domain and range assignment answer key

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Course: Algebra 1   >   Unit 8

• Intervals and interval notation
• What is the domain of a function?
• What is the range of a function?
• Worked example: domain and range from graph

Domain and range from graph

• (Choice A)   − 5 ≤ x ≤ 4 ‍   A − 5 ≤ x ≤ 4 ‍  
• (Choice B)   The x ‍   -values − 5 ‍   , 1 ‍   , 3 ‍   , and 4 ‍   B The x ‍   -values − 5 ‍   , 1 ‍   , 3 ‍   , and 4 ‍  
• (Choice C)   − 6 ≤ x ≤ 7 ‍   C − 6 ≤ x ≤ 7 ‍  
• (Choice D)   The x ‍   -values − 6 ‍   , 0 ‍   , 2 ‍   , 4 ‍   , and 7 ‍   D The x ‍   -values − 6 ‍   , 0 ‍   , 2 ‍   , 4 ‍   , and 7 ‍  

Domain And Range Worksheet and Answer Key

Students will practice Identifying the Domain and Range of a Mathematical Relation as well as classify a relation as a function based on its domain and range.

Example Questions

Directions Identify the Domain and Range of the Relations below:

{ (-1, 2), (2, 51), (1, 3), (8, 22), (9, 51) } { (-5, 6), (21, -51), (11, 93), (81, 202), (19, 51) }

For the Relation below to be a Function , X cannot be what values?

{ (12, 13), (-11, 22), (33, 101), (x, 22) }

Visual Aids

• Domain and Range
• Functions and Relations in Math

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Domain and Range Worksheets

This compilation of domain and range worksheet pdfs provides 8th grade and high school students with ample practice in determining the domain or the set of possible input values (x) and range, the resultant or output values (y) using a variety of exercises with ordered pairs presented on graphs and in table format. Find the domain and range of relations from mapping diagrams, from finite and infinite graphs and more. Get started with our free worksheets.

Write the Domain and Range | Relation - Ordered Pairs

State the domain and range of each relation represented as a set of ordered pairs in Part A and ordered pairs on a graph in Part B of these printable worksheets.

• Download the set

Write the Domain and Range | Relation - Mapping

Determine the domain and range in each of the relations presented in these relation mapping worksheets for grade 8 and high school students. Observe each relation and write the domain (x) and range (y) values in set notation.

Write the Domain and Range | Relation - Table

This batch presents the ordered pairs in tables with input and output columns. Identify the domain and range and write them in ascending order for each of the tables featured in these domain and range pdf worksheets.

Write the Domain and Range | Finite Graph

Observe the extent of the graph horizontally for the domain and the vertical extent of the graph for range and write the smallest and largest values of both in this set of identifying the domain and range from finite graphs worksheets. Use apt brackets to show if the interval is open or closed.

Write the Domain and Range | Infinite Graph

Bolster skills in identifying the domain and range of functions with infinite graphs. Analyze each graph, write the minimum and maximum points for both domain and range. If there is no endpoint, then it can be concluded it is infinite.

Write the Range | Function Rule - Level 1

In this set of pdf worksheets, the function rule is expressed as a linear function and the domain is also provided in each problem. Plug in the values of x in the function rule to determine the range.

Write the Range | Function Rule - Level 2

Substitute the input values or values of the domain in the given quadratic, polynomial, reciprocal or square root functions and determine the output values or range in this section of Level 2 exercises.

Write the Domain and Range | Function - Mixed Review

Test skills acquired with this printable domain and range revision worksheets that provide a mix of absolute, square root, quadratic and reciprocal functions f(x). Determine the domain (x) and plug in the possible x-values to find the range (y).

Related Worksheets

» Identifying Functions

» Function Tables

» Evaluating Functions

» Composition of Functions

» Inverse Function

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1.2: Domain and Range

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• Page ID 1293

Learning Objectives

• Find the domain of a function defined by an equation.
• Graph piecewise-defined functions.

If you’re in the mood for a scary movie, you may want to check out one of the five most popular horror movies of all time—I am Legend, Hannibal, The Ring, The Grudge, and The Conjuring. Figure \(\PageIndex{1}\) shows the amount, in dollars, each of those movies grossed when they were released as well as the ticket sales for horror movies in general by year. Notice that we can use the data to create a function of the amount each movie earned or the total ticket sales for all horror movies by year. In creating various functions using the data, we can identify different independent and dependent variables, and we can analyze the data and the functions to determine the domain and range. In this section, we will investigate methods for determining the domain and range of functions such as these.

Finding the Domain of a Function Defined by an Equation

In Functions and Function Notation, we were introduced to the concepts of domain and range . In this section, we will practice determining domains and ranges for specific functions. Keep in mind that, in determining domains and ranges, we need to consider what is physically possible or meaningful in real-world examples, such as tickets sales and year in the horror movie example above. We also need to consider what is mathematically permitted. For example, we cannot include any input value that leads us to take an even root of a negative number if the domain and range consist of real numbers. Or in a function expressed as a formula, we cannot include any input value in the domain that would lead us to divide by 0.

We can visualize the domain as a “holding area” that contains “raw materials” for a “function machine” and the range as another “holding area” for the machine’s products (Figure \(\PageIndex{2}\)).

We can write the domain and range in interval notation , which uses values within brackets to describe a set of numbers. In interval notation, we use a square bracket [ when the set includes the endpoint and a parenthesis ( to indicate that the endpoint is either not included or the interval is unbounded. For example, if a person has $100 to spend, he or she would need to express the interval that is more than 0 and less than or equal to 100 and write \(\left(0, 100\right]\). We will discuss interval notation in greater detail later.

Let’s turn our attention to finding the domain of a function whose equation is provided. Oftentimes, finding the domain of such functions involves remembering three different forms. First, if the function has no denominator or an even root, consider whether the domain could be all real numbers. Second, if there is a denominator in the function’s equation, exclude values in the domain that force the denominator to be zero. Third, if there is an even root, consider excluding values that would make the radicand negative.

Before we begin, let us review the conventions of interval notation:

• The smallest term from the interval is written first.
• The largest term in the interval is written second, following a comma.
• Parentheses, \((\) or \()\), are used to signify that an endpoint is not included, called exclusive.
• Brackets, \([\) or \(]\), are used to indicate that an endpoint is included, called inclusive.

See Figure \(\PageIndex{3}\) for a summary of interval notation.

Example \(\PageIndex{1}\): Finding the Domain of a Function as a Set of Ordered Pairs

Find the domain of the following function: \(\{(2, 10),(3, 10),(4, 20),(5, 30),(6, 40)\}\).

First identify the input values. The input value is the first coordinate in an ordered pair. There are no restrictions, as the ordered pairs are simply listed. The domain is the set of the first coordinates of the ordered pairs.

\[\{2,3,4,5,6\} \nonumber\]

Exercse \(\PageIndex{1}\)

Find the domain of the function:

\[\{(−5,4),(0,0),(5,−4),(10,−8),(15,−12)\} \nonumber\]

\(\{−5, 0, 5, 10, 15\}\)

How To: Given a function written in equation form, find the domain.

• Identify the input values.
• Identify any restrictions on the input and exclude those values from the domain.
• Write the domain in interval form, if possible.

Example \(\PageIndex{2}\): Finding the Domain of a Function

Find the domain of the function \(f(x)=x^2−1\).

The input value, shown by the variable x in the equation, is squared and then the result is lowered by one. Any real number may be squared and then be lowered by one, so there are no restrictions on the domain of this function. The domain is the set of real numbers.

In interval form, the domain of f is \((−\infty,\infty)\).

Exercse \(\PageIndex{2}\)

\[f(x)=5−x+x^3 \nonumber\]

\((−\infty,\infty)\)

Howto: Given a function written in an equation form that includes a fraction, find the domain

• Identify any restrictions on the input. If there is a denominator in the function’s formula, set the denominator equal to zero and solve for x . If the function’s formula contains an even root, set the radicand greater than or equal to 0, and then solve.
• Write the domain in interval form, making sure to exclude any restricted values from the domain.

Example \(\PageIndex{3}\): Finding the Domain of a Function Involving a Denominator

Find the domain of the function \(f(x)=\dfrac{x+1}{2−x}\).

When there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to 0 and solve for x.

\[ \begin{align*} 2−x=0 \\[4pt] −x &=−2 \\[4pt] x&=2 \end{align*}\]

Now, we will exclude 2 from the domain. The answers are all real numbers where \(x<2\) or \(x>2\). We can use a symbol known as the union, \(\cup\),to combine the two sets. In interval notation, we write the solution:\((−\infty,2)∪(2,\infty)\).

In interval form, the domain of f is \((−\infty,2)\cup(2,\infty)\).

Exercse \(\PageIndex{3}\)

\[f(x)=\dfrac{1+4x}{2x−1} \nonumber\]

\[(−\infty,\dfrac{1}{2})\cup(\dfrac{1}{2},\infty) \nonumber\]

How To: Given a function written in equation form including an even root, find the domain.

• Since there is an even root, exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for x.
• The solution(s) are the domain of the function. If possible, write the answer in interval form.

Example \(\PageIndex{4}\): Finding the Domain of a Function with an Even Root

\[f(x)=\sqrt{7-x} \nonumber .\]

When there is an even root in the formula, we exclude any real numbers that result in a negative number in the radicand.

Set the radicand greater than or equal to zero and solve for x.

\[ \begin{align*} 7−x&≥0 \\[4pt] −x&≥−7\\[4pt] x&≤7 \end{align*}\]

Now, we will exclude any number greater than 7 from the domain. The answers are all real numbers less than or equal to 7, or \(\left(−\infty,7\right]\).

Exercse \(\PageIndex{4}\)

Find the domain of the function

\[f(x)=\sqrt{5+2x}. \nonumber\]

\[\left[−2.5,\infty\right) \nonumber\]

Q&A: Can there be functions in which the domain and range do not intersect at all?

Yes. For example, the function \(f(x)=-\dfrac{1}{\sqrt{x}}\) has the set of all positive real numbers as its domain but the set of all negative real numbers as its range. As a more extreme example, a function’s inputs and outputs can be completely different categories (for example, names of weekdays as inputs and numbers as outputs, as on an attendance chart), in such cases the domain and range have no elements in common.

Using Notations to Specify Domain and Range

In the previous examples, we used inequalities and lists to describe the domain of functions. We can also use inequalities, or other statements that might define sets of values or data, to describe the behavior of the variable in set-builder notation. For example, \(\{x|10≤x<30\}\) describes the behavior of x in set-builder notation. The braces \(\{\}\) are read as “the set of,” and the vertical bar \(|\) is read as “such that,” so we would read\( \{x|10≤x<30\}\) as “the set of x-values such that 10 is less than or equal to x, and x is less than 30.”

Figure \(\PageIndex{4}\) compares inequality notation, set-builder notation, and interval notation.

To combine two intervals using inequality notation or set-builder notation, we use the word “or.” As we saw in earlier examples, we use the union symbol, \(\cup\),to combine two unconnected intervals. For example, the union of the sets\(\{2,3,5\}\) and \(\{4,6\}\) is the set \(\{2,3,4,5,6\}\). It is the set of all elements that belong to one or the other (or both) of the original two sets. For sets with a finite number of elements like these, the elements do not have to be listed in ascending order of numerical value. If the original two sets have some elements in common, those elements should be listed only once in the union set. For sets of real numbers on intervals, another example of a union is

\[\{x| |x|≥3\}=\left(−\infty,−3\right]\cup\left[3,\infty\right)\]

Set-Builder Notation and Interval Notation

Set-builder notation is a method of specifying a set of elements that satisfy a certain condition. It takes the form\(\{x|\text{ statement about x}\}\) which is read as, “the set of all x such that the statement about x is true.” For example,

\[\{x|4<x≤12\} \nonumber\]

Interval notation is a way of describing sets that include all real numbers between a lower limit that may or may not be included and an upper limit that may or may not be included. The endpoint values are listed between brackets or parentheses. A square bracket indicates inclusion in the set, and a parenthesis indicates exclusion from the set. For example,

\[\left(4,12\right] \nonumber\]

• Identify the intervals to be included in the set by determining where the heavy line overlays the real line.
• At the left end of each interval, use [ with each end value to be included in the set (solid dot) or ( for each excluded end value (open dot).
• At the right end of each interval, use ] with each end value to be included in the set (filled dot) or ) for each excluded end value (open dot).
• Use the union symbol \(\cup\) to combine all intervals into one set.

Example \(\PageIndex{5}\): Describing Sets on the Real-Number Line

Describe the intervals of values shown in Figure \(\PageIndex{5}\) using inequality notation, set-builder notation, and interval notation.

To describe the values, \(x\), included in the intervals shown, we would say, “\(x\) is a real number greater than or equal to 1 and less than or equal to 3, or a real number greater than 5.”

\[1≤x≤3 \text{ or }x>5 \nonumber\]

Set-builder Notation

\[\{x|1≤x≤3 \text{ or } x>5\}\nonumber\]

Interval notation

\[[1,3]\cup(5,\infty)\nonumber\]

Remember that, when writing or reading interval notation, using a square bracket means the boundary is included in the set. Using a parenthesis means the boundary is not included in the set.

Exercse \(\PageIndex{5}\)

Given Figure \(\PageIndex{6}\), specify the graphed set in

• set-builder notation
• interval notation

Values that are less than or equal to –2, or values that are greater than or equal to –1 and less than 3;

\(\{x|x≤−2 or −1≤x<3\}\)

\(\left(−∞,−2\right]\cup\left[−1,3\right)\)

Finding Domain and Range from Graphs

Another way to identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis. Keep in mind that if the graph continues beyond the portion of the graph we can see, the domain and range may be greater than the visible values. See Figure \(\PageIndex{7}\).

We can observe that the graph extends horizontally from −5 to the right without bound, so the domain is \(\left[−5,∞\right)\). The vertical extent of the graph is all range values 5 and below, so the range is \(\left(−∞,5\right]\). Note that the domain and range are always written from smaller to larger values, or from left to right for domain, and from the bottom of the graph to the top of the graph for range.

Example \(\PageIndex{6A}\): Finding Domain and Range from a Graph

Find the domain and range of the function f whose graph is shown in Figure 1.2.8.

We can observe that the horizontal extent of the graph is –3 to 1, so the domain of f is \(\left(−3,1\right]\).

The vertical extent of the graph is 0 to –4, so the range is \(\left[−4,0\right)\). See Figure \(\PageIndex{9}\).

Example \(\PageIndex{6B}\): Finding Domain and Range from a Graph of Oil Production

Find the domain and range of the function f whose graph is shown in Figure \(\PageIndex{10}\).

The input quantity along the horizontal axis is “years,” which we represent with the variable t for time. The output quantity is “thousands of barrels of oil per day,” which we represent with the variable b for barrels. The graph may continue to the left and right beyond what is viewed, but based on the portion of the graph that is visible, we can determine the domain as \(1973≤t≤2008\) and the range as approximately \(180≤b≤2010\).

In interval notation, the domain is \([1973, 2008]\), and the range is about \([180, 2010]\). For the domain and the range, we approximate the smallest and largest values since they do not fall exactly on the grid lines.

Exercse \(\PageIndex{6}\)

Given Figure \(\PageIndex{11}\), identify the domain and range using interval notation.

domain =\([1950,2002]\)

range = \([47,000,000,89,000,000]\)

Yes. For example, the domain and range of the cube root function are both the set of all real numbers.

Finding Domains and Ranges of the Toolkit Functions

We will now return to our set of toolkit functions to determine the domain and range of each.

For the constant function \( f(x)=c\), the domain consists of all real numbers; there are no restrictions on the input. The only output value is the constant \(c\), so the range is the set \(\{c\}\) that contains this single element. In interval notation, this is written as \([c,c]\), the interval that both begins and ends with \(c\).

For the identity function \(f(x)=x\), there is no restriction on \(x\). Both the domain and range are the set of all real numbers.

For the absolute value function \(f(x)=|x|\), there is no restriction on \(x\). However, because absolute value is defined as a distance from 0, the output can only be greater than or equal to 0.

For the quadratic function \(f(x)=x^2\), the domain is all real numbers since the horizontal extent of the graph is the whole real number line. Because the graph does not include any negative values for the range, the range is only nonnegative real numbers.

For the cubic function \(f(x)=x^3\), the domain is all real numbers because the horizontal extent of the graph is the whole real number line. The same applies to the vertical extent of the graph, so the domain and range include all real numbers.

For the reciprocal function \(f(x)=\dfrac{1}{x}\), we cannot divide by 0, so we must exclude 0 from the domain. Further, 1 divided by any value can never be 0, so the range also will not include 0. In set-builder notation, we could also write\(\{x| x≠0\}\),the set of all real numbers that are not zero.

For the reciprocal squared function \(f(x)=\dfrac{1}{x^2}\),we cannot divide by 0, so we must exclude 0 from the domain. There is also no x that can give an output of 0, so 0 is excluded from the range as well. Note that the output of this function is always positive due to the square in the denominator, so the range includes only positive numbers.

For the square root function \(f(x)=\sqrt{x}\), we cannot take the square root of a negative real number, so the domain must be 0 or greater. The range also excludes negative numbers because the square root of a positive number \(x\) is defined to be positive, even though the square of the negative number \(−\sqrt{x}\) also gives us \(x\).

For the cube root function \(f(x)=\sqrt[3]{x}\), the domain and range include all real numbers. Note that there is no problem taking a cube root, or any odd-integer root, of a negative number, and the resulting output is negative (it is an odd function).

• Exclude from the domain any input values that result in division by zero.
• Exclude from the domain any input values that have nonreal (or undefined) number outputs.
• Use the valid input values to determine the range of the output values.
• Look at the function graph and table values to confirm the actual function behavior.

Finding the Domain and Range Using Toolkit Functions

Find the domain and range of \(f(x)=2x^3−x\).

There are no restrictions on the domain, as any real number may be cubed and then subtracted from the result.

The domain is \((−\infty,\infty)\) and the range is also \((−\infty,\infty)\).

Example \(\PageIndex{7B}\): Finding the Domain and Range

Find the domain and range of \(f(x)=\frac{2}{x+1}\).

We cannot evaluate the function at −1 because division by zero is undefined. The domain is \((−\infty,−1)\cup(−1,\infty)\). Because the function is never zero, we exclude 0 from the range. The range is \((−\infty,0)\cup(0,\infty)\).

Example \(\PageIndex{7C}\): Finding the Domain and Range

Find the domain and range of \(f(x)=2 \sqrt{x+4}\).

We cannot take the square root of a negative number, so the value inside the radical must be nonnegative.

\(x+4≥0\) when \(x≥−4\)

The domain of \(f(x)\) is \([−4,\infty)\).

We then find the range. We know that \(f(−4)=0\), and the function value increases as \(x\) increases without any upper limit. We conclude that the range of f is \(\left[0,\infty\right)\).

Figure \(\PageIndex{19}\) represents the function \(f\).

Exercise \(\PageIndex{7}\)

Find the domain and range of

\(f(x)=\sqrt{−2−x}\).

domain: \(\left(−\infty,-2\right]\)

range: \(\left[0,\infty\right)\)

Graphing Piecewise-Defined Functions

Sometimes, we come across a function that requires more than one formula in order to obtain the given output. For example, in the toolkit functions, we introduced the absolute value function \(f(x)=|x|\). With a domain of all real numbers and a range of values greater than or equal to 0, absolute value can be defined as the magnitude , or modulus , of a real number value regardless of sign. It is the distance from 0 on the number line. All of these definitions require the output to be greater than or equal to 0.

If we input 0, or a positive value, the output is the same as the input.

\[ f(x)=x \; \text{ if } \; x≥0 \nonumber \]

If we input a negative value, the output is the opposite of the input.

\[ f(x) = -x \; \text { if } \; x < 0 \nonumber \]

Because this requires two different processes or pieces, the absolute value function is an example of a piecewise function . A piecewise function is a function in which more than one formula is used to define the output over different pieces of the domain.

We use piecewise functions to describe situations in which a rule or relationship changes as the input value crosses certain “boundaries.” For example, we often encounter situations in business for which the cost per piece of a certain item is discounted once the number ordered exceeds a certain value. Tax brackets are another real-world example of piecewise functions. For example, consider a simple tax system in which incomes up to $10,000 are taxed at 10%, and any additional income is taxed at 20%. The tax on a total income S would be \(0.1S\) if \(S≤$10,000\) and \($1000+0.2(S−$10,000)\) if \(S>$10,000\).

Piecewise Function

A piecewise function is a function in which more than one formula is used to define the output. Each formula has its own domain, and the domain of the function is the union of all these smaller domains. We notate this idea like this:

\[f(x)= \begin{cases} \text{formula 1} & \text{if x is in domain 1} \\ \text{formula 2} &\text{if x is in domain 2} \\ \text{formula 3} &\text{if x is in domain 3}\end{cases} \nonumber \]

In piecewise notation, the absolute value function is

\[|x|= \begin{cases} x & \text{if $x \geq 0$} \\ -x &\text{if $x<0$} \end{cases} \nonumber \]

• Identify the intervals for which different rules apply.
• Determine formulas that describe how to calculate an output from an input in each interval.
• Use braces and if-statements to write the function.

Example \(\PageIndex{8A}\): Writing a Piecewise Function

A museum charges $5 per person for a guided tour with a group of 1 to 9 people or a fixed $50 fee for a group of 10 or more people. Write a function relating the number of people, \(n\), to the cost, \(C\).

Two different formulas will be needed. For \(n\)-values under 10, \(C=5n\). For values of n that are 10 or greater, \(C=50\).

\[C(n)= \begin{cases} 5n & \text{if $n < 10$} \\ 50 &\text{if $n\geq10$} \end{cases} \nonumber \]

The function is represented in Figure \(\PageIndex{20}\). The graph is a diagonal line from \(n=0\) to \(n=10\) and a constant after that. In this example, the two formulas agree at the meeting point where \(n=10\), but not all piecewise functions have this property.

Example \(\PageIndex{8B}\): Working with a Piecewise Function

A cell phone company uses the function below to determine the cost, C, in dollars for g gigabytes of data transfer.

\[C(g)= \begin{cases} 25 & \text{if $0<g<2$} \\ 25+10(g-2) &\text{if $g\geq2$} \end{cases} \nonumber \]

Find the cost of using 1.5 gigabytes of data and the cost of using 4 gigabytes of data.

To find the cost of using 1.5 gigabytes of data, \(C(1.5)\), we first look to see which part of the domain our input falls in. Because 1.5 is less than 2, we use the first formula.

\[C(1.5)=$25 \nonumber \]

To find the cost of using 4 gigabytes of data, C(4), we see that our input of 4 is greater than 2, so we use the second formula.

\[C(4)=25+10(4−2)=$45 \nonumber \]

The function is represented in Figure \(\PageIndex{21}\). We can see where the function changes from a constant to a shifted and stretched identity at \(g=2\). We plot the graphs for the different formulas on a common set of axes, making sure each formula is applied on its proper domain.

• Indicate on the x-axis the boundaries defined by the intervals on each piece of the domain.
• For each piece of the domain, graph on that interval using the corresponding equation pertaining to that piece. Do not graph two functions over one interval because it would violate the criteria of a function.

Example \(\PageIndex{8C}\): Graphing a Piecewise Function

Sketch a graph of the function.

\[f(x)= \begin{cases} x^2 & \text{if $x \leq 1$} \\ 3 &\text{if $1<x\leq2$} \\ x &\text{if $x>2$} \end{cases} \nonumber \]

Each of the component functions is from our library of toolkit functions, so we know their shapes. We can imagine graphing each function and then limiting the graph to the indicated domain. At the endpoints of the domain, we draw open circles to indicate where the endpoint is not included because of a less-than or greater-than inequality; we draw a closed circle where the endpoint is included because of a less-than-or-equal-to or greater-than-or-equal-to inequality.

Figure \(\PageIndex{20}\) shows the three components of the piecewise function graphed on separate coordinate systems.

(a)\( f(x)=x^2\) if \(x≤1\); (b) \(f(x)=3\) if \(1< x≤2\); (c) \(f(x)=x\) if \(x>2\)

Now that we have sketched each piece individually, we combine them in the same coordinate plane. See Figure \(\PageIndex{21}\).

Note that the graph does pass the vertical line test even at \(x=1\) and \(x=2\) because the points \((1,3)\) and \((2,2)\) are not part of the graph of the function, though \((1,1)\) and \((2, 3)\) are.

Exercise \(\PageIndex{8}\)

Graph the following piecewise function.

\[f(x)= \begin{cases} x^3 & \text{if $x < -1$} \\ -2 &\text{if $-1<x<4$} \\ \sqrt{x} &\text{if $x>4$} \end{cases} \nonumber \]

No. Each value corresponds to one equation in a piecewise formula.

Key Concepts

• The domain of a function includes all real input values that would not cause us to attempt an undefined mathematical operation, such as dividing by zero or taking the square root of a negative number.
• The domain of a function can be determined by listing the input values of a set of ordered pairs.
• The domain of a function can also be determined by identifying the input values of a function written as an equation.
• Interval values represented on a number line can be described using inequality notation, set-builder notation, and interval notation.
• For many functions, the domain and range can be determined from a graph.
• An understanding of toolkit functions can be used to find the domain and range of related functions.
• A piecewise function is described by more than one formula.
• A piecewise function can be graphed using each algebraic formula on its assigned subdomain.

1 The Numbers: Where Data and the Movie Business Meet. “Box Office History for Horror Movies.” http://www.the-numbers.com/market/genre/Horror . Accessed 3/24/2014 2 www.eia.gov/dnav/pet/hist/Lea...s=MCRFPAK2&f=A.

a method of describing a set that includes all numbers between a lower limit and an upper limit; the lower and upper values are listed between brackets or parentheses, a square bracket indicating inclusion in the set, and a parenthesis indicating exclusion

a function in which more than one formula is used to define the output

a method of describing a set by a rule that all of its members obey; it takes the form {x| statement about x}

1.1 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.

1.2 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 ]

1.3 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 ) .

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

1.5 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

1.6 Absolute Value Functions

| x − 2 | ≤ 3 | x − 2 | ≤ 3

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

f ( 0 ) = 1 , f ( 0 ) = 1 , so the graph intersects the vertical axis at ( 0 , 1 ) . ( 0 , 1 ) . f ( x ) = 0 f ( x ) = 0 when x = − 5 x = − 5 and x = 1 x = 1 so the graph intersects the horizontal axis at ( − 5 , 0 ) ( − 5 , 0 ) and ( 1 , 0 ) . ( 1 , 0 ) .

- 8 ≤ x ≤ 4 - 8 ≤ x ≤ 4

k ≤ 1 k ≤ 1 or k ≥ 7 ; k ≥ 7 ; in interval notation, this would be ( − ∞ , 1 ] ∪ [ 7 , ∞ ) ( − ∞ , 1 ] ∪ [ 7 , ∞ )

1.7 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.

a. 3; b. 5.6

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 ]

1.1 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

• ⓐ f ( − 2 ) = 14 ; f ( − 2 ) = 14 ;
• ⓑ x = 3 x = 3
• ⓐ f ( 5 ) = 10 ; f ( 5 ) = 10 ;
• ⓑ 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.

1.2 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 .

1.3 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 , 50 ) , ( − 3 , 50 ) , local minimum: ( 3 , − 50 ) ( 3 , − 50 )

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

a. –3000; b. –1250

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

1.4 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 ;
• ⓒ g ( f ) ( x ) ) = 6 x 2 − 2 ; g ( f ) ( 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

• ⓐ Text ( 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 , ∞ ) ; c. ( 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 f ( x ) = x g ( x ) = 2 x + 6 g ( x ) = 2 x + 6

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

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

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

1.5 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.

1.6 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.

First determine the boundary points by finding the solution(s) of the equation. Use the boundary points to form possible solution intervals. Choose a test value in each interval to determine which values satisfy the inequality.

| x + 4 | = 1 2 | x + 4 | = 1 2

| f ( x ) − 8 | < 0.03 | f ( x ) − 8 | < 0.03

{ 1 , 11 } { 1 , 11 }

{ - 9 4 , 13 4 } { - 9 4 , 13 4 }

{ 10 3 , 20 3 } { 10 3 , 20 3 }

{ 11 5 , 29 5 } { 11 5 , 29 5 }

{ 5 2 , 7 2 } { 5 2 , 7 2 }

No solution

{ − 57 , 27 } { − 57 , 27 }

( 0 , − 8 ) ; ( − 6 , 0 ) , ( 4 , 0 ) ( 0 , − 8 ) ; ( − 6 , 0 ) , ( 4 , 0 )

( 0 , − 7 ) ; ( 0 , − 7 ) ; no x x -intercepts

( − ∞ , − 8 ) ∪ ( 12 , ∞ ) ( − ∞ , − 8 ) ∪ ( 12 , ∞ )

− 4 3 ≤ x ≤ 4 − 4 3 ≤ x ≤ 4

( − ∞ , − 8 3 ] ∪ [ 6 , ∞ ) ( − ∞ , − 8 3 ] ∪ [ 6 , ∞ )

( − ∞ , − 8 3 ] ∪ [ 16 , ∞ ) ( − ∞ , − 8 3 ] ∪ [ 16 , ∞ )

range: [ 0 , 20 ] [ 0 , 20 ]

x - x - intercepts:

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

1.7 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

• ⓐ f ( g ( x ) ) = x f ( g ( x ) ) = x and g ( f ( x ) ) = x . g ( f ( x ) ) = x .
• ⓑ 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 )

Absolute Maximum: 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 ; ( f ∘ g ) ( x ) = 1 x + 2 ; ( g ∘ f ) ( 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

x = − 22 , x = 14 x = − 22 , x = 14

( − 5 3 , 3 ) ( − 5 3 , 3 )

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

The function is one-to-one.

The function is not 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

x = − 7 x = − 7 and x = 10 x = 10

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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Access for free at https://openstax.org/books/precalculus/pages/1-introduction-to-functions

• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: Precalculus
• Publication date: Oct 23, 2014
• Location: Houston, Texas
• Book URL: https://openstax.org/books/precalculus/pages/1-introduction-to-functions
• Section URL: https://openstax.org/books/precalculus/pages/chapter-1

© Dec 8, 2021 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.