MCV4U, Grade 12 Calculus and Vectors Reviews of Functions

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Reviews of Functions

Functions

Given two sets and , a set with elements that are ordered pairs , where is an element of and is an element of , is a relation from to . A relation from to defines a relationship between those two sets. A function is a special type of relation in which each element of the first set is related to exactly one element of the second set. The element of the first set is called the input; the element of the second set is called the output. Functions are used all the time in mathematics to describe relationships between two sets. For any function, when we know the input, the output is determined, so we say that the output is a function of the input. For example, the area of a square is determined by its side length, so we say that the area (the output) is a function of its side length (the input). The velocity of a ball thrown in the air can be described as a function of the amount of time the ball is in the air. The cost of mailing a package is a function of the weight of the package. Since functions have so many uses, it is important to have precise definitions and terminology to study them.

Definition

A function consists of a set of inputs, a set of outputs, and a rule for assigning each input to exactly one output. The set of inputs is called the domain of the function. The set of outputs is called the range of the function.

For example, consider the function , where the domain is the set of all real numbers and the rule is to square the input. Then, the input is assigned to the output . Since every nonnegative real number has a real-value square root, every nonnegative number is an element of the range of this function. Since there is no real number with a square that is negative, the negative real numbers are not elements of the range. We conclude that the range is the set of nonnegative real numbers.

For a general function with domain , we often use to denote the input and to denote the output associated with . When doing so, we refer to as the independent variable and as the dependent variable, because it depends on . Using function notation, we write , and we read this equation as “ equals of .” For the squaring function described earlier, we write .

The concept of a function can be visualized:

An image with three items. The first item is text that reads “Input, x”. An arrow points from the first item to the second item, which is a box with the label “function”. An arrow points from the second item to the third item, which is text that reads “Output, f(x)”. — **Figure 1.** A function can be visualized as an input/output device.

An image with two items. The first item is a bubble labeled domain. Within the bubble are the numbers 1, 2, 3, and 4. An arrow with the label “f” points from the first item to the second item, which is a bubble labeled “range”. Within this bubble are the numbers 2, 4, and 6. An arrow points from the 1 in the domain bubble to the 6 in the range bubble. An arrow points from the 1 in the domain bubble to the 6 in the range bubble. An arrow points from the 2 in the domain bubble to the 4 in the range bubble. An arrow points from the 3 in the domain bubble to the 2 in the range bubble. An arrow points from the 4 in the domain bubble to the 2 in the range bubble. — **Figure 2.** A function maps every element in the domain to exactly one element in the range. Although each input can be sent to only one output, two different inputs can be sent to the same output.

An image of a graph. The y axis runs from 0 to 3 and has the label “dependent variable, y = f(x)”. The x axis runs from 0 to 5 and has the label “independent variable, x”. There are three points on the graph. The first point is at (1, 2) and has the label “(1, f(1)) = (1, 2)”. The second point is at (2, 1) and has the label “(2, f(2))=(2,1)”. The third point is at (3, 2) and has the label “(3, f(3)) = (3,2)”. There is text along the y axis that reads “range = {1, 2}” and text along the x axis that reads “domain = {1,2,3}”. — **Figure 3.** In this case, a graph of a function has a domain of $\{1,2,3\}$ and a range of $\{1,2\}$ . The independent variable is and the dependent variable is .

We can also visualize a function by plotting points in the coordinate plane where . The graph of a function is the set of all these points. For example, consider the function , where the domain is the set $D=\{1,2,3\}$ and the rule is . In (Figure), we plot a graph of this function.

An image of a graph. The y axis runs from 0 to 5. The x axis runs from 0 to 5. There are three points on the graph at (1, 2), (2, 1), and (3, 0). There is text along the y axis that reads “range = {0,1,2}” and text along the x axis that reads “domain = {1,2,3}”. — **Figure 4.** Here we see a graph of the function with domain $\{1,2,3\}$ and rule . The graph consists of the points for all in the domain.

Every function has a domain. However, sometimes a function is described by an equation, as in , with no specific domain given. In this case, the domain is taken to be the set of all real numbers for which is a real number. For example, since any real number can be squared, if no other domain is specified, we consider the domain of to be the set of all real numbers. On the other hand, the square root function $f(x)=\sqrt{x}$ only gives a real output if is nonnegative. Therefore, the domain of the function $f(x)=\sqrt{x}$ is the set of nonnegative real numbers, sometimes called the natural domain.

For the functions and $f(x)=\sqrt{x}$ , the domains are sets with an infinite number of elements. Clearly we cannot list all these elements. When describing a set with an infinite number of elements, it is often helpful to use set-builder or interval notation. When using set-builder notation to describe a subset of all real numbers, denoted , we write $\{x|x \, \text{has some property}\}$ .

We read this as the set of real numbers such that has some property. For example, if we were interested in the set of real numbers that are greater than one but less than five, we could denote this set using set-builder notation by writing $\{x|1<x<5\}$ .

A set such as this, which contains all numbers greater than and less than , can also be denoted using the interval notation . Therefore, $(1,5)=\{x|1<x<5\}$ .

The numbers 1 and 5 are called the endpoints of this set. If we want to consider the set that includes the endpoints, we would denote this set by writing $[1,5]=\{x|1\le x\le 5\}$ .

We can use similar notation if we want to include one of the endpoints, but not the other. To denote the set of nonnegative real numbers, we would use the set-builder notation $\{x|0\le x\}$ .

The smallest number in this set is zero, but this set does not have a largest number. Using interval notation, we would use the symbol $\infty$ , which refers to positive infinity, and we would write the set as $[0,\infty)=\{x|0\le x\}$ .

It is important to note that $\infty$ is not a real number. It is used symbolically here to indicate that this set includes all real numbers greater than or equal to zero. Similarly, if we wanted to describe the set of all nonpositive numbers, we could write $(−\infty ,0]=\{x|x\le 0\}$ .

Here, the notation $−\infty$ refers to negative infinity, and it indicates that we are including all numbers less than or equal to zero, no matter how small. The set

(−\infty ,\infty)=\{x|x \, \text{is any real number}\}

refers to the set of all real numbers.

Some functions are defined using different equations for different parts of their domain. These types of functions are known as piecewise-defined functions. For example, suppose we want to define a function with a domain that is the set of all real numbers such that for $x\ge 2$ and for . We denote this function by writing $f(x)=\begin{cases} 3x+1, & x \ge 2 \\ x^2, & x < 2 \end{cases}$ .

When evaluating this function for an input , the equation to use depends on whether $x\ge 2$ or . For example, since , we use the fact that for $x\ge 2$ and see that . On the other hand, for , we use the fact that for and see that .

Finding Domain and Range

For each of the following functions, determine the i. domain and ii. range.

$f(x)=\sqrt{3x+2}-1$
$f(x)=\frac{3}{x-2}$

Solution

Consider .
1. Since is a real number for any real number , the domain of is the interval $(−\infty ,\infty)$ .
2. Since $(x-4)^2\ge 0$ , we know $f(x)=(x-4)^2+5\ge 5$ . Therefore, the range must be a subset of $\{y|y\ge 5\}$ . To show that every element in this set is in the range, we need to show that for a given in that set, there is a real number such that . Solving this equation for , we see that we need such that.This equation is satisfied as long as there exists a real number such that $x-4=±\sqrt{y-5}$ .Since $y\ge 5$ , the square root is well-defined. We conclude that for $x=4±\sqrt{y-5}$ , , and therefore the range is $\{y|y\ge 5\}$ .
Consider .
1. To find the domain of , we need the expression $3x+2\ge 0$ . Solving this inequality, we conclude that the domain is $\{x|x\ge -2/3\}$ .
2. To find the range of , we note that since $\sqrt{3x+2}\ge 0$ , $f(x)=\sqrt{3x+2}-1\ge -1$ . Therefore, the range of must be a subset of the set $\{y|y\ge -1\}$ . To show that every element in this set is in the range of , we need to show that for all in this set, there exists a real number in the domain such that . Let $y\ge -1$ . Then, if and only if $\sqrt{3x+2}-1=y$ .Solving this equation for , we see that must solve the equation $\sqrt{3x+2}=y+1$ .Since $y\ge -1$ , such an could exist. Squaring both sides of this equation, we have .
  Therefore, we need,which implies $x=\frac{1}{3}(y+1)^2-\frac{2}{3}$ .We just need to verify that is in the domain of . Since the domain of consists of all real numbers greater than or equal to , and $\frac{1}{3}(y+1)^2-\frac{2}{3}\ge -\frac{2}{3}$ ,there does exist an in the domain of . We conclude that the range of is $\{y|y\ge -1\}$ .
Consider .
1. Since is defined when the denominator is nonzero, the domain is $\{x|x\ne 2\}$ .
2. To find the range of , we need to find the values of such that there exists a real number in the domain with the property that $\frac{3}{x-2}=y$ .Solving this equation for , we find that $x=\frac{3}{y}+2$ .Therefore, as long as $y\ne 0$ , there exists a real number in the domain such that . Thus, the range is $\{y|y\ne 0\}$ .

Representing Functions

Typically, a function is represented using one or more of the following tools:

A table
A graph
A formula

We can identify a function in each form, but we can also use them together. For instance, we can plot on a graph the values from a table or create a table from a formula.

Tables

Functions described using a table of values arise frequently in real-world applications. Consider the following simple example. We can describe temperature on a given day as a function of time of day. Suppose we record the temperature every hour for a 24-hour period starting at midnight. We let our input variable be the time after midnight, measured in hours, and the output variable be the temperature hours after midnight, measured in degrees Fahrenheit. We record our data in (Figure).

Hours after Midnight	Temperature $(\text{°}F)$	Hours after Midnight	Temperature $(\text{°}F)$
0	58	12	84
1	54	13	85
2	53	14	85
3	52	15	83
4	52	16	82
5	55	17	80
6	60	18	77
7	64	19	74
8	72	20	69
9	75	21	65
10	78	22	60
11	80	23	58

We can see from the table that temperature is a function of time, and the temperature decreases, then increases, and then decreases again. However, we cannot get a clear picture of the behavior of the function without graphing it.

Graphs

Given a function described by a table, we can provide a visual picture of the function in the form of a graph. Graphing the temperatures listed in (Figure) can give us a better idea of their fluctuation throughout the day. (Figure) shows the plot of the temperature function.

An image of a graph. The y axis runs from 0 to 90 and has the label “Temperature in Fahrenheit”. The x axis runs from 0 to 24 and has the label “hours after midnight”. There are 24 points on the graph, one at each increment of 1 on the x-axis. The first point is at (0, 58) and the function decreases until x = 4, where the point is (4, 52) and is the minimum value of the function. After x=4, the function increases until x = 13, where the point is (13, 85) and is the maximum of the function along with the point (14, 85). After x = 14, the function decreases until the last point on the graph, which is (23, 58). — Figure 5. The graph of the data from (Table) shows temperature as a function of time.

From the points plotted on the graph in Figure 5, we can visualize the general shape of the graph. It is often useful to connect the dots in the graph, which represent the data from the table. In this example, although we cannot make any definitive conclusion regarding what the temperature was at any time for which the temperature was not recorded, given the number of data points collected and the pattern in these points, it is reasonable to suspect that the temperatures at other times followed a similar pattern, as we can see in Figure 6.

Algebraic Formulas

Sometimes we are not given the values of a function in table form, rather we are given the values in an explicit formula. Formulas arise in many applications. For example, the area of a circle of radius is given by the formula $A(r)=\pi r^2$ . When an object is thrown upward from the ground with an initial velocity ft/s, its height above the ground from the time it is thrown until it hits the ground is given by the formula . When dollars are invested in an account at an annual interest rate compounded continuously, the amount of money after years is given by the formula $A(t)=Pe^{rt}$ . Algebraic formulas are important tools to calculate function values. Often we also represent these functions visually in graph form.

Given an algebraic formula for a function , the graph of is the set of points , where is in the domain of and is in the range. To graph a function given by a formula, it is helpful to begin by using the formula to create a table of inputs and outputs. If the domain of consists of an infinite number of values, we cannot list all of them, but because listing some of the inputs and outputs can be very useful, it is often a good way to begin.

When creating a table of inputs and outputs, we typically check to determine whether zero is an output. Those values of where are called the zeros of a function. For example, the zeros of are . The zeros determine where the graph of intersects the -axis, which gives us more information about the shape of the graph of the function. The graph of a function may never intersect the -axis, or it may intersect multiple (or even infinitely many) times.

Another point of interest is the -intercept, if it exists. The -intercept is given by .

Since a function has exactly one output for each input, the graph of a function can have, at most, one -intercept. If is in the domain of a function , then has exactly one -intercept. If is not in the domain of , then has no -intercept. Similarly, for any real number , if is in the domain of , there is exactly one output , and the line intersects the graph of exactly once. On the other hand, if is not in the domain of , is not defined and the line does not intersect the graph of . This property is summarized in the vertical line test.

Rule: Vertical Line Test

Given a function , every vertical line that may be drawn intersects the graph of no more than once. If any vertical line intersects a set of points more than once, the set of points does not represent a function.

We can use this test to determine whether a set of plotted points represents the graph of a function ((Figure)).

An image of two graphs. The first graph is labeled “a” and is of the function “y = f(x)”. Three vertical lines run through 3 points on the function, each vertical line only passing through the function once. The second graph is labeled “b” and is of the relation “y not equal to f(x)”. Two vertical lines run through the relation, one line intercepting the relation at 3 points and the other line intercepting the relation at 3 different points. — **Figure 7.** (a) The set of plotted points represents the graph of a function because every vertical line intersects the set of points, at most, once. (b) The set of plotted points does not represent the graph of a function because some vertical lines intersect the set of points more than once.

Finding Zeros and -Intercepts of a Function

Consider the function .

Find all zeros of .
Find the -intercept (if any).
Sketch a graph of .

Solution

To find the zeros, solve . We discover that has one zero at .
The -intercept is given by .
Given that is a linear function of the form that passes through the points and , we can sketch the graph of ((Figure)).Figure 8. The function is a line with -intercept and -intercept .

Finding Zeros and -Intercepts of a Function

Consider the function .

Find all zeros of .
Find the -intercept (if any).
Sketch a graph of .

Solution

To find the zeros, solve . We discover that has one zero at .
The -intercept is given by .
Given that is a linear function of the form that passes through the points and , we can sketch the graph of ((Figure)).Figure 8. The function is a line with -intercept and -intercept .

Using Zeros and -Intercepts to Sketch a Graph

Consider the function $f(x)=\sqrt{x+3}+1$ .

Find all zeros of .
Find the -intercept (if any).
Sketch a graph of .

Solution

To find the zeros, solve $\sqrt{x+3}+1=0$ . This equation implies $\sqrt{x+3}=-1$ . Since $\sqrt{x+3}\ge 0$ for all , this equation has no solutions, and therefore has no zeros.
The -intercept is given by $(0,f(0))=(0,\sqrt{3}+1)$ .
To graph this function, we make a table of values. Since we need $x+3\ge 0$ , we need to choose values of $x\ge -3$ . We choose values that make the square-root function easy to evaluate.
$\mathbf{x}$ -3-21 $\mathbf{f(x)}$ 123

Making use of the table and knowing that, since the function is a square root, the graph of should be similar to the graph of $y=\sqrt{x}$ , we sketch the graph ((Figure)).

An image of a graph. The y axis runs from -2 to 4 and the x axis runs from -3 to 2. The graph is of the function “f(x) = (square root of x + 3) + 1”, which is an increasing curved function that starts at the point (-3, 1). There are 3 points plotted on the function at (-3, 1), (-2, 2), and (1, 3). The function has a y intercept at (0, 1 + square root of 3). — **Figure 9.** The graph of $f(x)=\sqrt{x+3}+1$ has a -intercept but no -intercepts.

**Figure 9.** The graph of $f(x)=\sqrt{x+3}+1$ has a -intercept but no -intercepts.

Find the zeros of .

Solution

Hint

Factor the polynomial.

Finding the Height of a Free-Falling Object

If a ball is dropped from a height of 100 ft, its height at time is given by the function , where is measured in feet and is measured in seconds. The domain is restricted to the interval , where is the time when the ball is dropped and is the time when the ball hits the ground.

Create a table showing the height when and . Using the data from the table, determine the domain for this function. That is, find the time when the ball hits the ground.
Sketch a graph of .

Solution

Height as a Function of Time $\mathbf{t}$ 00.511.522.5 $\mathbf{s(t)}$ 100968464360Since the ball hits the ground when , the domain of this function is the interval .

Note that for this function and the function graphed in (Figure), the values of are getting smaller as is getting larger. A function with this property is said to be decreasing. On the other hand, for the function $f(x)=\sqrt{x+3}+1$ graphed in (Figure), the values of are getting larger as the values of are getting larger. A function with this property is said to be increasing. It is important to note, however, that a function can be increasing on some interval or intervals and decreasing over a different interval or intervals. For example, using our temperature function in (Figure), we can see that the function is decreasing on the interval , increasing on the interval , and then decreasing on the interval . We make the idea of a function increasing or decreasing over a particular interval more precise in the next definition.

Definition

We say that a function is increasing on the interval if for all $x_1, x_2\in I$ , $f(x_1)\le f(x_2) \, \text{when} \, x_1<x_2$ .

We say is strictly increasing on the interval if for all $x_1,x_2\in I$ , $f(x_1)<f(x_2) \, \text{when} \, x_1<x_2$ .

We say that a function is decreasing on the interval if for all $x_1, x_2\in I$ , $f(x_1)\ge f(x_2) \, \text{if} \, x_1<x_2$ .

We say that a function is strictly decreasing on the interval if for all $x_1, x_2 \in I$ , $f(x_1)>f(x_2) \, \text{if} \, x_1<x_2$ .

For example, the function is increasing on the interval $(−\infty ,\infty)$ because whenever . On the other hand, the function is decreasing on the interval $(−\infty ,\infty)$ because ((Figure)).

An image of two graphs. The first graph is labeled “a” and is of the function “f(x) = 3x”, which is an increasing straight line that passes through the origin. The second graph is labeled “b” and is of the function “f(x) = -x cubed”, which is curved function that decreases until the function hits the origin where it becomes level, then decreases again after the origin. — **Figure 10.** (a) The function is increasing on the interval $(−\infty ,\infty)$ . (b) The function is decreasing on the interval $(−\infty ,\infty)$ .

f(x)=3x — **Figure 10.** (a) The function is increasing on the interval $(−\infty ,\infty)$ . (b) The function is decreasing on the interval $(−\infty ,\infty)$ .

Combining Functions

Now that we have reviewed the basic characteristics of functions, we can see what happens to these properties when we combine functions in different ways, using basic mathematical operations to create new functions. For example, if the cost for a company to manufacture items is described by the function and the revenue created by the sale of items is described by the function , then the profit on the manufacture and sale of items is defined as . Using the difference between two functions, we created a new function.

Alternatively, we can create a new function by composing two functions. For example, given the functions and , the composite function $f\circ g$ is defined such that $(f\circ g)(x)=f(g(x))=(g(x))^2=(3x+1)^2$ .

The composite function $g\circ f$ is defined such that $(g\circ f)(x)=g(f(x))=3f(x)+1=3x^2+1$ .

Note that these two new functions are different from each other.

Combining Functions with Mathematical Operators

To combine functions using mathematical operators, we simply write the functions with the operator and simplify. Given two functions and we can define four new functions:

\begin{array}{cccc}(f+g)(x)=f(x)+g(x)\hfill & & & \text{Sum}\hfill \\ (f-g)(x)=f(x)-g(x)\hfill & & & \text{Difference}\hfill \\ (f·g)(x)=f(x)g(x)\hfill & & & \text{Product}\hfill \\ \Big(\frac{f}{g}\Big)(x)=\frac{f(x)}{g(x)} \, \text{for} \, g(x)\ne 0\hfill & & & \text{Quotient}\hfill \end{array}

Combining Functions Using Mathematical Operations

Given the functions and , find each of the following functions and state its domain.

$\Big(\frac{f}{g}\Big)(x)$

Solution

. The domain of this function is the interval $(−\infty ,\infty )$ .
. The domain of this function is the interval $(−\infty ,\infty)$ .
. The domain of this function is the interval $(−\infty ,\infty )$ .
$\Big(\frac{f}{g}\Big)(x)=\frac{2x-3}{x^2-1}$ . The domain of this function is $\{x|x\ne \text{±}1\}$ .

For and , find and state its domain.

Solution

$\Big(\frac{f}{g}\Big)(x)=\frac{x^2+3}{2x-5}$ . The domain is $\{x|x\ne \frac{5}{2}\}$ .

Hint

The new function is a quotient of two functions. For what values of is the denominator zero?

Function Composition

When we compose functions, we take a function of a function. For example, suppose the temperature on a given day is described as a function of time (measured in hours after midnight) as in (Figure). Suppose the cost , to heat or cool a building for 1 hour, can be described as a function of the temperature . Combining these two functions, we can describe the cost of heating or cooling a building as a function of time by evaluating . We have defined a new function, denoted $C\circ T$ , which is defined such that $(C\circ T)(t)=C(T(t))$ for all in the domain of . This new function is called a composite function. We note that since cost is a function of temperature and temperature is a function of time, it makes sense to define this new function $(C\circ T)(t)$ . It does not make sense to consider $(T\circ C)(t)$ , because temperature is not a function of cost.

Definition

Consider the function with domain and range , and the function with domain and range . If is a subset of , then the composite function $(g\circ f)(x)$ is the function with domain such that $(g\circ f)(x)=g(f(x))$ .

A composite function $g\circ f$ can be viewed in two steps. First, the function maps each input in the domain of to its output in the range of . Second, since the range of is a subset of the domain of , the output is an element in the domain of , and therefore it is mapped to an output in the range of . In (Figure), we see a visual image of a composite function.

An image with three items. The first item is a blue bubble that has two labels: “domain of f” and “domain of g of f”. This item contains the numbers 1, 2, and 3. The second item is two bubbles: an orange bubble labeled “domain of g” and a blue bubble that is completely contained within the orange bubble and is labeled “range of f”. The blue bubble contains the numbers 0 and 1, which are thus also contained within the larger orange bubble. The orange bubble contains two numbers not contained within the smaller blue bubble, which are 2 and 3. The third item is two bubbles: an orange bubble labeled “range of g” and a blue bubble that is completely contained within the orange bubble and is labeled “range of g of f”. The blue bubble contains the numbers 4 and 5, which are thus also contained within the larger orange bubble. The orange bubble contains one number not contained within the smaller blue bubble, which is the number 3. The first item points has a blue arrow with the label “f” that points to the blue bubble in the second item. The orange bubble in the second item has an orange arrow labeled “g” that points the orange bubble in the third item. The first item has a blue arrow labeled “g of f” which points to the blue bubble in the third item. There are three blue arrows pointing from numbers in the first item to the numbers contained in the blue bubble of the second item. The first blue arrow points from the 1 to the 0, the second blue arrow points from the 2 to the 1, and the third blue arrow points from the 3 to the 0. There are 4 orange arrows pointing from the numbers contained in the orange bubble in the second item, including those also contained in the blue bubble of the second item, to the numbers contained in the orange bubble of the third item, including the numbers in the blue bubble of the third item. The first orange arrow points from 2 to 3, the second orange arrow points from 3 to 5, the third orange arrow points from 0 to 4, and the fourth orange arrow points from 1 to 5. — **Figure 11.** For the composite function $g\circ f$ , we have $(g\circ f)(1)=4, \, (g\circ f)(2)=5,$ and $(g\circ f)(3)=4$ .

Compositions of Functions Defined by Formulas

Consider the functions and .

Find $(g\circ f)(x)$ and state its domain and range.
Evaluate $(g\circ f)(4)$ and $(g\circ f)(-1/2)$ .
Find $(f\circ g)(x)$ and state its domain and range.
Evaluate $(f\circ g)(4)$ and $(f\circ g)(-1/2)$ .

Solution

We can find the formula for $(g\circ f)(x)$ in two different ways. We could write $(g\circ f)(x)=g(f(x))=g(x^2+1)=\frac{1}{x^2+1}$ .Alternatively, we could write $(g\circ f)(x)=g(f(x))=\frac{1}{f(x)}=\frac{1}{x^2+1}$ .Since $x^2+1\ne 0$ for all real numbers , the domain of $(g\circ f)(x)$ is the set of all real numbers. Since $0<1/(x^2+1)\le 1$ , the range is, at most, the interval . To show that the range is this entire interval, we let and solve this equation for to show that for all in the interval , there exists a real number such that . Solving this equation for , we see that , which implies that $x=±\sqrt{\frac{1}{y}-1}$ .If is in the interval , the expression under the radical is nonnegative, and therefore there exists a real number such that . We conclude that the range of $g\circ f$ is the interval .
$(g\circ f)(4)=g(f(4))=g(4^2+1)=g(17)=\frac{1}{17}$
$(g\circ f)(-\frac{1}{2})=g(f(-\frac{1}{2}))=g((-\frac{1}{2})^2+1)=g(\frac{5}{4})=\frac{4}{5}$
We can find a formula for $(f\circ g)(x)$ in two ways. First, we could write $(f\circ g)(x)=f(g(x))=f(\frac{1}{x})=(\frac{1}{x})^2+1$ .Alternatively, we could write $(f\circ g)(x)=f(g(x))=(g(x))^2+1=(\frac{1}{x})^2+1$ .The domain of $f\circ g$ is the set of all real numbers such that $x\ne 0$ . To find the range of , we need to find all values for which there exists a real number $x\ne 0$ such that $(\frac{1}{x})^2+1=y$ .Solving this equation for , we see that we need to satisfy $(\frac{1}{x})^2=y-1$ ,which simplifies to $\frac{1}{x}=±\sqrt{y-1}$ .Finally, we obtain $x=±\frac{1}{\sqrt{y-1}}$ .Since $1/\sqrt{y-1}$ is a real number if and only if is the set $\{y|y\ge 1\}$ .
$(f\circ g)(4)=f(g(4))=f(\frac{1}{4})=(\frac{1}{4})^2+1=\frac{17}{16}$
$(f\circ g)(-\frac{1}{2})=f(g(-\frac{1}{2}))=f(-2)=(-2)^2+1=5$

In (Figure), we can see that $(f\circ g)(x)\ne (g\circ f)(x)$ . This tells us, in general terms, that the order in which we compose functions matters.

Let Let $g(x)=\sqrt{x}.$ Find $(f\circ g)(x)$ .

Solution

$(f\circ g)(x)=2-5\sqrt{x}$ .

Composition of Functions Defined by Tables

Consider the functions and described by (Figure) and (Figure).

$\mathbf{x}$	-3	-2	-1	0	1	2	3	4
$\mathbf{f(x)}$	0	4	2	4	-2	0	-2	4

$\mathbf{x}$	-4	-2	0	2	4
$\mathbf{g(x)}$	1	0	3	0	5

Evaluate $(g\circ f)(3)$ and $(g\circ f)(0)$ .
State the domain and range of $(g\circ f)(x)$ .
Evaluate $(f\circ f)(3)$ and $(f\circ f)(1)$ .
State the domain and range of $(f\circ f)(x)$ .

Solution

$(g\circ f)(3)=g(f(3))=g(-2)=0$
$(g\circ f)(0)=g(4)=5$
The domain of $g\circ f$ is the set $\{-3,-2,-1,0,1,2,3,4\}$ . Since the range of is the set $\{-2,0,2,4\}$ , the range of $g\circ f$ is the set $\{0,3,5\}$ .
$(f\circ f)(3)=f(f(3))=f(-2)=4$
$(f\circ f)(1)=f(f(1))=f(-2)=4$
The domain of $f\circ f$ is the set $\{-3,-2,-1,0,1,2,3,4\}$ . Since the range of is the set $\{-2,0,2,4\}$ , the range of $f\circ f$ is the set $\{0,4\}$ .

Application Involving a Composite Function

A store is advertising a sale of $20\%$ off all merchandise. Caroline has a coupon that entitles her to an additional $15\%$ off any item, including sale merchandise. If Caroline decides to purchase an item with an original price of dollars, how much will she end up paying if she applies her coupon to the sale price? Solve this problem by using a composite function.

Solution

Since the sale price is $20\%$ off the original price, if an item is dollars, its sale price is given by . Since the coupon entitles an individual to $15\%$ off the price of any item, if an item is dollars, the price, after applying the coupon, is given by . Therefore, if the price is originally dollars, its sale price will be and then its final price after the coupon will be .

If items are on sale for $10\%$ off their original price, and a customer has a coupon for an additional $30\%$ off, what will be the final price for an item that is originally dollars, after applying the coupon to the sale price?

Solution

Hint

The sale price of an item with an original price of dollars is . The coupon price for an item that is dollars is .

Symmetry of Functions

The graphs of certain functions have symmetry properties that help us understand the function and the shape of its graph. For example, consider the function shown in (Figure)(a). If we take the part of the curve that lies to the right of the -axis and flip it over the -axis, it lays exactly on top of the curve to the left of the -axis. In this case, we say the function has symmetry about the -axis. On the other hand, consider the function shown in (Figure)(b). If we take the graph and rotate it 180° about the origin, the new graph will look exactly the same. In this case, we say the function has symmetry about the origin.

An image of two graphs. The first graph is labeled “(a), symmetry about the y-axis” and is of the curved function “f(x) = (x to the 4th) - 2(x squared) - 3”. The x axis runs from -3 to 4 and the y axis runs from -4 to 5. This function decreases until it hits the point (-1, -4), which is minimum of the function. Then the graph increases to the point (0,3), which is a local maximum. Then the the graph decreases until it hits the point (1, -4), before it increases again. The second graph is labeled “(b), symmetry about the origin” and is of the curved function “f(x) = x cubed - 4x”. The x axis runs from -3 to 4 and the y axis runs from -4 to 5. The graph of the function starts at the x intercept at (-2, 0) and increases until the approximate point of (-1.2, 3.1). The function then decreases, passing through the origin, until it hits the approximate point of (1.2, -3.1). The function then begins to increase again and has another x intercept at (2, 0). — **Figure 12.** (a) A graph that is symmetric about the -axis. (b) A graph that is symmetric about the origin.

If we are given the graph of a function, it is easy to see whether the graph has one of these symmetry properties. But without a graph, how can we determine algebraically whether a function has symmetry? Looking at (Figure) again, we see that since is symmetric about the -axis, if the point is on the graph, the point is on the graph. In other words, . If a function has this property, we say is an even function, which has symmetry about the -axis. For example, is even because.

In contrast, looking at (Figure) again, if a function is symmetric about the origin, then whenever the point is on the graph, the point is also on the graph. In other words, . If has this property, we say is an odd function, which has symmetry about the origin. For example, is odd because.

Definition

If for all in the domain of , then is an even function. An even function is symmetric about the -axis.

If for all in the domain of , then is an odd function. An odd function is symmetric about the origin.

Even and Odd Functions

Determine whether each of the following functions is even, odd, or neither.

$f(x)=\large{\frac{3x}{x^2+1}}$

Solution

To determine whether a function is even or odd, we evaluate and compare it to and .

. Therefore, is even.
. Now, $f(−x)\ne f(x)$ . Furthermore, noting that , we see that $f(−x)\ne −f(x)$ . Therefore, is neither even nor odd.
. Therefore, is odd.

Determine whether is even, odd, or neither.

Solution

is odd.

Hint

Compare with and .

One symmetric function that arises frequently is the absolute value function, written as . The absolute value function is defined as

f(x)=\begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases}

Some students describe this function by stating that it “makes everything positive.” By the definition of the absolute value function, we see that if , then . Therefore, it is more accurate to say that for all nonzero inputs, the output is positive, but if , the output . We conclude that the range of the absolute value function is $\{y|y\ge 0\}$ . In (Figure), we see that the absolute value function is symmetric about the -axis and is therefore an even function.

An image of a graph. The x axis runs from -3 to 3 and the y axis runs from -4 to 4. The graph is of the function “f(x) = absolute value of x”. The graph starts at the point (-3, 3) and decreases in a straight line until it hits the origin. Then the graph increases in a straight line until it hits the point (3, 3). — **Figure 13.** The graph of is symmetric about the -axis.

f(x)=|x| — **Figure 13.** The graph of is symmetric about the -axis.

Working with the Absolute Value Function

Find the domain and range of the function .

Solution

Since the absolute value function is defined for all real numbers, the domain of this function is $(−\infty ,\infty )$ . Since $|x-3|\ge 0$ for all , the function $f(x)=2|x-3|+4\ge 4$ . Therefore, the range is, at most, the set $\{y|y\ge 4\}$ . To see that the range is, in fact, this whole set, we need to show that for $y\ge 4$ there exists a real number such that.

A real number satisfies this equation as long as $|x-3|=\frac{1}{2}(y-4)$ .

Since $y\ge 4$ , we know $y-4\ge 0$ , and thus the right-hand side of the equation is nonnegative, so it is possible that there is a solution. Furthermore, $|x-3|= \begin{cases} x-3, & \text{if} \, x \ge 3 \\ -(x-3), & \text{if} \, x < 3 \end{cases}$ .

Therefore, we see there are two solutions: $x=±\frac{1}{2}(y-4)+3$ .

The range of this function is $\{y|y\ge 4\}$ .

For the function , find the domain and range.

Solution

Domain = $(−\infty ,\infty )$ , range = $\{y|y\ge -4\}$ .

Hint

$|x+2|\ge 0$ for all real numbers .

Key Concepts

A function is a mapping from a set of inputs to a set of outputs with exactly one output for each input.
If no domain is stated for a function , the domain is considered to be the set of all real numbers for which the function is defined.
When sketching the graph of a function , each vertical line may intersect the graph, at most, once.
A function may have any number of zeros, but it has, at most, one -intercept.
To define the composition $g\circ f$ , the range of must be contained in the domain of .
Even functions are symmetric about the -axis whereas odd functions are symmetric about the origin.

Key Equations

Composition of two functions
$(g\circ f)(x)=g(f(x))$
Absolute value function
$f(x) = |x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases}$