ) = x + 1 x 3 3 = x + 1 3 x , g ( f ( x ) ) = x 3 + 1 x

( f ∘ g ) ( x ) = 1 2 x + 4 − 4 = x 2 , ( g ∘ f ) ( x ) = 2 x − 4 ( f ∘ g ) ( x ) = 1 2 x + 4 − 4 = x 2 , ( g ∘ f ) ( x ) = 2 x − 4

f ( g ( h ( x ) ) ) = ( 1 x + 3 ) 2 + 1 f ( g ( h ( x ) ) ) = ( 1 x + 3 ) 2 + 1

• ⓐ ( g ∘ f ) ( x ) = − 3 2 − 4 x ( g ∘ f ) ( x ) = − 3 2 − 4 x
• ⓑ ( − ∞ , 1 2 ) ( − ∞ , 1 2 )
• ⓐ ( 0 , 2 ) ∪ ( 2 , ∞ ) ; ( 0 , 2 ) ∪ ( 2 , ∞ ) ;
• ⓑ ( − ∞ , − 2 ) ∪ ( 2 , ∞ ) ; ( − ∞ , − 2 ) ∪ ( 2 , ∞ ) ;
• ⓒ ( 0 , ∞ ) ( 0 , ∞ )

( 1 , ∞ ) ( 1 , ∞ )

sample: f ( x ) = x 3 g ( x ) = x − 5 f ( x ) = x 3 g ( x ) = x − 5

sample: f ( x ) = 4 x g ( x ) = ( x + 2 ) 2 f ( x ) = 4 x g ( x ) = ( x + 2 ) 2

sample: f ( x ) = x 3 g ( x ) = 1 2 x − 3 f ( x ) = x 3 g ( x ) = 1 2 x − 3

sample: f ( x ) = x 4 g ( x ) = 3 x − 2 x + 5 f ( x ) = x 4 g ( x ) = 3 x − 2 x + 5

sample: f ( x ) = x g ( x ) = 2 x + 6 f ( x ) = x g ( x ) = 2 x + 6

sample: f ( x ) = x 3 g ( x ) = ( x − 1 ) f ( x ) = x 3 g ( x ) = ( x − 1 )

sample: f ( x ) = x 3 g ( x ) = 1 x − 2 f ( x ) = x 3 g ( x ) = 1 x − 2

sample: f ( x ) = x g ( x ) = 2 x − 1 3 x + 4 f ( x ) = x g ( x ) = 2 x − 1 3 x + 4

f ( g ( 0 ) ) = 27 , g ( f ( 0 ) ) = − 94 f ( g ( 0 ) ) = 27 , g ( f ( 0 ) ) = − 94

f ( g ( 0 ) ) = 1 5 , g ( f ( 0 ) ) = 5 f ( g ( 0 ) ) = 1 5 , g ( f ( 0 ) ) = 5

18 x 2 + 60 x + 51 18 x 2 + 60 x + 51

g ∘ g ( x ) = 9 x + 20 g ∘ g ( x ) = 9 x + 20

( f ∘ g ) ( 6 ) = 6 ( f ∘ g ) ( 6 ) = 6 ; ( g ∘ f ) ( 6 ) = 6 ( g ∘ f ) ( 6 ) = 6

( f ∘ g ) ( 11 ) = 11 , ( g ∘ f ) ( 11 ) = 11 ( f ∘ g ) ( 11 ) = 11 , ( g ∘ f ) ( 11 ) = 11

A ( t ) = π ( 25 t + 2 ) 2 A ( t ) = π ( 25 t + 2 ) 2 and A ( 2 ) = π ( 25 4 ) 2 = 2500 π A ( 2 ) = π ( 25 4 ) 2 = 2500 π square inches

A ( 5 ) = π ( 2 ( 5 ) + 1 ) 2 = 121 π A ( 5 ) = π ( 2 ( 5 ) + 1 ) 2 = 121 π square units

• ⓐ N ( T ( t ) ) = 23 ( 5 t + 1.5 ) 2 − 56 ( 5 t + 1.5 ) + 1 N ( T ( t ) ) = 23 ( 5 t + 1.5 ) 2 − 56 ( 5 t + 1.5 ) + 1
• ⓑ 3.38 hours

3.5 Section Exercises

A horizontal shift results when a constant is added to or subtracted from the input. A vertical shifts results when a constant is added to or subtracted from the output.

A horizontal compression results when a constant greater than 1 is multiplied by the input. A vertical compression results when a constant between 0 and 1 is multiplied by the output.

For a function f , f , substitute ( − x ) ( − x ) for ( x ) ( x ) in f ( x ) . f ( x ) . Simplify. If the resulting function is the same as the original function, f ( − x ) = f ( x ) , f ( − x ) = f ( x ) , then the function is even. If the resulting function is the opposite of the original function, f ( − x ) = − f ( x ) , f ( − x ) = − f ( x ) , then the original function is odd. If the function is not the same or the opposite, then the function is neither odd nor even.

g ( x ) = | x - 1 | − 3 g ( x ) = | x - 1 | − 3

g ( x ) = 1 ( x + 4 ) 2 + 2 g ( x ) = 1 ( x + 4 ) 2 + 2

The graph of f ( x + 43 ) f ( x + 43 ) is a horizontal shift to the left 43 units of the graph of f . f .

The graph of f ( x - 4 ) f ( x - 4 ) is a horizontal shift to the right 4 units of the graph of f . f .

The graph of f ( x ) + 8 f ( x ) + 8 is a vertical shift up 8 units of the graph of f . f .

The graph of f ( x ) − 7 f ( x ) − 7 is a vertical shift down 7 units of the graph of f . f .

The graph of f ( x + 4 ) − 1 f ( x + 4 ) − 1 is a horizontal shift to the left 4 units and a vertical shift down 1 unit of the graph of f . f .

decreasing on ( − ∞ , − 3 ) ( − ∞ , − 3 ) and increasing on ( − 3 , ∞ ) ( − 3 , ∞ )

decreasing on ( 0 , ∞ ) ( 0 , ∞ )

g ( x ) = f ( x - 1 ) , h ( x ) = f ( x ) + 1 g ( x ) = f ( x - 1 ) , h ( x ) = f ( x ) + 1

f ( x ) = | x - 3 | − 2 f ( x ) = | x - 3 | − 2

f ( x ) = x + 3 − 1 f ( x ) = x + 3 − 1

f ( x ) = ( x - 2 ) 2 f ( x ) = ( x - 2 ) 2

f ( x ) = | x + 3 | − 2 f ( x ) = | x + 3 | − 2

f ( x ) = − x f ( x ) = − x

f ( x ) = − ( x + 1 ) 2 + 2 f ( x ) = − ( x + 1 ) 2 + 2

f ( x ) = − x + 1 f ( x ) = − x + 1

The graph of g g is a vertical reflection (across the x x -axis) of the graph of f . f .

The graph of g g is a vertical stretch by a factor of 4 of the graph of f . f .

The graph of g g is a horizontal compression by a factor of 1 5 1 5 of the graph of f . f .

The graph of g g is a horizontal stretch by a factor of 3 of the graph of f . f .

The graph of g g is a horizontal reflection across the y y -axis and a vertical stretch by a factor of 3 of the graph of f . f .

g ( x ) = | − 4 x | g ( x ) = | − 4 x |

g ( x ) = 1 3 ( x + 2 ) 2 − 3 g ( x ) = 1 3 ( x + 2 ) 2 − 3

g ( x ) = 1 2 ( x - 5 ) 2 + 1 g ( x ) = 1 2 ( x - 5 ) 2 + 1

The graph of the function f ( x ) = x 2 f ( x ) = x 2 is shifted to the left 1 unit, stretched vertically by a factor of 4, and shifted down 5 units.

The graph of f ( x ) = | x | f ( x ) = | x | is stretched vertically by a factor of 2, shifted horizontally 4 units to the right, reflected across the horizontal axis, and then shifted vertically 3 units up.

The graph of the function f ( x ) = x 3 f ( x ) = x 3 is compressed vertically by a factor of 1 2 . 1 2 .

The graph of the function is stretched horizontally by a factor of 3 and then shifted vertically downward by 3 units.

The graph of f ( x ) = x f ( x ) = x is shifted right 4 units and then reflected across the vertical line x = 4. x = 4.

3.6 Section Exercises

Isolate the absolute value term so that the equation is of the form | A | = B . | A | = B . Form one equation by setting the expression inside the absolute value symbol, A , A , equal to the expression on the other side of the equation, B . B . Form a second equation by setting A A equal to the opposite of the expression on the other side of the equation, − B . − B . Solve each equation for the variable.

The graph of the absolute value function does not cross the x x -axis, so the graph is either completely above or completely below the x x -axis.

The distance from x to 8 can be represented using the absolute value statement: ∣ x − 8 ∣ = 4.

∣ x − 10 ∣ ≥ 15

There are no x-intercepts.

(−4, 0) and (2, 0)

( 0 , − 4 ) , ( 4 , 0 ) , ( − 2 , 0 ) ( 0 , − 4 ) , ( 4 , 0 ) , ( − 2 , 0 )

( 0 , 7 ) , ( 25 , 0 ) , ( − 7 , 0 ) ( 0 , 7 ) , ( 25 , 0 ) , ( − 7 , 0 )

range: [ – 400 , 100 ] [ – 400 , 100 ]

There is no solution for a a that will keep the function from having a y y -intercept. The absolute value function always crosses the y y -intercept when x = 0. x = 0.

| p − 0.08 | ≤ 0.015 | p − 0.08 | ≤ 0.015

| x − 5.0 | ≤ 0.01 | x − 5.0 | ≤ 0.01

3.7 Section Exercises

Each output of a function must have exactly one output for the function to be one-to-one. If any horizontal line crosses the graph of a function more than once, that means that y y -values repeat and the function is not one-to-one. If no horizontal line crosses the graph of the function more than once, then no y y -values repeat and the function is one-to-one.

Yes. For example, f ( x ) = 1 x f ( x ) = 1 x is its own inverse.

Given a function y = f ( x ) , y = f ( x ) , solve for x x in terms of y . y . Interchange the x x and y . y . Solve the new equation for y . y . The expression for y y is the inverse, y = f − 1 ( x ) . y = f − 1 ( x ) .

f − 1 ( x ) = x − 3 f − 1 ( x ) = x − 3

f − 1 ( x ) = 2 − x f − 1 ( x ) = 2 − x

f − 1 ( x ) = − 2 x x − 1 f − 1 ( x ) = − 2 x x − 1

domain of f ( x ) : [ − 7 , ∞ ) ; f − 1 ( x ) = x − 7 f ( x ) : [ − 7 , ∞ ) ; f − 1 ( x ) = x − 7

domain of f ( x ) : [ 0 , ∞ ) ; f − 1 ( x ) = x + 5 f ( x ) : [ 0 , ∞ ) ; f − 1 ( x ) = x + 5

a. f ( g ( x ) ) = x f ( g ( x ) ) = x and g ( f ( x ) ) = x . g ( f ( x ) ) = x . b. This tells us that f f and g g are inverse functions

  f ( g ( x ) ) = x , g ( f ( x ) ) = x   f ( g ( x ) ) = x , g ( f ( x ) ) = x

not one-to-one

[ 2 , 10 ] [ 2 , 10 ]

f − 1 ( x ) = ( 1 + x ) 1 / 3 f − 1 ( x ) = ( 1 + x ) 1 / 3

f − 1 ( x ) = 5 9 ( x − 32 ) . f − 1 ( x ) = 5 9 ( x − 32 ) . Given the Fahrenheit temperature, x , x , this formula allows you to calculate the Celsius temperature.

t ( d ) = d 50 , t ( d ) = d 50 , t ( 180 ) = 180 50 . t ( 180 ) = 180 50 . The time for the car to travel 180 miles is 3.6 hours.

Review Exercises

f ( − 3 ) = − 27 ; f ( − 3 ) = − 27 ; f ( 2 ) = − 2 ; f ( 2 ) = − 2 ; f ( − a ) = − 2 a 2 − 3 a ; f ( − a ) = − 2 a 2 − 3 a ; − f ( a ) = 2 a 2 − 3 a ; − f ( a ) = 2 a 2 − 3 a ; f ( a + h ) = − 2 a 2 + 3 a − 4 a h + 3 h − 2 h 2 f ( a + h ) = − 2 a 2 + 3 a − 4 a h + 3 h − 2 h 2

x = − 1.8 x = − 1.8 or or  x = 1.8 or  x = 1.8

− 64 + 80 a − 16 a 2 − 1 + a = − 16 a + 64 − 64 + 80 a − 16 a 2 − 1 + a = − 16 a + 64

( − ∞ , − 2 ) ∪ ( − 2 , 6 ) ∪ ( 6 , ∞ ) ( − ∞ , − 2 ) ∪ ( − 2 , 6 ) ∪ ( 6 , ∞ )

increasing ( 2 , ∞ ) ; ( 2 , ∞ ) ; decreasing ( − ∞ , 2 ) ( − ∞ , 2 )

increasing ( − 3 , 1 ) ; ( − 3 , 1 ) ; constant ( − ∞ , − 3 ) ∪ ( 1 , ∞ ) ( − ∞ , − 3 ) ∪ ( 1 , ∞ )

local minimum ( − 2 , − 3 ) ; ( − 2 , − 3 ) ; local maximum ( 1 , 3 ) ( 1 , 3 )

( − 1.8 , 10 ) ( − 1.8 , 10 )

( f ∘ g ) ( x ) = 17 − 18 x ; ( g ∘ f ) ( x ) = − 7 − 18 x ( f ∘ g ) ( x ) = 17 − 18 x ; ( g ∘ f ) ( x ) = − 7 − 18 x

( f ∘ g ) ( x ) = 1 x + 2 ; ( g ∘ f ) ( x ) = 1 x + 2 ( f ∘ g ) ( x ) = 1 x + 2 ; ( g ∘ f ) ( x ) = 1 x + 2

( f ∘ g ) ( x ) = 1 + x 1 + 4 x ,   x ≠ 0 ,   x ≠ − 1 4 ( f ∘ g ) ( x ) = 1 + x 1 + 4 x ,   x ≠ 0 ,   x ≠ − 1 4

( f ∘ g ) ( x ) = 1 x , x > 0 ( f ∘ g ) ( x ) = 1 x , x > 0

sample: g ( x ) = 2 x − 1 3 x + 4 ; f ( x ) = x g ( x ) = 2 x − 1 3 x + 4 ; f ( x ) = x

f ( x ) = | x − 3 | f ( x ) = | x − 3 |

f ( x ) = 1 2 | x + 2 | + 1 f ( x ) = 1 2 | x + 2 | + 1

f ( x ) = − 3 | x − 3 | + 3 f ( x ) = − 3 | x − 3 | + 3

f − 1 ( x ) = x - 9 10 f − 1 ( x ) = x - 9 10

f − 1 ( x ) = x - 1 f − 1 ( x ) = x - 1

The function is one-to-one.

Practice Test

The relation is a function.

The graph is a parabola and the graph fails the horizontal line test.

2 a 2 − a 2 a 2 − a

− 2 ( a + b ) + 1 − 2 ( a + b ) + 1

f − 1 ( x ) = x + 5 3 f − 1 ( x ) = x + 5 3

( − ∞ , − 1.1 ) and  ( 1.1 , ∞ ) ( − ∞ , − 1.1 ) and  ( 1.1 , ∞ )

( 1.1 , − 0.9 ) ( 1.1 , − 0.9 )

f ( 2 ) = 2 f ( 2 ) = 2

f ( x ) = { | x | if x ≤ 2 3 if x > 2 f ( x ) = { | x | if x ≤ 2 3 if x > 2

x = 2 x = 2

f − 1 ( x ) = − x − 11 2 f − 1 ( x ) = − x − 11 2

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• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: College Algebra
• Publication date: Feb 13, 2015
• Location: Houston, Texas
• Book URL: https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites
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Eureka Math Grade 3 Module 2 Lesson 12 Answer Key

Engage ny eureka math 3rd grade module 2 lesson 12 answer key, eureka math grade 3 module 2 lesson 12 problem set answer key.

Eureka Math Grade 3 Module 2 Lesson 12 Exit Ticket Answer Key

Answer: The golf ball weighs 46grams

b. Round the weight of the golf ball to the nearest ten grams. Model your thinking on the number line.

Answer: The weight 46g is between 40g and 50g The nearest tens is 50g Label the numberline and mark on the number 50.

c. The golf ball weighs about _________________.

Answer: The golf ball weighs about 50g

d. Explain how you used the halfway point on the number line to round to the nearest ten grams.

Answer: I used the half point as 45g and rounded of it to the nearest tens.As the weight of golfball is 46g it is near to 50g.

Eureka Math Grade 3 Module 2 Lesson 12 Homework Answer Key

Explanation: The Gym class ends at 10:27am 10:27 is between 10:20 and 10:30am The nearest tens is 10:30am Therefore, Gym class ends at about 10:30 a.m

Answer: The liquid in the beaker is between 20ml and 30ml As the liquid is near to 20ml So, There are about 20milliliters in the beaker.

Explanation: As shown n the picture Mrs.Santos weighs 53 kg 53 is between 50 and 60kg If we round of 53 to the nearest 10kg, the weight of Mrs.Santos wiil be about 50kg.

Answer: The chimp’s weight is 58 kilograms. The chimp weighs about 60 kilograms.

Explanation: As shown in the above picture the weight of the chimp is 58kilograms The weight of the chimp is 58 is between 50kg and 60kg If we round of to the nearest tens the weight of the chimp is 60kilograms.

Eureka Math Grade 3 Module 2 Answer Key

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Pic Answer - AI Solver: Snap, Ask, Learn! Step into a world of effortless learning with Pic Answer - AI Solver, where each photo brings you closer to understanding. This app goes beyond traditional math solvers, offering a unique blend of AI-driven solutions and interactive chat for a variety of academic subjects. Why Choose Pic Answer? • Instant Photo Answers: Just snap a picture of your homework and watch as our AI provides accurate solutions in seconds. • All-Subject AI Chat: Whether you need help in math, science, or literature, our AI chat is ready to assist with personalized guidance. • Comprehensive Learning Made Simple: Our focus isn't just on answers, but on helping you understand the concepts behind them. • Intuitive and User-Friendly: We've designed our app to be easy and enjoyable for learners of all ages and abilities. Key Features: • AI-Powered Scanner: Simply snap a picture of the problem, and our app swiftly analyzes and resolves it with precision. • Real-Time AI Chat Assistance: Engage in interactive conversations for in-depth explanations and study tips. • Broad Subject Coverage: From complex math equations to science experiments and literary analysis, we've got you covered. Your Educational Companion: • Always Evolving: We constantly enhance our app with the latest AI advancements and user feedback. • Perfect for Everyone: Whether you're a student, teacher, or just curious, our app is tailored to fit your learning needs. The app includes optional subscriptions to unlock pro features. terms and conditions: http://techconsolidated.org/terms.html Download "Pic Answer - AI Solver" today and revolutionize the way you learn!

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This app is horrible. I expected to be accurate it wasn’t. For starters, I downloaded this app so it could help me with my math. I wanted to be sure is was accurate so I took a picture of a old math problem and see if it would work. I looked at the answer key with the answer. For example, let’s say the answer from this app was 26. The answer key said it was 78. It was confusing so it has a respond feature and I typed this is not accurate. Then it gave me a complete different answer and the new answer was 12! I was puzzled. So I solved the question on my own and the answer key was correct. I took a another photo and the same answer, 12. I searched up on google the problem and it was 78. I paid $6.99 for this trash.

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It’s a good app, but most of the time you have to re-take a photo, or the answer is wrong. 40% of the time the app doesn’t want to work, or closes on me when I’m using it. It’s a good app for easy, quick, answers.. but not so much big ones. I do like that it works for questions with pictures.. and it has a specific mode for math questions. Would I recommend you download it.. yes. It is a good app, maybe just needs a couple of updates/bug fixes.

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