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• Module 1: Algebra Essentials
• Module 2: Polynomial and Rational Expressions
• Module 3: The Rectangular Coordinate System and Equations of Lines
• Module 4: Equations and Inequalities
• Module 5: Function Basics
• Module 6: Algebraic Operations on Functions
• Module 7: Linear and Absolute Value Functions
• Module 8: Quadratic Functions
• Module 9: Power and Polynomial Functions
• Module 10: Rational and Radical Functions
• Module 11: Exponential and Logarithmic Functions
• Module 12: Exponential and Logarithmic Equations and Models
• Module 13: Systems of Equations and Inequalities
• Module 14: Solve Systems With Matrices
• Module 15: Conic Sections
• Module 16: Sequences and Series
• Module 17: Probability and Counting Principles

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4.1: Introduction to Functions

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Our development of the function concept is a modern one, but quite quick, particularly in light of the fact that today’s definition took over 300 years to reach its present state. We begin with the definition of a relation.

We use the notation (2, 4) to denote what is called an ordered pair. If you think of the positions taken by the ordered pairs (4, 2) and (2, 4) in the coordinate plane (see Figure \(\PageIndex{1}\)), then it is immediately apparent why order is important. The ordered pair (4, 2) is simply not the same as the ordered pair (2, 4).

Figure \(\PageIndex{1}\)

The first element of an ordered pair is called its abscissa. The second element of an ordered pair is called its ordinate. Thus, for example, the abscissa of (4, 2) is 4, while the ordinate of (4, 2) is 2.

A collection of ordered pairs is called a relation . (2)

For example, the collection of ordered pairs \[R=\{(0,1),(0,2),(3,4)\} \nonumber \] is a relation.

The domain of a relation is the collection of all abscissas of each ordered pair.

Thus, the domain of the relation R in (2) is \[\text { Domain }=\{0,3\} \nonumber \]

Note that we list each abscissa only once.

The range of a relation is the collection of all ordinates of each ordered pair.

Thus, the range of the relation R in (2) is \[\text { Range }=\{1,2,4\} \nonumber \]

Let’s look at an example.

Example \(\PageIndex{1}\)

Consider the relation T defined by \[T=\{(1,2),(3,2),(4,5)\} \nonumber \]

What are the domain and range of this relation?

The domain is the collection of abscissas of each ordered pair. Hence, the domain of T is \[\text { Domain }=\{1,3,4\} \nonumber \]

The range is the collection of ordinates of each ordered pair. Hence, the range of T is \[\text { Range }=\{2,5\} \nonumber \]

Note that we list each ordinate in the range only once.

In Example \(\PageIndex{1}\), the relation is described by listing the ordered pairs. This is not the only way that one can describe a relation. For example, a graph certainly represents a collection of ordered pairs.

Example \(\PageIndex{2}\)

Consider the graph of the relation S shown in Figure \(\PageIndex{2}\).

Figure \(\PageIndex{2}\) The graph of a relation.

What are the domain and range of the relation S?

There are five ordered pairs (points) plotted in Figure \(\PageIndex{2}\). They are \[S=\{(1,2),(2,1),(2,4),(3,3),(4,4)\} \nonumber \]

Therefore, the relation S has Domain = {1, 2, 3, 4} and Range = {1, 2, 3, 4}. In the case of the range, note how we’ve sorted the ordinates of each ordered pair in ascending order, taking care not to list any ordinate more than once.

A function is a very special type of relation. We begin with a formal definition.

A relation is a function if and only if each object in its domain is paired with one and only one object in its range.

This is not an easy definition, so let’s take our time and consider a few examples. Consider, if you will, the relation R in (2), repeated here again for convenience.

\[R=\{(0,1),(0,2),(3,4)\} \nonumber \]

The domain is {0, 3} and the range is {1, 2, 4}. Note that the number 0 in the domain of R is paired with two numbers from the range, namely, 1 and 2. Therefore, R is not a function.

There is a construct, called a mapping diagram, which can be helpful in determining whether a relation is a function. To craft a mapping diagram, first list the domain on the left, then the range on the right, then use arrows to indicate the ordered pairs in your relation, as shown in Figure \(\PageIndex{3}\).

Figure \(\PageIndex{3}\) A mapping diagram for R.

It’s clear from the mapping diagram in Figure \(\PageIndex{3}\) that the number 0 in the domain is being paired (mapped) with two different range objects, namely, 1 and 2. Thus, R is not a function.

Let’s look at another example.

Example \(\PageIndex{3}\)

Is the relation described in Example \(\PageIndex{1}\) a function?

First, let’s repeat the listing of the relation T here for convenience.

\[T=\{(1,2),(3,2),(4,5)\} \nonumber \]

Next, construct a mapping diagram for the relation T. List the domain on the left, the range on the right, then use arrows to indicate the pairings, as shown in Figure \(\PageIndex{4}\).

From the mapping diagram in Figure \(\PageIndex{4}\), we can see that each domain object on the left is paired (mapped) with exactly one range object on the right. Hence, the relation T is a function.

Figure \(\PageIndex{4}\). A mapping diagram for T.

Example \(\PageIndex{4}\)

Is the relation of Example \(\PageIndex{2}\), pictured in Figure \(\PageIndex{2}\), a function?

First, we repeat the graph of the relation from Example \(\PageIndex{2}\) here for convenience. This is shown in Figure \(\PageIndex{5}\)(a). Next, we list the ordered pairs of the relation S.

\[S=\{(1,2),(2,1),(2,4),(3,3),(4,4)\} \nonumber \]

Then we create a mapping diagram by first listing the domain on the left, the range on the right, then using arrows to indicate the pairings, as shown in Figure \(\PageIndex{5}\)(b).

Figure \(\PageIndex{5}\) A graph of the relation S and its corresponding mapping diagram

Each object in the domain of S gets mapped to exactly one range object with one exception. The domain object 2 is paired with two range objects, namely, 1 and 4. Consequently, S is not a function.

This is a good point to summarize what we’ve learned about functions thus far.

A function consists of three parts:

• a set of objects which mathematicians call the domain ,
• a second set of objects which mathematicians call the range ,
• and a rule that describes how to assign a unique range object to each object in the domain.

The rule can take many forms. For example, we can use sets of ordered pairs, graphs, and mapping diagrams to describe the function. In the sections that follow, we will explore other ways of describing a function, for example, through the use of equations and simple word descriptions.

Function Notation

We’ve used the word “mapping” several times in the previous examples. This is not a word to be taken lightly; it is an important concept. In the case of the mapping diagram in Figure \(\PageIndex{5}\)(b), we would say that the number 1 in the domain of S is “mapped” (or “sent”) to the number 2 in the range of S.

There are a number of different notations we could use to indicate that the number 1 in the domain is “mapped” or “sent” to the number 2 in the range. One possible notation is

\[S : 1 \longrightarrow 2 \nonumber \]

which we would read as follows: “The relation S maps (sends) 1 to 2.” In a similar vein, we see in Figure \(\PageIndex{5}\)(b) that the domain objects 3 and 4 are mapped (sent) to the range objects 3 and 4, respectively. In symbols, we would write

\[\begin{array}{l}{S : 3 \longrightarrow 3, \text { and }} \\ {S : 4 \longrightarrow 4}\end{array} \nonumber \]

A difficulty arises when we examine what happens to the domain object 2. There are two possibilities, either

\[S : 2 \longrightarrow 1 \nonumber \] or \[S : 2 \longrightarrow 4 \nonumber \]

Which should we choose? The 1? Or the 4? Thus, S is not well-defined and is not a function, since we don’t know which range object to pair with the domain object 1.

The idea of mapping gives us an alternative way to describe a function. We could say that a function is a rule that assigns a unique object in its range to each object in its domain. Take for example, the function that maps each real number to its square. If we name the function f, then f maps 5 to 25, 6 to 36, −7 to 49, and so on. In symbols, we would write

\[f : 5 \longrightarrow 25, \quad f : 6 \longrightarrow 36, \quad \text { and } \quad f :-7 \longrightarrow 49 \nonumber \]

In general, we could write

\[f : x \longrightarrow x^{2} \nonumber \]

Note that each real number x gets mapped to a unique number in the range of f, namely, \(x^{2}\). Consequently, the function f is well defined. We’ve succeeded in writing a rule that completely defines the function f.

As another example, let’s define a function that takes a real number, doubles it, then adds 3. If we name the function g, then g would take the number 7, double it, then add 3. That is,

\[g : 7 \longrightarrow 2(7)+3 \nonumber \]

Simplifying, \(g : 7 \longrightarrow 17\). Similarly, g would take the number 9, double it, then add 3. That is,

\[g : 9 \longrightarrow 2(9)+3 \nonumber \]

Simplifying, \(g : 9 \longrightarrow 21\). In general, g takes a real number x, doubles it, then adds three. In symbols, we would write

\[g : x \longrightarrow 2 x+3 \nonumber \]

Notice that each real number x is mapped by g to a unique number in its range. Therefore, we’ve again defined a rule that completely defines the function g.

It is helpful to think of a function as a machine. The machine receives input, processes it according to some rule, then outputs a result. Something goes in (input), then something comes out (output). In the case of the function described by the rule \(f : x \longrightarrow x^{2}\), the “f-machine” receives input x, then applies its “square rule” to the input and outputs \(x^{2}\), as shown in Figure \(\PageIndex{6}\)(a). In the case of the function described by the rule \(g : x \longrightarrow 2 x+3\), the “g-machine” receives input x, then applies the rules “double,” then “add 3,” in that order, then outputs \(2x + 3\), as shown in Figure \(\PageIndex{6}\)(b).

Figure \(\PageIndex{6}\) Function machines.

Example \(\PageIndex{5}\)

Suppose that f is defined by the following rule. For each real number x,

\[f : x \longrightarrow x^{2}-2 x-3 \nonumber \]

Where does f map the number −3? Is f a function?

We substitute −3 for x in the rule for f and obtain

\[f :-3 \longrightarrow(-3)^{2}-2(-3)-3 \nonumber \]

Simplifying,

\[f :-3 \longrightarrow 9+6-3 \nonumber \]

\[f :-3 \longrightarrow 12 \nonumber \]

Thus, f maps (sends) the number −3 to the number 12. It should be clear that each real number x will be mapped (sent) to a unique real number, as defined by the rule \(f : x \longrightarrow x^{2}-2 x-3\). Therefore, f is a function.

Example \(\PageIndex{6}\)

Suppose that g is defined by the following rule. For each real number x that is greater than or equal to zero,

\[g : x \longrightarrow \pm \sqrt{x} \nonumber \]

Where does g map the number 4? Is g a function?

Again, we substitute 4 for x in the rule for g and obtain

\[g : 4 \longrightarrow \pm \sqrt{4} \nonumber \]

\[g : 4 \longrightarrow \pm 2 \nonumber \]

Thus, g maps (sends) the number 4 to two different objects in its range, namely, 2 and −2. Consequently, g is not well-defined and is not a function.

Let’s look at another example

Example \(\PageIndex{7}\)

Suppose that we have functions f and g, defined by

\[f : x \longrightarrow x^{4}+11 \quad \text { and } \quad g : x \longrightarrow(x+2)^{2} \nonumber \]

Where does g send 5?

In this example, we see a clear advantage of function notation. Because our functions have distinct names, we can simply reference the name of the function we want our readers to use. In this case, we are asked where the function g sends the number 5, so we substitute 5 for x in

\[g : x \longrightarrow(x+2)^{2} \nonumber \]

\[g : 5 \longrightarrow(5+2)^{2} \nonumber \]

Simplifying, \(g : 5 \longrightarrow 49\).

Modern Notation

Function notation is relatively new, with some of the earliest symbolism first occurring in the 17th century. In a letter to Leibniz (1698), Johann Bernoulli wrote “For denoting any function of a variable quantity x, I rather prefer to use the capital letter having the same name X or the Greek \(\xi\), for it appears at once of what variable it is a function; this relieves the memory.”

Mathematicians are fond of the notation \[f : x \longrightarrow x^{2}-2 x \nonumber \]

because it conveys a sense of what a function does; namely, it “maps” or “sends” the number x to the number \(x^{2}-2 x\). This is what functions do, they pair each object in their domain with a unique object in their range. Equivalently, functions “send” each object in their domain to a unique object in their range.

However, in common computational situations, the “arrow” notation can be a bit clumsy, so mathematicians tend to favor a slightly different notation. Instead of writing

\[f : x \longrightarrow x^{2}-2 x \nonumber \]

mathematicians tend to favor the notation

\[f(x)=x^{2}-2 x \nonumber \]

It is important to understand from the outset that these two different notations are equivalent; they represent the same function f, one that pairs each real number x in its domain with the real number \(x^{2}-2 x\) in its range.

The first notation, \(f : x \longrightarrow x^{2}-2 x\), conveys the sense that the function f is a mapping. If we read this notation aloud, we should pronounce it as “f sends (or maps) x to \(x^{2}-2 x\).” The second notation, \(f(x) = x^{2}-2 x\), is pronounced “f of x equals \(x^{2}-2 x\).”

The phrase “f of x” is unfortunate, as our readers might recall being trained from a very early age to pair the word “of” with the operation of multiplication. For example, 1/2 of 12 is 6, as in \(1 / 2 \times 12=6\). However, in the context of function notation, even though f(x) is read aloud as “f of x,” it does not mean “f times x.” Indeed, if we remind ourselves that the notation \(f(x)=x^{2}-2 x\) is equivalent to the notation \(f : x \longrightarrow x^{2}-2 x\), then even though we might say “f of x,” we should be thinking “f sends x” or “f maps x.” We should not be thinking “f times x.”

Now, let’s see how each of these notations operates on the number 5. In the first case, using the “arrow” notation,

To find where f sends 5, we substitute 5 for x as follows.

\[f : 5 \longrightarrow(5)^{2}-2(5) \nonumber \]

Simplifying,\(f : 5 \longrightarrow 15\). Now, because both notations are equivalent, to compute f(5), we again substitute 5 for x in

\[f(5)=(5)^{2}-2(5) \nonumber \]

Simplifying, \(f(5)=15\). This result is read aloud as “f of 5 equals 15,” but we want to be thinking “f sends 5 to 15.”

Let’s look at examples that use this modern notation.

Example \(\PageIndex{8}\)

Given \(f(x)=x^{3}+3 x^{2}-5,\) determine \(f(-2)\)

Simply substitute −2 for x. That is,

\[\begin{aligned} f(-2) &=(-2)^{3}+3(-2)^{2}-5 \\ &=-8+3(4)-5 \\ &=-8+12-5 \\ &=-1 \end{aligned} \nonumber \]

Thus, \(f(−2) = −1\). Again, even though this is pronounced “f of −2 equals −1,” we still should be thinking “f sends −2 to −1.”

Example \(\PageIndex{9}\)

Given \[f(x)=\frac{x+3}{2 x-5} \nonumber \] determine f(6).

Simply substitute 6 for x. That is, \[\begin{aligned} f(6) &=\frac{6+3}{2(6)-5} \\ &=\frac{9}{12-5} \\ &=\frac{9}{7} \end{aligned} \nonumber \]

Thus, \(f(6) = 9/7\). Again, even though this is pronounced “f of 6 equals 9/7,” we should still be thinking “f sends 6 to 9/7.”

Example \(\PageIndex{10}\)

Given \(f(x)=5 x-3,\) determine \(f(a+2)\).

If we’re thinking in terms of mapping notation, then \[f : x \longrightarrow 5 x-3 \nonumber \]

Think of this mapping as a “machine.” Whatever we put into the machine, it first is multiplied by 5, then 3 is subtracted from the result, as shown in Figure \(\PageIndex{7}\). For example, if we put a 4 into the machine, then the function rule requires that we multiply input 4 by 5, then subtract 3 from the result. That is,

\[f : 4 \longrightarrow 5(4)-3 \nonumber \]

Simplifying, \(f : 4 \longrightarrow 17\)

Figure \(\PageIndex{7}\). The multiply by 5 then subtract 3 machine.

Similarly, if we put an a + 2 into the machine, then the function rule requires that we multiply the input a + 2 by 5, then subtract 3 from the result. That is,

\[f : a+2 \longrightarrow 5(a+2)-3 \nonumber \]

Using modern function notation, we would write

\[f(a+2)=5(a+2)-3 \nonumber \]

Note that this is again a simple substitution, where we replace each occurrence of x in the formula \(f(x) = 5x − 3\) with the expression a + 2. Finally, use the distributive property to first multiply by 5, then subtract 3.

\[\begin{aligned} f(a+2) &=5 a+10-3 \\ &=5 a+7 \end{aligned} \nonumber \]

We will often need to substitute the result of one function evaluation into a second function for evaluation. Let’s look at an example.

Example \(\PageIndex{11}\)

Given two functions defined by \(f(x) = 3x + 2\) and \(g(x) = 5 − 4x\), find f(g(2)).

The nested parentheses in the expression f(g(2)) work in the same manner that they do with nested expressions. The rule is to work the innermost grouping symbols first, proceeding outward as you work. We’ll first evaluate g(2), then evaluate f at the result.

We begin. First, evaluate g(2) by substituting 2 for x in the defining equation \(g(x) = 5 − 4x\). Note that \(g(2) = 5 − 4(2)\), then simplify.

\[f(g(2))=f(5-4(2))=f(5-8)=f(-3) \nonumber \]

To complete the evaluation, we substitute −3 for x in the defining equation \(f(x) = 3x + 2\), then simplify.

\[f(-3)=3(-3)+2=-9+2=-7 \nonumber \]

Hence, \(f(g(2))=-7\).

It is conventional to arrange the work in one contiguous block, as follows.

\[\begin{aligned} f(g(2)) &=f(5-4(2)) \\ &=f(-3) \\ &=3(-3)+2 \\ &=-7 \end{aligned} \nonumber \]

You can shorten the task even further if you are willing to do the function substitution and simplification in your head. First, evaluate g at 2, then f at the result.

\[f(g(2))=f(-3)=-7 \nonumber \]

Let’s look at another example of this unique way of combining functions.

Example \(\PageIndex{12}\)

Given \(f(x) = 5x + 2\) and \(g(x) = 3 − 2x\), evaluate \(g(f(a))\) and simplify the result.

We work the inner function evaluation in the expression \(g(f(a))\) first. Thus, to evaluate f(a), we substitute a for x in the definition \(f(x) = 5x + 2\) to get

\[g(f(a))=g(5 a+2) \nonumber \]

Now we need to evaluate \(g(5a + 2)\). To do this, we substitute \(5a + 2\) for x in the definition \(g(x) = 3 − 2x\) to get

\[g(5 a+2)=3-2(5 a+2) \nonumber \]

We can expand this last result and simplify. Thus,

\[g(f(a))=3-10 a-4=-10 a-1 \nonumber \]

Again, it is conventional to arrange the work in one continuous block, as follows.

\[\begin{aligned} g(f(a)) &=g(5 a+2) \\ &=3-2(5 a+2) \\ &=3-10 a-4 \\ &=-10 a-1 \end{aligned} \nonumber \]

Hence, \(g(f(a))=-10 a-1\).

Extracting the Domain of a Function

We’ve seen that the domain of a relation or function is the set of all the first coordinates of its ordered pairs. However, if a functional relationship is defined by an equation such as \(f(x) = 3x − 4\), then it is not practical to list all ordered pairs defined by this relationship. For any real x-value, you get an ordered pair. For example, if x = 5, then \(f(5) = 3(5) − 4 = 11\), leading to the ordered pair (5, f(5)) or (5, 11). As you can see, the number of such ordered pairs is infinite. For each new x-value, we get another function value and another ordered pair.

Therefore, it is easier to turn our attention to the values of x that yield real number responses in the equation \(f(x) = 3x − 4\). This leads to the following key idea.

If a function is defined by an equation, then the domain of the function is the set of “permissible x-values,” the values that produce a real number response defined by the equation.

We sometimes like to say that the domain of a function is the set of “OK x-values to use in the equation.” For example, if we define a function with the rule \(f(x) = 3x − 4\), it is immediately apparent that we can use any value we want for x in the rule \(f(x) = 3x − 4\). Thus, the domain of f is all real numbers. We can write that the domain \(D=\mathbb{R}\), or we can use interval notation and write that the domain \(D=(-\infty, \infty)\).

It is not the case that x can be any real number in the function defined by the rule \(f(x)=\sqrt{x}\). It is not possible to take the square root of a negative number.2 Therefore, x must either be zero or a positive real number. In set-builder notation, we can describe the domain with \(D=\{x : x \geq 0\}\). In interval notation, we write \(D=[0, \infty)\).

We must also be aware of the fact that we cannot divide by zero. If we define a function with the rule \(f(x)=x /(x-3)\), we immediately see that x = 3 will put a zero in the denominator. Division by zero is not defined. Therefore, 3 is not in the domain of f. No other x-value will cause a problem. The domain of f is best described with set-builder notation as \(D=\{x : x \neq 3\}\).

Functions Without Formulae

In the previous section, we defined functions by means of a formula, for example, as in

\[f(x)=\frac{x+3}{2-3 x} \nonumber \]

Euler would be pleased with this definition, for as we have said previously, Euler thought of functions as analytic expressions.

However, it really isn’t necessary to provide an expression or formula to define a function. There are other forms we can use to express a functional relationship: a graph, a table, or even a narrative description. The only thing that is really important is the requirement that the function be well-defined, and by “well-defined,” we mean that each object in the function’s domain is paired with one and only one object in its range.

As an example, let’s look at a special function \(\pi\) on the natural numbers,3 which returns the number of primes less than or equal to a given natural number. For example, the primes less than or equal to the number 23 are 2, 3, 5, 7, 11, 13, 17, 19, and 23, nine numbers in all. Therefore, the number of primes less than or equal to 23 is nine. In symbols, we would write

\[\pi(23)=9 \nonumber \]

Note the absence of a formula in the definition of this function. Indeed, the definition is descriptive in nature, so we might write

\[\pi(n)=\text { number of primes less than or equal to } n \nonumber \]

The important thing is not how we define this special function \(π\), but the fact that it is well-defined; that is, for each natural number \(n\), there are a fixed number of primes less than or equal to \(n\). Thus, each natural number in the domain of \(π\) is paired with one and only one number in its range.

Now, just because our function doesn’t provide an expression for calculating the number of primes less than or equal to a given natural number n, it doesn’t stop mathematicians from seeking such a formula. Euclid of Alexandria (325-265 BC), a Greek mathematician, proved that the number of primes is infinite, but it was the German mathematician and scientist, Johann Carl Friedrich Gauss (1777-1855), who first proposed that the number of primes less than or equal to n can be approximated by the formula

\[\pi(n) \approx \frac{n}{\ln n} \nonumber \]

where ln n is the “natural logarithm” of n (to be explained in Chapter 9). This approximation gets better and better with larger and larger values of n. The formula was refined by Gauss, who did not provide a proof, and the problem became known as the Prime Number Theorem. It was not until 1896 that Jacques Salomon Hadamard (1865-1963) and Charles Jean Gustave Nicolas Baron de la Vallee Poussin (1866-1962), working independently, provided a proof of the Prime Number Theorem.

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Course: Algebra 1   >   Unit 8

What is a function.

• Worked example: Evaluating functions from equation
• Evaluate functions
• Worked example: Evaluating functions from graph
• Evaluating discrete functions
• Evaluate functions from their graph
• Worked example: evaluating expressions with function notation
• Evaluate function expressions

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On this page

• Introduction to Geometry
• Functions and Graphs

1. Introduction to Functions

• 2. Functions from Verbal Statements
• 3. Rectangular Coordinates
• 4. The Graph of a Function
• 4a. Domain and Range of a Function
• 4b. Domain and Range interactive applet
• 4c. Comparison calculator BMI - BAI
• 5. Graphing Using a Computer Algebra System
• 5a. Online graphing calculator (1): Plot your own graph (JSXGraph)
• 5b. Online graphing calculator (2): Plot your own graph (SVG)
• 6. Graphs of Functions Defined by Tables of Data
• 7. Continuous and Discontinuous Functions
• 8. Split Functions
• 9. Even and Odd Functions

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Get help with your math queries:

In everyday life, many quantities depend on one or more changing variables. For example:

(a) Plant growth depends on sunlight and rainfall

(b) Speed depends on distance travelled and time taken

(c) Voltage depends on current and resistance

(d) Test marks depend on attitude, listening in lectures and doing tutorials (among many other variables!!)

A function is a rule that relates how one quantity depends on other quantities.

A particular electrical circuit has a power source and an 8 ohms (Ω) resistor. The voltage in that circult is given by:

V = voltage (in volts, V) I = current (in amperes, A)

So if I = 4 amperes, then the voltage is V = 8 × 4 = 32 volts.

If I increases, so does the voltage, V .

If I decreases, so does the voltage, V .

We say voltage is a function of current (when resistance is constant). We get only one value of V for each value of I .

A bicycle covers a distance in 20 seconds. The speed of the bicycle is given by

`s=d/20=0.05d`

s = speed (in ms −1 , or meters per second, m/s) d = distance (in meters, m)

If the distance covered by the bike is 10 m, then the speed is `s = 0.05 × 10 = 0.5\ "m/s"`.

If d increases, the speed goes up .

If d decreases, the speed goes down .

We say speed is a function of distance (when time is constant). We get only one value of s for each value of d .

Definition of a Function

We have 2 quantities (called "variables") and we observe there is a relationship between them. If we find that for every value of the first variable there is only one value of the second variable, then we say:

"The second variable is a function of the first variable."

The first variable is the independent variable (usually written as x ), and the second variable is the dependent variable (usually written as y ).

The independent variable and the dependent variable are real numbers . (We'll learn about numbers which are not real later, in Complex Numbers .)

We know the equation for the area, A , of a circle from primary school:

A = πr 2 , where r is the radius of the circle

This is a function as each value of the independent variable r gives us one value of the dependent variable A .

General Cases

We use x for the independent variable and y for the dependent variable for general cases. This is very common in math. Please realize these general quantities can represent millions of relationships between real quantities.

In the equation

`y = 3x + 1`,

y is a function of x , since for each value of x , there is only one value of y .

If we substitute `x = 5`, we get `y = 16` and no other value.

The values of y we get depend on the values chosen for x .

Therefore, x is the independent variable and y is the dependent variable.

The force F required to accelerate an object of mass 5 kg by an acceleration of a ms -2 is given by: `F = 5a`.

Here, F is a function of the acceleration, a.

The dependent variable is F and the independent variable is a .

Function Notation

We normally write functions as: `f(x)` and read this as "function f of x ".

We can use other letters for functions, like g ( x ) or y ( x ).

When we are solving real problems, we use meaningful letters like

P ( t ) for power at time t , F ( t ) for force at time t , h ( x ) for height of an object, x horizontal units from a fixed point.

We often come across functions like: y = 2 x 2 + 5 x + 3 in math.

We can write this using function notation:

y = f ( x ) = 2 x 2 + 5 x + 3

Function notation is all about substitution.

The value of this function f ( x ) when `x = 0` is written as `f(0)`. We calculate its value by substituting as follows:

f (0) = 2(0) 2 + 5(0) + 3 = 0 + 0 + 3 = 3

Function Notation: In General

In general, the value of any function f ( x ) when x = a is written as f ( a ).

If we have `f(x) = 4x + 10`, the value of `f(x)` for `x = 3` is written:

`f(3) = 4 × 3 + 10 = 22`

In other words, when `x = 3`, the value of the function f ( x ) is `22`.

Mathematical Notation

Mathematics is often confusing because of the way it is written.

We write `5(10)` and it means `5 × 10= 50`.

But if we write `a(10)`, this could mean, depending on the situation,

"function a of `10`" (that is, the value of the function a when the independent variable is `10`)

Or it could mean multiplication, as in:

`a × 10 = 10a`.

You have to be careful with this.

Also, be careful when substituting letters or expressions into functions.

See a discussion on this: Towards more meaningful math notation .

This example involves some fixed constant, d .

If `h(x) = dx^3+ 5x` then value of `h(x)` for `x = 10` is:

`h(10) = d(10)^3+ 5(10)` `= 1000d + 50`

We leave the d there because we don't know anything about its value.

This example involves the value of a function when the independent variable contains a constant.

If the height of an object at time t is given by

h ( t ) = 10 t 2 − 2 t , then

a. The height at time `t = 4` is

h (4) = 10(4) 2 − 2(4) = 10 ×16 − 8 = 152

b. The height at time t = b is

h ( b ) = 10 b 2 − 2 b

c. The height at time `t = 3b` is

h (3 b ) = 10(3 b ) 2 − 2(3 b ) = 10 × 9 b 2 − 6 b = 90 b 2 − 6 b

d. The height at time `t = b + 1` is

h ( b + 1) = 10( b + 1) 2 − 2( b + 1) = 10 × ( b 2 + 2 b + 1) − 2 b − 2 = 10 b 2 + 20 b + 10 − 2 b − 2 = 10 b 2 + 18 b + 8

Evaluate the following functions:

(1) Given `f(x) = 3x + 20`, find

a. `f(-4)` b. `f(10)`

a. f (−4)

= 3(−4) + 20 = −12 + 20 = 8

= 3(10) + 20 = 30 + 20 = 50

(2) Given that the height of a particular object at time t is

h ( t ) = 50 t − 4.9 t 2 , find

a. `h(2)` b. `h(5)`

= 50(2) − 4.9(2) 2 = 100 − 19.6 = 80.4

= 50(5) − 4.9(5) 2 = 250 − 122.5 = 127.5

(3) The voltage, V , in a particular circuit is a function of time t , and is given by:

V ( t ) = 3 t − 1.02 t

Find the voltage at time

a. `t = 4` b. `t = c + 10`

= 3(4) − 1.02 4 = 12 − 1.08243216 = 10.9175678

b. V ( c + 10)

= 3( c + 10) − 1.02 c + 10 = 3 c + 30 − 1.02 c + 10

(4) If F ( t ) = 3 t − t 2 for t ≤ 2 , find F (2) and F (3) .

F ( t ) = 3 t − t 2

F (2) = 3(2) − (2) 2

= 6 − 4

F (3) is not defined since `t ≤ 2`.

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1.1 Review of Functions

Learning objectives.

• 1.1.1 Use functional notation to evaluate a function.
• 1.1.2 Determine the domain and range of a function.
• 1.1.3 Draw the graph of a function.
• 1.1.4 Find the zeros of a function.
• 1.1.5 Recognize a function from a table of values.
• 1.1.6 Make new functions from two or more given functions.
• 1.1.7 Describe the symmetry properties of a function.

In this section, we provide a formal definition of a function and examine several ways in which functions are represented—namely, through tables, formulas, and graphs. We study formal notation and terms related to functions. We also define composition of functions and symmetry properties. Most of this material will be a review for you, but it serves as a handy reference to remind you of some of the algebraic techniques useful for working with functions.

Given two sets A A and B , B , a set with elements that are ordered pairs ( x , y ) , ( x , y ) , where x x is an element of A A and y y is an element of B , B , is a relation from A A to B . B . A relation from A A to B B 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.

A function f f 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 f , f , where the domain is the set of all real numbers and the rule is to square the input. Then, the input x = 3 x = 3 is assigned to the output 3 2 = 9 . 3 2 = 9 . 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 f f with domain D , D , we often use x x to denote the input and y y to denote the output associated with x . x . When doing so, we refer to x x as the independent variable and y y as the dependent variable , because it depends on x . x . Using function notation, we write y = f ( x ) , y = f ( x ) , and we read this equation as “ y “ y equals f f of x . ” x . ” For the squaring function described earlier, we write f ( x ) = x 2 . f ( x ) = x 2 .

The concept of a function can be visualized using Figure 1.2 , Figure 1.3 , and Figure 1.4 .

Visit this applet link to see more about graphs of functions.

We can also visualize a function by plotting points ( x , y ) ( x , y ) in the coordinate plane where y = f ( x ) . y = f ( x ) . The graph of a function is the set of all these points. For example, consider the function f , f , where the domain is the set D = { 1 , 2 , 3 } D = { 1 , 2 , 3 } and the rule is f ( x ) = 3 − x . f ( x ) = 3 − x . In Figure 1.5 , we plot a graph of this function.

Every function has a domain. However, sometimes a function is described by an equation, as in f ( x ) = x 2 , f ( x ) = x 2 , with no specific domain given. In this case, the domain is taken to be the set of all real numbers x x for which f ( x ) f ( x ) is a real number. For example, since any real number can be squared, if no other domain is specified, we consider the domain of f ( x ) = x 2 f ( x ) = x 2 to be the set of all real numbers. On the other hand, the square root function f ( x ) = x f ( x ) = x only gives a real output if x x is nonnegative. Therefore, the domain of the function f ( x ) = x f ( x ) = x is the set of nonnegative real numbers, sometimes called the natural domain .

For the functions f ( x ) = x 2 f ( x ) = x 2 and f ( x ) = x , f ( x ) = 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 ℝ , ℝ , we write

We read this as the set of real numbers x x such that x x 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

A set such as this, which contains all numbers greater than a a and less than b , b , can also be denoted using the interval notation ( a , b ) . ( a , b ) . Therefore,

The numbers 1 1 and 5 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

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

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 ∞ , ∞ , which refers to positive infinity, and we would write the set as

It is important to note that ∞ ∞ 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

Here, the notation − ∞ − ∞ 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

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 f f with a domain that is the set of all real numbers such that f ( x ) = 3 x + 1 f ( x ) = 3 x + 1 for x ≥ 2 x ≥ 2 and f ( x ) = x 2 f ( x ) = x 2 for x < 2 . x < 2 . We denote this function by writing

When evaluating this function for an input x , x , the equation to use depends on whether x ≥ 2 x ≥ 2 or x < 2 . x < 2 . For example, since 5 > 2 , 5 > 2 , we use the fact that f ( x ) = 3 x + 1 f ( x ) = 3 x + 1 for x ≥ 2 x ≥ 2 and see that f ( 5 ) = 3 ( 5 ) + 1 = 16 . f ( 5 ) = 3 ( 5 ) + 1 = 16 . On the other hand, for x = −1 , x = −1 , we use the fact that f ( x ) = x 2 f ( x ) = x 2 for x < 2 x < 2 and see that f ( −1 ) = 1 . f ( −1 ) = 1 .

Example 1.1

Evaluating functions.

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

• f ( −2 ) f ( −2 )
• f ( 2 ) f ( 2 )
• f ( a + h ) f ( a + h )

Substitute the given value for x in the formula for f ( x ) . f ( x ) .

• f ( −2 ) = 3 ( −2 ) 2 + 2 ( −2 ) − 1 = 12 − 4 − 1 = 7 f ( −2 ) = 3 ( −2 ) 2 + 2 ( −2 ) − 1 = 12 − 4 − 1 = 7
• f ( 2 ) = 3 ( 2 ) 2 + 2 2 − 1 = 6 + 2 2 − 1 = 5 + 2 2 f ( 2 ) = 3 ( 2 ) 2 + 2 2 − 1 = 6 + 2 2 − 1 = 5 + 2 2
• f ( a + h ) = 3 ( a + h ) 2 + 2 ( a + h ) − 1 = 3 ( a 2 + 2 a h + h 2 ) + 2 a + 2 h − 1 = 3 a 2 + 6 a h + 3 h 2 + 2 a + 2 h − 1 f ( a + h ) = 3 ( a + h ) 2 + 2 ( a + h ) − 1 = 3 ( a 2 + 2 a h + h 2 ) + 2 a + 2 h − 1 = 3 a 2 + 6 a h + 3 h 2 + 2 a + 2 h − 1

Checkpoint 1.1

For f ( x ) = x 2 − 3 x + 5 , f ( x ) = x 2 − 3 x + 5 , evaluate f ( 1 ) f ( 1 ) and f ( a + h ) . f ( a + h ) .

Example 1.2

Finding domain and range.

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

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

Checkpoint 1.2

Find the domain and range for f ( x ) = 4 − 2 x + 5 . f ( x ) = 4 − 2 x + 5 .

Representing Functions

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

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.

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 x x be the time after midnight, measured in hours, and the output variable y y be the temperature x x hours after midnight, measured in degrees Fahrenheit. We record our data in Table 1.1 .

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.

Given a function f f described by a table, we can provide a visual picture of the function in the form of a graph. Graphing the temperatures listed in Table 1.1 can give us a better idea of their fluctuation throughout the day. Figure 1.6 shows the plot of the temperature function.

From the points plotted on the graph in Figure 1.6 , 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 1.7 .

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 r r is given by the formula A ( r ) = π r 2 . A ( r ) = π r 2 . When an object is thrown upward from the ground with an initial velocity v 0 v 0 ft/s, its height above the ground from the time it is thrown until it hits the ground is given by the formula s ( t ) = −16 t 2 + v 0 t . s ( t ) = −16 t 2 + v 0 t . When P P dollars are invested in an account at an annual interest rate r r compounded continuously, the amount of money after t t years is given by the formula A ( t ) = P e r t . A ( t ) = P e r t . 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 f , f , the graph of f f is the set of points ( x , f ( x ) ) , ( x , f ( x ) ) , where x x is in the domain of f f and f ( x ) f ( x ) 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 f f 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 x x where f ( x ) = 0 f ( x ) = 0 are called the zeros of a function . For example, the zeros of f ( x ) = x 2 − 4 f ( x ) = x 2 − 4 are x = ± 2 . x = ± 2 . The zeros determine where the graph of f f intersects the x x -axis, which gives us more information about the shape of the graph of the function. The graph of a function may never intersect the x -axis, or it may intersect multiple (or even infinitely many) times.

Another point of interest is the y y -intercept, if it exists. The y y -intercept is given by ( 0 , f ( 0 ) ) . ( 0 , f ( 0 ) ) .

Since a function has exactly one output for each input, the graph of a function can have, at most, one y y -intercept. If x = 0 x = 0 is in the domain of a function f , f , then f f has exactly one y y -intercept. If x = 0 x = 0 is not in the domain of f , f , then f f has no y y -intercept. Similarly, for any real number c , c , if c c is in the domain of f , f , there is exactly one output f ( c ) , f ( c ) , and the line x = c x = c intersects the graph of f f exactly once. On the other hand, if c c is not in the domain of f , f ( c ) f , f ( c ) is not defined and the line x = c x = c does not intersect the graph of f . f . This property is summarized in the vertical line test .

Rule: Vertical Line Test

Given a function f , f , every vertical line that may be drawn intersects the graph of f f 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 1.8 ).

Example 1.3

Finding zeros and y y -intercepts of a function.

Consider the function f ( x ) = −4 x + 2 . f ( x ) = −4 x + 2 .

• Find all zeros of f . f .
• Find the y y -intercept (if any).
• Sketch a graph of f . f .
• To find the zeros, solve f ( x ) = −4 x + 2 = 0 . f ( x ) = −4 x + 2 = 0 . We discover that f f has one zero at x = 1 / 2 . x = 1 / 2 .
• The y y -intercept is given by ( 0 , f ( 0 ) ) = ( 0 , 2 ) . ( 0 , f ( 0 ) ) = ( 0 , 2 ) .

Example 1.4

Using zeros and y y -intercepts to sketch a graph.

Consider the function f ( x ) = x + 3 + 1 . f ( x ) = x + 3 + 1 .

• To find the zeros, solve x + 3 + 1 = 0 . x + 3 + 1 = 0 . This equation implies x + 3 = −1 . x + 3 = −1 . Since x + 3 ≥ 0 x + 3 ≥ 0 for all x , x , this equation has no solutions, and therefore f f has no zeros.
• The y y -intercept is given by ( 0 , f ( 0 ) ) = ( 0 , 3 + 1 ) . ( 0 , f ( 0 ) ) = ( 0 , 3 + 1 ) .

Making use of the table and knowing that, since the function is a square root, the graph of f f should be similar to the graph of y = x , y = x , we sketch the graph ( Figure 1.10 ).

Checkpoint 1.3

Find the zeros of f ( x ) = x 3 − 5 x 2 + 6 x . f ( x ) = x 3 − 5 x 2 + 6 x .

Example 1.5

Finding the height of a free-falling object.

If a ball is dropped from a height of 100 100 ft, its height s s at time t t is given by the function s ( t ) = −16 t 2 + 100 , s ( t ) = −16 t 2 + 100 , where s s is measured in feet and t t is measured in seconds. The domain is restricted to the interval [ 0 , c ] , [ 0 , c ] , where t = 0 t = 0 is the time when the ball is dropped and t = c t = c is the time when the ball hits the ground.

• Create a table showing the height s ( t ) s ( t ) when t = 0 , 0.5 , 1 , 1.5 , 2 , and 2.5 . t = 0 , 0.5 , 1 , 1.5 , 2 , and 2.5 . Using the data from the table, determine the domain for this function. That is, find the time c c when the ball hits the ground.
• Sketch a graph of s . s .

Since the ball hits the ground when t = 2.5 , t = 2.5 , the domain of this function is the interval [ 0 , 2.5 ] . [ 0 , 2.5 ] .

Note that for this function and the function f ( x ) = −4 x + 2 f ( x ) = −4 x + 2 graphed in Figure 1.9 , the values of f ( x ) f ( x ) are getting smaller as x x is getting larger. A function with this property is said to be decreasing. On the other hand, for the function f ( x ) = x + 3 + 1 f ( x ) = x + 3 + 1 graphed in Figure 1.10 , the values of f ( x ) f ( x ) are getting larger as the values of x x 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 1.6 , we can see that the function is decreasing on the interval ( 0 , 4 ) , ( 0 , 4 ) , increasing on the interval ( 4 , 14 ) , ( 4 , 14 ) , and then decreasing on the interval ( 14 , 23 ) . ( 14 , 23 ) . We make the idea of a function increasing or decreasing over a particular interval more precise in the next definition.

We say that a function f f is increasing on the interval I I if for all x 1 , x 2 ∈ I , x 1 , x 2 ∈ I ,

We say f f is strictly increasing on the interval I I if for all x 1 , x 2 ∈ I , x 1 , x 2 ∈ I ,

We say that a function f f is decreasing on the interval I I if for all x 1 , x 2 ∈ I , x 1 , x 2 ∈ I ,

We say that a function f f is strictly decreasing on the interval I I if for all x 1 , x 2 ∈ I , x 1 , x 2 ∈ I ,

For example, the function f ( x ) = 3 x f ( x ) = 3 x is increasing on the interval ( − ∞ , ∞ ) ( − ∞ , ∞ ) because 3 x 1 < 3 x 2 3 x 1 < 3 x 2 whenever x 1 < x 2 . x 1 < x 2 . On the other hand, the function f ( x ) = − x 3 f ( x ) = − x 3 is decreasing on the interval ( − ∞ , ∞ ) ( − ∞ , ∞ ) because − x 1 3 > − x 2 3 − x 1 3 > − x 2 3 whenever x 1 < x 2 x 1 < x 2 ( Figure 1.11 ).

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 x x items is described by the function C ( x ) C ( x ) and the revenue created by the sale of x x items is described by the function R ( x ) , R ( x ) , then the profit on the manufacture and sale of x x items is defined as P ( x ) = R ( x ) − C ( x ) . P ( x ) = R ( x ) − C ( x ) . 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 f ( x ) = x 2 f ( x ) = x 2 and g ( x ) = 3 x + 1 , g ( x ) = 3 x + 1 , the composite function f ∘ g f ∘ g is defined such that

The composite function g ∘ f g ∘ f is defined such that

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 f f and g , g , we can define four new functions:

Example 1.6

Combining functions using mathematical operations.

Given the functions f ( x ) = 2 x − 3 f ( x ) = 2 x − 3 and g ( x ) = x 2 − 1 , g ( x ) = x 2 − 1 , find each of the following functions and state its domain.

• ( f + g ) ( x ) ( f + g ) ( x )
• ( f − g ) ( x ) ( f − g ) ( x )
• ( f · g ) ( x ) ( f · g ) ( x )
• ( f g ) ( x ) ( f g ) ( x )
• ( f + g ) ( x ) = ( 2 x − 3 ) + ( x 2 − 1 ) = x 2 + 2 x − 4 . ( f + g ) ( x ) = ( 2 x − 3 ) + ( x 2 − 1 ) = x 2 + 2 x − 4 . The domain of this function is the interval ( − ∞ , ∞ ) . ( − ∞ , ∞ ) .
• ( f − g ) ( x ) = ( 2 x − 3 ) − ( x 2 − 1 ) = − x 2 + 2 x − 2 . ( f − g ) ( x ) = ( 2 x − 3 ) − ( x 2 − 1 ) = − x 2 + 2 x − 2 . The domain of this function is the interval ( − ∞ , ∞ ) . ( − ∞ , ∞ ) .
• ( f · g ) ( x ) = ( 2 x − 3 ) ( x 2 − 1 ) = 2 x 3 − 3 x 2 − 2 x + 3 . ( f · g ) ( x ) = ( 2 x − 3 ) ( x 2 − 1 ) = 2 x 3 − 3 x 2 − 2 x + 3 . The domain of this function is the interval ( − ∞ , ∞ ) . ( − ∞ , ∞ ) .
• ( f g ) ( x ) = 2 x − 3 x 2 − 1 . ( f g ) ( x ) = 2 x − 3 x 2 − 1 . The domain of this function is { x | x ≠ ± 1 } . { x | x ≠ ± 1 } .

Checkpoint 1.4

For f ( x ) = x 2 + 3 f ( x ) = x 2 + 3 and g ( x ) = 2 x − 5 , g ( x ) = 2 x − 5 , find ( f / g ) ( x ) ( f / g ) ( x ) and state its domain.

Function Composition

When we compose functions, we take a function of a function. For example, suppose the temperature T T on a given day is described as a function of time t t (measured in hours after midnight) as in Table 1.1 . Suppose the cost C , C , to heat or cool a building for 1 hour, can be described as a function of the temperature T . T . Combining these two functions, we can describe the cost of heating or cooling a building as a function of time by evaluating C ( T ( t ) ) . C ( T ( t ) ) . We have defined a new function, denoted C ∘ T , C ∘ T , which is defined such that ( C ∘ T ) ( t ) = C ( T ( t ) ) ( C ∘ T ) ( t ) = C ( T ( t ) ) for all t t in the domain of T . T . 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 ∘ T ) ( t ) . ( C ∘ T ) ( t ) . It does not make sense to consider ( T ∘ C ) ( t ) , ( T ∘ C ) ( t ) , because temperature is not a function of cost.

Consider the function f f with domain A A and range B , B , and the function g g with domain D D and range E . E . If B B is a subset of D , D , then the composite function ( g ∘ f ) ( x ) ( g ∘ f ) ( x ) is the function with domain A A such that

A composite function g ∘ f g ∘ f can be viewed in two steps. First, the function f f maps each input x x in the domain of f f to its output f ( x ) f ( x ) in the range of f . f . Second, since the range of f f is a subset of the domain of g , g , the output f ( x ) f ( x ) is an element in the domain of g , g , and therefore it is mapped to an output g ( f ( x ) ) g ( f ( x ) ) in the range of g . g . In Figure 1.12 , we see a visual image of a composite function.

Example 1.7

Compositions of functions defined by formulas.

Consider the functions f ( x ) = x 2 + 1 f ( x ) = x 2 + 1 and g ( x ) = 1 / x . g ( x ) = 1 / x .

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

In Example 1.7 , we can see that ( f ∘ g ) ( x ) ≠ ( g ∘ f ) ( x ) . ( f ∘ g ) ( x ) ≠ ( g ∘ f ) ( x ) . This tells us, in general terms, that the order in which we compose functions matters.

Checkpoint 1.5

Let f ( x ) = 2 − 5 x . f ( x ) = 2 − 5 x . Let g ( x ) = x . g ( x ) = x . Find ( f ∘ g ) ( x ) . ( f ∘ g ) ( x ) .

Example 1.8

Composition of functions defined by tables.

Consider the functions f f and g g described by Table 1.4 and Table 1.5 .

• Evaluate ( g ∘ f ) ( 3 ) , ( g ∘ f ) ( 0 ) . ( g ∘ f ) ( 3 ) , ( g ∘ f ) ( 0 ) .
• State the domain and range of ( g ∘ f ) ( x ) . ( g ∘ f ) ( x ) .
• Evaluate ( f ∘ f ) ( 3 ) , ( f ∘ f ) ( 1 ) . ( f ∘ f ) ( 3 ) , ( f ∘ f ) ( 1 ) .
• State the domain and range of ( f ∘ f ) ( x ) . ( f ∘ f ) ( x ) .
• ( g ∘ f ) ( 3 ) = g ( f ( 3 ) ) = g ( −2 ) = 0 ( g ∘ f ) ( 3 ) = g ( f ( 3 ) ) = g ( −2 ) = 0 ( g ∘ f ) ( 0 ) = g ( 4 ) = 5 ( g ∘ f ) ( 0 ) = g ( 4 ) = 5
• The domain of g ∘ f g ∘ f is the set { −3 , −2 , −1 , 0 , 1 , 2 , 3 , 4 } . { −3 , −2 , −1 , 0 , 1 , 2 , 3 , 4 } . Since the range of f f is the set { −2 , 0 , 2 , 4 } , { −2 , 0 , 2 , 4 } , the range of g ∘ f g ∘ f is the set { 0 , 3 , 5 } . { 0 , 3 , 5 } .
• ( f ∘ f ) ( 3 ) = f ( f ( 3 ) ) = f ( −2 ) = 4 ( f ∘ f ) ( 3 ) = f ( f ( 3 ) ) = f ( −2 ) = 4 ( f ∘ f ) ( 1 ) = f ( f ( 1 ) ) = f ( −2 ) = 4 ( f ∘ f ) ( 1 ) = f ( f ( 1 ) ) = f ( −2 ) = 4
• The domain of f ∘ f f ∘ f is the set { −3 , −2 , −1 , 0 , 1 , 2 , 3 , 4 } . { −3 , −2 , −1 , 0 , 1 , 2 , 3 , 4 } . Since the range of f f is the set { −2 , 0 , 2 , 4 } , { −2 , 0 , 2 , 4 } , the range of f ∘ f f ∘ f is the set { 0 , 4 } . { 0 , 4 } .

Example 1.9

Application involving a composite function.

A store is advertising a sale of 20 % 20 % off all merchandise. Caroline has a coupon that entitles her to an additional 15 % 15 % off any item, including sale merchandise. If Caroline decides to purchase an item with an original price of x x 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.

Since the sale price is 20 % 20 % off the original price, if an item is x x dollars, its sale price is given by f ( x ) = 0.80 x . f ( x ) = 0.80 x . Since the coupon entitles an individual to 15 % 15 % off the price of any item, if an item is y y dollars, the price, after applying the coupon, is given by g ( y ) = 0.85 y . g ( y ) = 0.85 y . Therefore, if the price is originally x x dollars, its sale price will be f ( x ) = 0.80 x f ( x ) = 0.80 x and then its final price after the coupon will be g ( f ( x ) ) = 0.85 ( 0.80 x ) = 0.68 x . g ( f ( x ) ) = 0.85 ( 0.80 x ) = 0.68 x .

Checkpoint 1.6

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

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 f ( x ) = x 4 − 2 x 2 − 3 f ( x ) = x 4 − 2 x 2 − 3 shown in Figure 1.13 (a). If we take the part of the curve that lies to the right of the y -axis and flip it over the y -axis, it lays exactly on top of the curve to the left of the y -axis. In this case, we say the function has symmetry about the y -axis . On the other hand, consider the function f ( x ) = x 3 − 4 x f ( x ) = x 3 − 4 x shown in Figure 1.13 (b). If we take the graph and rotate it 180 ° 180 ° about the origin, the new graph will look exactly the same. In this case, we say the function has symmetry 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 f f has symmetry? Looking at Figure 1.13 again, we see that since f f is symmetric about the y y -axis, if the point ( x , y ) ( x , y ) is on the graph, the point ( − x , y ) ( − x , y ) is on the graph. In other words, f ( − x ) = f ( x ) . f ( − x ) = f ( x ) . If a function f f has this property, we say f f is an even function, which has symmetry about the y -axis. For example, f ( x ) = x 2 f ( x ) = x 2 is even because

In contrast, looking at Figure 1.13 again, if a function f f is symmetric about the origin, then whenever the point ( x , y ) ( x , y ) is on the graph, the point ( − x , − y ) ( − x , − y ) is also on the graph. In other words, f ( − x ) = − f ( x ) . f ( − x ) = − f ( x ) . If f f has this property, we say f f is an odd function, which has symmetry about the origin. For example, f ( x ) = x 3 f ( x ) = x 3 is odd because

If f ( x ) = f ( − x ) f ( x ) = f ( − x ) for all x x in the domain of f , f , then f f is an even function . An even function is symmetric about the y -axis.

If f ( − x ) = − f ( x ) f ( − x ) = − f ( x ) for all x x in the domain of f , f , then f f is an odd function . An odd function is symmetric about the origin.

Example 1.10

Even and odd functions.

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

• f ( x ) = −5 x 4 + 7 x 2 − 2 f ( x ) = −5 x 4 + 7 x 2 − 2
• f ( x ) = 2 x 5 − 4 x + 5 f ( x ) = 2 x 5 − 4 x + 5
• f ( x ) = 3 x x 2 + 1 f ( x ) = 3 x x 2 + 1

To determine whether a function is even or odd, we evaluate f ( − x ) f ( − x ) and compare it to f ( x ) and − f ( x ) . − f ( x ) .

• f ( − x ) = −5 ( − x ) 4 + 7 ( − x ) 2 − 2 = −5 x 4 + 7 x 2 − 2 = f ( x ) . f ( − x ) = −5 ( − x ) 4 + 7 ( − x ) 2 − 2 = −5 x 4 + 7 x 2 − 2 = f ( x ) . Therefore, f f is even.
• f ( − x ) = 2 ( − x ) 5 − 4 ( − x ) + 5 = −2 x 5 + 4 x + 5 . f ( − x ) = 2 ( − x ) 5 − 4 ( − x ) + 5 = −2 x 5 + 4 x + 5 . Now, f ( − x ) ≠ f ( x ) . f ( − x ) ≠ f ( x ) . Furthermore, noting that − f ( x ) = −2 x 5 + 4 x − 5 , − f ( x ) = −2 x 5 + 4 x − 5 , we see that f ( − x ) ≠ − f ( x ) . f ( − x ) ≠ − f ( x ) . Therefore, f f is neither even nor odd.
• f ( − x ) = 3 ( − x ) / ( ( − x ) 2 + 1 ) = −3 x / ( x 2 + 1 ) = − [ 3 x / ( x 2 + 1 ) ] = − f ( x ) . f ( − x ) = 3 ( − x ) / ( ( − x ) 2 + 1 ) = −3 x / ( x 2 + 1 ) = − [ 3 x / ( x 2 + 1 ) ] = − f ( x ) . Therefore, f f is odd.

Checkpoint 1.7

Determine whether f ( x ) = 4 x 3 − 5 x f ( x ) = 4 x 3 − 5 x is even, odd, or neither.

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

Some students describe this function by stating that it “makes everything positive.” By the definition of the absolute value function, we see that if x < 0 , x < 0 , then | x | = − x > 0 , | x | = − x > 0 , and if x > 0 , x > 0 , then | x | = x > 0 . | x | = x > 0 . However, for x = 0 , | x | = 0 . x = 0 , | x | = 0 . Therefore, it is more accurate to say that for all nonzero inputs, the output is positive, but if x = 0 , x = 0 , the output | x | = 0 . | x | = 0 . We conclude that the range of the absolute value function is { y | y ≥ 0 } . { y | y ≥ 0 } . In Figure 1.14 , we see that the absolute value function is symmetric about the y -axis and is therefore an even function.

Example 1.11

Working with the absolute value function.

Find the domain and range of the function f ( x ) = 2 | x − 3 | + 4 . f ( x ) = 2 | x − 3 | + 4 .

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

A real number x x satisfies this equation as long as

Since y ≥ 4 , y ≥ 4 , we know y − 4 ≥ 0 , y − 4 ≥ 0 , and thus the right-hand side of the equation is nonnegative, so it is possible that there is a solution. Furthermore,

Therefore, we see there are two solutions:

The range of this function is { y | y ≥ 4 } . { y | y ≥ 4 } .

Checkpoint 1.8

For the function f ( x ) = | x + 2 | − 4 , f ( x ) = | x + 2 | − 4 , find the domain and range.

Section 1.1 Exercises

For the following exercises, (a) determine the domain and the range of each relation, and (b) state whether the relation is a function.

For the following exercises, find the values for each function, if they exist, then simplify.

a. f ( 0 ) f ( 0 ) b. f ( 1 ) f ( 1 ) c. f ( 3 ) f ( 3 ) d. f ( − x ) f ( − x ) e. f ( a ) f ( a ) f. f ( a + h ) f ( a + h )

f ( x ) = 5 x − 2 f ( x ) = 5 x − 2

f ( x ) = 4 x 2 − 3 x + 1 f ( x ) = 4 x 2 − 3 x + 1

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

f ( x ) = | x − 7 | + 8 f ( x ) = | x − 7 | + 8

f ( x ) = 6 x + 5 f ( x ) = 6 x + 5

f ( x ) = x − 2 3 x + 7 f ( x ) = x − 2 3 x + 7

f ( x ) = 9 f ( x ) = 9

For the following exercises, find the domain, range, and all zeros/intercepts, if any, of the functions.

f ( x ) = x x 2 − 16 f ( x ) = x x 2 − 16

g ( x ) = 8 x − 1 g ( x ) = 8 x − 1

h ( x ) = 3 x 2 + 4 h ( x ) = 3 x 2 + 4

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

f ( x ) = 1 x − 9 f ( x ) = 1 x − 9

g ( x ) = 3 x − 4 g ( x ) = 3 x − 4

f ( x ) = 4 | x + 5 | f ( x ) = 4 | x + 5 |

g ( x ) = 7 x − 5 g ( x ) = 7 x − 5

For the following exercises, sketch the graph with the aid of the tables given:

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

f ( x ) = 3 x − 6 f ( x ) = 3 x − 6

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

f ( x ) = 2 | x | f ( x ) = 2 | x |

f ( x ) = − x 2 f ( x ) = − x 2

f ( x ) = x 3 f ( x ) = x 3

For the following exercises, use the vertical line test to determine whether each of the given graphs represents a function. Assume that a graph continues at both ends if it extends beyond the given grid. If the graph represents a function, then determine the following for each graph:

• Domain and range
• x x -intercept, if any (estimate where necessary)
• y y -Intercept, if any (estimate where necessary)
• The intervals for which the function is increasing
• The intervals for which the function is decreasing
• The intervals for which the function is constant
• Symmetry about any axis and/or the origin
• Whether the function is even, odd, or neither

For the following exercises, for each pair of functions, find a. f + g f + g b. f − g f − g c. f · g f · g d. f / g . f / g . Determine the domain of each of these new functions.

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

f ( x ) = x − 8 , g ( x ) = 5 x 2 f ( x ) = x − 8 , g ( x ) = 5 x 2

f ( x ) = 3 x 2 + 4 x + 1 , g ( x ) = x + 1 f ( x ) = 3 x 2 + 4 x + 1 , g ( x ) = x + 1

f ( x ) = 9 − x 2 , g ( x ) = x 2 − 2 x − 3 f ( x ) = 9 − x 2 , g ( x ) = x 2 − 2 x − 3

f ( x ) = x , g ( x ) = x − 2 f ( x ) = x , g ( x ) = x − 2

f ( x ) = 6 + 1 x , g ( x ) = 1 x f ( x ) = 6 + 1 x , g ( x ) = 1 x

For the following exercises, for each pair of functions, find a. ( f ∘ g ) ( x ) ( f ∘ g ) ( x ) and b. ( g ∘ f ) ( x ) ( g ∘ f ) ( x ) Simplify the results. Find the domain of each of the results.

f ( x ) = 3 x , g ( x ) = x + 5 f ( x ) = 3 x , g ( x ) = x + 5

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

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

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

f ( x ) = x , g ( x ) = x + 9 f ( x ) = x , g ( x ) = x + 9

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

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

The table below lists the NBA championship winners for the years 2001 to 2012.

• Consider the relation in which the domain values are the years 2001 to 2012 and the range is the corresponding winner. Is this relation a function? Explain why or why not.
• Consider the relation where the domain values are the winners and the range is the corresponding years. Is this relation a function? Explain why or why not.

[T] The area A A of a square depends on the length of the side s . s .

• Write a function A ( s ) A ( s ) for the area of a square.
• Find and interpret A ( 6.5 ) . A ( 6.5 ) .
• Find the exact and the two-significant-digit approximation to the length of the sides of a square with area 56 square units.

[T] The volume of a cube depends on the length of the sides s . s .

• Write a function V ( s ) V ( s ) for the volume of a cube.
• Find and interpret V ( 11.8 ) . V ( 11.8 ) .

[T] A rental car company rents cars for a flat fee of $20 and an hourly charge of $10.25. Therefore, the total cost C C to rent a car is a function of the hours t t the car is rented plus the flat fee.

• Write the formula for the function that models this situation.
• Find the total cost to rent a car for 2 days and 7 hours.
• Determine how long the car was rented if the bill is $430.

[T] A vehicle has a 20-gal tank and gets 15 mpg. The number of miles N that can be driven depends on the amount of gas x in the tank.

• Write a formula that models this situation.
• Determine the number of miles the vehicle can travel on (i) a full tank of gas and (ii) 3/4 of a tank of gas.
• Determine the domain and range of the function.
• Determine how many times the driver had to stop for gas if she has driven a total of 578 mi.

[T] The volume V of a sphere depends on the length of its radius as V = ( 4 / 3 ) π r 3 . V = ( 4 / 3 ) π r 3 . Because Earth is not a perfect sphere, we can use the mean radius when measuring from the center to its surface. The mean radius is the average distance from the physical center to the surface, based on a large number of samples. Find the volume of Earth with mean radius 6.371 × 10 6 6.371 × 10 6 m.

[T] A certain bacterium grows in culture in a circular region. The radius of the circle, measured in centimeters, is given by r ( t ) = 6 − [ 5 / ( t 2 + 1 ) ] , r ( t ) = 6 − [ 5 / ( t 2 + 1 ) ] , where t is time measured in hours since a circle of a 1-cm radius of the bacterium was put into the culture.

• Express the area of the bacteria as a function of time.
• Find the exact and approximate area of the bacterial culture in 3 hours.
• Express the circumference of the bacteria as a function of time.
• Find the exact and approximate circumference of the bacteria in 3 hours.

[T] An American tourist visits Paris and must convert U.S. dollars to Euros, which can be done using the function E ( x ) = 0.79 x , E ( x ) = 0.79 x , where x is the number of U.S. dollars and E ( x ) E ( x ) is the equivalent number of Euros. Since conversion rates fluctuate, when the tourist returns to the United States 2 weeks later, the conversion from Euros to U.S. dollars is D ( x ) = 1.245 x , D ( x ) = 1.245 x , where x is the number of Euros and D ( x ) D ( x ) is the equivalent number of U.S. dollars.

• Find the composite function that converts directly from U.S. dollars to U.S. dollars via Euros. Did this tourist lose value in the conversion process?
• Use (a) to determine how many U.S. dollars the tourist would get back at the end of her trip if she converted an extra $200 when she arrived in Paris.

[T] The manager at a skateboard shop pays his workers a monthly salary S of $750 plus a commission of $8.50 for each skateboard they sell.

• Write a function y = S ( x ) y = S ( x ) that models a worker’s monthly salary based on the number of skateboards x he or she sells.
• Find the monthly salary when a worker sells 25, 40, or 55 skateboards.
• Use the INTERSECT feature on a graphing calculator to determine the number of skateboards that must be sold for a worker to earn a monthly income of $1400. ( Hint : Find the intersection of the function and the line y = 1400 .) y = 1400 .)

[T] Use a graphing calculator to graph the half-circle y = 25 − ( x − 4 ) 2 . y = 25 − ( x − 4 ) 2 . Then, use the INTERCEPT feature to find the value of both the x x - and y y -intercepts.

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• Authors: Gilbert Strang, Edwin “Jed” Herman
• Publisher/website: OpenStax
• Book title: Calculus Volume 1
• Publication date: Mar 30, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/calculus-volume-1/pages/1-introduction
• Section URL: https://openstax.org/books/calculus-volume-1/pages/1-1-review-of-functions

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Relations & Functions

Relation : a set of ordered pairs

Domain : the set of x -coordinates

Range : the set of y -coordinates

When writing the domain and range,

do not repeat values.

DIX & ROY

Y: y-coordinate

X: x-coordinate

Domain: { }

Relations and Functions

Given the relation:

{(2, -6), (1, 4), (2, 4), (0,0), (1, -6), (3, 0)}

State the domain:

D: {0,1, 2, 3}

State the range:

R: {-6, 0, 4}

{(1, -6), (-3, 2), (2, 5), (0,1), (5, -3), (3, 0)}

• A function is a relation in which the members of the domain ( x -values) DO NOT repeat.
• So, for every x -value there is only one y -value that corresponds to it.
• y -values can be repeated.

Relations can be written in several ways: ordered pairs, table, graph, or mapping.

Group of ordered pairs

Each input has ONLY ONE output

{(3, 4), (7, 2), (0, -1), (-2, 2), (-5, 0), (3, 3)}

Do the ordered pairs represent a function?

No, 3 is repeated in the domain.

{(4, 1), (5, 2), (8, 2), (9, 8)}

Yes, no x -coordinate is repeated.

{(2, -6), (1, 4), (2, 4), (0, 0), (1, -6), (3, 0)}

Create two ovals with the domain on the left and the range on the right.

Elements are not repeated.

Connect elements of the domain with the corresponding elements in the range by drawing an arrow.

Graphs of a Function

Vertical Line Test:

If a vertical line is passed over the graph and it intersects the graph in exactly one point, the graph represents a function.

Does the graph represent a function? Name the domain and range.

D: all reals

R: all reals

Identify the…

{-6,-2,0,5,7}

{-3,-2,4,5,7}

Let’s Try Some Practice

Solve for the Domain

Range: (1, 4, 7)

Solve for the Range

Domain: (2, 5, 8)

Rule: y = 2 x�Domain: (-2, 0, 2, 4)

Rule: y = x + 5�Domain: (-4, 2, 8)

Solve for the Missing Part of the Relation

Rule: y = 5x + 2 Domain: (1,3,5)

What would the " range " values be if the "domain" values are ( 0, 3, 6, 9 )?

Rule: 10 less than a “x”�Domain: (-4, 2, 8)

Rule: y = x - 4�Domain:

Range: (9, 15, 21, 27)

Rule: y = 2 times “x” plus 6�Domain:

Range: (22, 26, 34)

Solve for the Missing Part of the Function’s Relation

Rule: y = 3 x�Domain:

Equations working with Tables

Rule: �Domain: (-4, 2, 8)

Range: (5, 11, 17)

Rule:�Domain: (3, 6, 9, 12)

Range: (12, 24, 36, 48)

Solve for the Function’s Unknown Rule

What is the Rule?

What is the relation rule to the following data?

Domain: (2,4,6,8)

Range: (2,3,4,5)

Story Problem: Disney Trip

Your family has decided to take a road trip to Disney World. From Fords NJ to Orlando FL it is about 1200 miles. Your family travels at an average speed of 60 miles per hour. Your distance “ d ” (in miles) from Disney is given by:

d=1200 - 60t , where “ t ” is the time (in hours) spent driving. Graph the function and identify its domain and range.

• How long will it take to get to Disney World from Fords?
• How far away from Disney are you if you have already driven 10 hours?

Story Problem: Submarine

A submarine is designed to explore the ocean floor at an elevation of -13,000 feet (13,000 feet below sea level). The submarine ascends to the surface at an average rate of 650 feet per minute. The elevation “ e ” (in feet) of the submarine is given by the function:

e =650 ( t ) - 13,000

where “ t ” is time (in minutes) since the submarine began to ascend.

• Graph the function to show how long it will take to reach the surface (sea level).
• How long does it take for the submarine to reach the surface?
• How long does it take for the submarine to reach the halfway point to the surface?
• How long would it take for the submarine to ascend if it was only 1625 ft below sea level?
• What elevation would the submarine be at after one hour?

d=1200-60(0)

d=1200-60(5)

d=1200-60(10)

d=1200-60(15)

d=1200-60(20)

• How long does it take for the submarine to reach the surface? 20 minutes
• How long does it take for the submarine to reach the halfway point to the surface? 10 minutes
• How long would it take for the submarine to ascend if it was only 1625 ft below sea level? 2.5 minutes
• What elevation would the submarine be at after one hour? 0 feet (sea-level)

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Functions: Quadratic Functions and Equations - Math - 10th Grade

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Mathematics > Number Theory

Title: extended genus fields of abelian extensions of rational function fields.

Abstract: In this paper we obtain the extended genus field of a finite abelian extension of a global rational function field. We first study the case of a cyclic extension of prime power degree. Next, we use that the extended genus fields of a composite of two cyclotomic extensions of a global rational function field is equal to the composite of their respective extended genus fields, to obtain our main result. This result is that the extended genus field of a general finite abelian extension of a global rational function field, is given explicitly in terms of the field and of the extended genus field of its "cyclotomic projection".

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