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 [latex]x[/latex] be the time after midnight, measured in hours, and the output variable [latex]y[/latex] be the temperature [latex]x[/latex] hours after midnight, measured in degrees Fahrenheit. We record our data in (Figure) .

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 [latex]f[/latex] 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.

From the points plotted on the graph in (Figure) , 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) .

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 [latex]r[/latex] is given by the formula [latex]A(r)=\pi r^2[/latex]. When an object is thrown upward from the ground with an initial velocity [latex]v_0[/latex] ft/s, its height above the ground from the time it is thrown until it hits the ground is given by the formula [latex]s(t)=-16t^2+v_0t[/latex]. When [latex]P[/latex] dollars are invested in an account at an annual interest rate [latex]r[/latex] compounded continuously, the amount of money after [latex]t[/latex] years is given by the formula [latex]A(t)=Pe^{rt}[/latex]. 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 [latex]f[/latex], the graph of [latex]f[/latex] is the set of points [latex](x,f(x))[/latex], where [latex]x[/latex] is in the domain of [latex]f[/latex] and [latex]f(x)[/latex] 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 [latex]f[/latex] 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 [latex]x[/latex] where [latex]f(x)=0[/latex] are called the zeros of a function . For example, the zeros of [latex]f(x)=x^2-4[/latex] are [latex]x=±2[/latex]. The zeros determine where the graph of [latex]f[/latex] intersects the [latex]x[/latex]-axis, which gives us more information about the shape of the graph of the function. The graph of a function may never intersect the [latex]x[/latex]-axis, or it may intersect multiple (or even infinitely many) times.

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

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

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

Using Zeros and y-Intercepts to Sketch a Graph

Consider the function [latex]f(x)=\sqrt{x+3}+1[/latex].

1. Find all zeros of [latex]f[/latex].
2. Find the [latex]y[/latex]-intercept (if any).
3. Sketch a graph of [latex]f[/latex].

Solution

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

   [latex]\mathbf{x}[/latex]	-3	-2	1
   [latex]\mathbf{f(x)}[/latex]	1	2	3

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

Note that for this function and the function [latex]f(x)=-4x+2[/latex] graphed in (Figure) , the values of [latex]f(x)[/latex] are getting smaller as [latex]x[/latex] is getting larger. A function with this property is said to be decreasing. On the other hand, for the function [latex]f(x)=\sqrt{x+3}+1[/latex] graphed in (Figure) , the values of [latex]f(x)[/latex] are getting larger as the values of [latex]x[/latex] 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 [latex](0,4)[/latex], increasing on the interval [latex](4,14)[/latex], and then decreasing on the interval [latex](14,23)[/latex]. 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 [latex]f[/latex] is increasing on the interval [latex]I[/latex] if for all [latex]x_1, x_2\in I[/latex],

[latex]f(x_1)\le f(x_2) \, \text{when} \, x_1

We say [latex]f[/latex] is strictly increasing on the interval [latex]I[/latex] if for all [latex]x_1,x_2\in I[/latex],

[latex]f(x_1)

We say that a function [latex]f[/latex] is decreasing on the interval [latex]I[/latex] if for all [latex]x_1, x_2\in I[/latex],

[latex]f(x_1)\ge f(x_2) \, \text{if} \, x_1

We say that a function [latex]f[/latex] is strictly decreasing on the interval [latex]I[/latex] if for all [latex]x_1, x_2 \in I[/latex],

[latex]f(x_1) \symbol{"3E} f(x_2) \, \text{if} \, x_1

For example, the function [latex]f(x)=3x[/latex] is increasing on the interval [latex](-\infty ,\infty)[/latex] because [latex]3x_1<3x_2[/latex] whenever [latex]x_1 (Figure) ).

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 [latex]x[/latex] items is described by the function [latex]C(x)[/latex] and the revenue created by the sale of [latex]x[/latex] items is described by the function [latex]R(x)[/latex], then the profit on the manufacture and sale of [latex]x[/latex] items is defined as [latex]P(x)=R(x)-C(x)[/latex]. 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 [latex]f(x)=x^2[/latex] and [latex]g(x)=3x+1[/latex], the composite function [latex]f\circ g[/latex] is defined such that

[latex](f\circ g)(x)=f(g(x))=(g(x))^2=(3x+1)^2[/latex].

The composite function [latex]g\circ f[/latex] is defined such that

[latex](g\circ f)(x)=g(f(x))=3f(x)+1=3x^2+1[/latex].

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 [latex]f[/latex] and [latex]g,[/latex] we can define four new functions:

[latex]\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\cdot 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}[/latex]

Combining Functions Using Mathematical Operations

Given the functions [latex]f(x)=2x-3[/latex] and [latex]g(x)=x^2-1[/latex], find each of the following functions and state its domain.

1. [latex](f+g)(x)[/latex]
2. [latex](f-g)(x)[/latex]
3. [latex](f\cdot g)(x)[/latex]
4. [latex]\Big(\frac{f}{g}\Big)(x)[/latex]

Solution

1. [latex](f+g)(x)=(2x-3)+(x^2-1)=x^2+2x-4[/latex]. The domain of this function is the interval [latex](-\infty ,\infty )[/latex].
2. [latex](f-g)(x)=(2x-3)-(x^2-1)=-x^2+2x-2[/latex]. The domain of this function is the interval [latex](-\infty ,\infty)[/latex].
3. [latex](f\cdot g)(x)=(2x-3)(x^2-1)=2x^3-3x^2-2x+3[/latex]. The domain of this function is the interval [latex](-\infty ,\infty )[/latex].
4. [latex]\Big(\frac{f}{g}\Big)(x)=\frac{2x-3}{x^2-1}[/latex]. The domain of this function is [latex]\{x|x\ne \pm 1\}[/latex].

For [latex]f(x)=x^2+3[/latex] and [latex]g(x)=2x-5[/latex], find [latex](f/g)(x)[/latex] and state its domain.

Hint

The new function [latex](f/g)(x)[/latex] is a quotient of two functions. For what values of [latex]x[/latex] is the denominator zero?

Solution

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

Function Composition

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

Definition

Consider the function [latex]f[/latex] with domain [latex]A[/latex] and range [latex]B[/latex], and the function [latex]g[/latex] with domain [latex]D[/latex] and range [latex]E[/latex]. If [latex]B[/latex] is a subset of [latex]D[/latex], then the composite function [latex](g\circ f)(x)[/latex] is the function with domain [latex]A[/latex] such that

[latex](g\circ f)(x)=g(f(x))[/latex].

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

Compositions of Functions Defined by Formulas

Consider the functions [latex]f(x)=x^2+1[/latex] and [latex]g(x)=1/x[/latex].

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

Solution

1. We can find the formula for [latex](g\circ f)(x)[/latex] in two different ways. We could write
   [latex](g\circ f)(x)=g(f(x))=g(x^2+1)=\frac{1}{x^2+1}[/latex].

   Alternatively, we could write

   [latex](g\circ f)(x)=g(f(x))=\frac{1}{f(x)}=\frac{1}{x^2+1}[/latex].

   Since [latex]x^2+1\ne 0[/latex] for all real numbers [latex]x[/latex], the domain of [latex](g\circ f)(x)[/latex] is the set of all real numbers. Since [latex]0<1/(x^2+1)\le 1[/latex], the range is, at most, the interval [latex](0,1][/latex]. To show that the range is this entire interval, we let [latex]y=1/(x^2+1)[/latex] and solve this equation for [latex]x[/latex] to show that for all [latex]y[/latex] in the interval [latex](0,1][/latex], there exists a real number [latex]x[/latex] such that [latex]y=1/(x^2+1)[/latex]. Solving this equation for [latex]x[/latex], we see that [latex]x^2+1=1/y[/latex], which implies that

   [latex]x=±\sqrt{\frac{1}{y}-1}[/latex].

   If [latex]y[/latex] is in the interval [latex](0,1][/latex], the expression under the radical is nonnegative, and therefore there exists a real number [latex]x[/latex] such that [latex]1/(x^2+1)=y[/latex]. We conclude that the range of [latex]g\circ f[/latex] is the interval [latex](0,1][/latex].

2. [latex](g\circ f)(4)=g(f(4))=g(4^2+1)=g(17)=\frac{1}{17}[/latex]
   [latex](g\circ f)(-\frac{1}{2})=g(f(-\frac{1}{2}))=g((-\frac{1}{2})^2+1)=g(\frac{5}{4})=\frac{4}{5}[/latex]
3. We can find a formula for [latex](f\circ g)(x)[/latex] in two ways. First, we could write
   [latex](f\circ g)(x)=f(g(x))=f(\frac{1}{x})=(\frac{1}{x})^2+1[/latex].

   Alternatively, we could write

   [latex](f\circ g)(x)=f(g(x))=(g(x))^2+1=(\frac{1}{x})^2+1[/latex].

   The domain of [latex]f\circ g[/latex] is the set of all real numbers [latex]x[/latex] such that [latex]x\ne 0[/latex]. To find the range of [latex]f[/latex], we need to find all values [latex]y[/latex] for which there exists a real number [latex]x\ne 0[/latex] such that

   [latex](\frac{1}{x})^2+1=y[/latex].

   Solving this equation for [latex]x[/latex], we see that we need [latex]x[/latex] to satisfy

   [latex](\frac{1}{x})^2=y-1[/latex],

   which simplifies to

   [latex]\frac{1}{x}=±\sqrt{y-1}[/latex].

   Finally, we obtain

   [latex]x=±\frac{1}{\sqrt{y-1}}[/latex].

   Since [latex]1/\sqrt{y-1}[/latex] is a real number if and only if [latex]y \symbol{"3E} 1[/latex], the range of [latex]f[/latex] is the set [latex]\{y|y\ge 1\}[/latex].

4. [latex](f\circ g)(4)=f(g(4))=f(\frac{1}{4})=(\frac{1}{4})^2+1=\frac{17}{16}[/latex]
   [latex](f\circ g)(-\frac{1}{2})=f(g(-\frac{1}{2}))=f(-2)=(-2)^2+1=5[/latex]

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

Let [latex]f(x)=2-5x.[/latex] Let [latex]g(x)=\sqrt{x}.[/latex] Find [latex](f\circ g)(x)[/latex].

Solution

[latex](f\circ g)(x)=2-5\sqrt{x}[/latex].

Composition of Functions Defined by Tables

Consider the functions [latex]f[/latex] and [latex]g[/latex] described by (Figure) and (Figure) .

[latex]\mathbf{x}[/latex]	-3	-2	-1	0	1	2	3	4
[latex]\mathbf{f(x)}[/latex]	0	4	2	4	-2	0	-2	4

[latex]\mathbf{x}[/latex]	-4	-2	0	2	4
[latex]\mathbf{g(x)}[/latex]	1	0	3	0	5

1. Evaluate [latex](g\circ f)(3)[/latex] and [latex](g\circ f)(0)[/latex].
2. State the domain and range of [latex](g\circ f)(x)[/latex].
3. Evaluate [latex](f\circ f)(3)[/latex] and [latex](f\circ f)(1)[/latex].
4. State the domain and range of [latex](f\circ f)(x)[/latex].

Solution

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

Application Involving a Composite Function

A store is advertising a sale of [latex]20\%[/latex] off all merchandise. Caroline has a coupon that entitles her to an additional [latex]15\%[/latex] off any item, including sale merchandise. If Caroline decides to purchase an item with an original price of [latex]x[/latex] 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 [latex]20\%[/latex] off the original price, if an item is [latex]x[/latex] dollars, its sale price is given by [latex]f(x)=0.80x[/latex]. Since the coupon entitles an individual to [latex]15\%[/latex] off the price of any item, if an item is [latex]y[/latex] dollars, the price, after applying the coupon, is given by [latex]g(y)=0.85y[/latex]. Therefore, if the price is originally [latex]x[/latex] dollars, its sale price will be [latex]f(x)=0.80x[/latex] and then its final price after the coupon will be [latex]g(f(x))=0.85(0.80x)=0.68x[/latex].

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

Hint

The sale price of an item with an original price of [latex]x[/latex] dollars is [latex]f(x)=0.90x[/latex]. The coupon price for an item that is [latex]y[/latex] dollars is [latex]g(y)=0.70y[/latex].

Solution

[latex](g\circ f)(x)=0.63x[/latex]

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 [latex]f(x)=x^4-2x^2-3[/latex] shown in (Figure) (a). If we take the part of the curve that lies to the right of the [latex]y[/latex]-axis and flip it over the [latex]y[/latex]-axis, it lays exactly on top of the curve to the left of the [latex]y[/latex]-axis. In this case, we say the function has symmetry about the [latex]y[/latex]-axis . On the other hand, consider the function [latex]f(x)=x^3-4x[/latex] 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.

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 [latex]f[/latex] has symmetry? Looking at (Figure) again, we see that since [latex]f[/latex] is symmetric about the [latex]y[/latex]-axis, if the point [latex](x,y)[/latex] is on the graph, the point [latex](-x,y)[/latex] is on the graph. In other words, [latex]f(-x)=f(x)[/latex]. If a function [latex]f[/latex] has this property, we say [latex]f[/latex] is an even function , which has symmetry about the [latex]y[/latex]-axis. For example, [latex]f(x)=x^2[/latex] is even because

[latex]f(-x)=(-x)^2=x^2=f(x)[/latex].

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

[latex]f(-x)=(-x)^3=-x^3=-f(x)[/latex].

Definition

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

If [latex]f(-x)=-f(x)[/latex] for all [latex]x[/latex] in the domain of [latex]f[/latex], then [latex]f[/latex] 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.

1. [latex]f(x)=-5x^4+7x^2-2[/latex]
2. [latex]f(x)=2x^5-4x+5[/latex]
3. [latex]f(x)=\large{\frac{3x}{x^2+1}}[/latex]

Solution

To determine whether a function is even or odd, we evaluate [latex]f(-x)[/latex] and compare it to [latex]f(x)[/latex] and [latex]-f(x)[/latex].

1. [latex]f(-x)=-5(-x)^4+7(-x)^2-2=-5x^4+7x^2-2=f(x)[/latex]. Therefore, [latex]f[/latex] is even.
2. [latex]f(-x)=2(-x)^5-4(-x)+5=-2x^5+4x+5[/latex]. Now, [latex]f(-x)\ne f(x)[/latex]. Furthermore, noting that [latex]-f(x)=-2x^5+4x-5[/latex], we see that [latex]f(-x)\ne -f(x)[/latex]. Therefore, [latex]f[/latex] is neither even nor odd.
3. [latex]f(-x)=3(-x)/((-x)^2+1)=-3x/(x^2+1)=-[3x/(x^2+1)]=-f(x)[/latex]. Therefore, [latex]f[/latex] is odd.

Determine whether [latex]f(x)=4x^3-5x[/latex] is even, odd, or neither.

Hint

Compare [latex]f(-x)[/latex] with [latex]f(x)[/latex] and [latex]-f(x)[/latex].

Solution

[latex]f(x)[/latex] is odd.

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

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

Some students describe this function by stating that it “makes everything positive.” By the definition of the absolute value function, we see that if [latex]x<0[/latex], then [latex]|x|=-x \symbol{"3E} 0[/latex], and if [latex]x \symbol{"3E} 0[/latex], then [latex]|x|=x \symbol{"3E} 0[/latex]. However, for [latex]x=0, \, |x|=0[/latex]. Therefore, it is more accurate to say that for all nonzero inputs, the output is positive, but if [latex]x=0[/latex], the output [latex]|x|=0[/latex]. We conclude that the range of the absolute value function is [latex]\{y|y\ge 0\}[/latex]. In (Figure) , we see that the absolute value function is symmetric about the [latex]y[/latex]-axis and is therefore an even function.

Working with the Absolute Value Function

Find the domain and range of the function [latex]f(x)=2|x-3|+4[/latex].

Solution

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

[latex]2|x-3|+4=y[/latex].

A real number [latex]x[/latex] satisfies this equation as long as

[latex]|x-3|=\frac{1}{2}(y-4)[/latex].

Since [latex]y\ge 4[/latex], we know [latex]y-4\ge 0[/latex], and thus the right-hand side of the equation is nonnegative, so it is possible that there is a solution. Furthermore,

[latex]|x-3|= \begin{cases} x-3, & \text{if} \, x \ge 3 \\ -(x-3), & \text{if} \, x < 3 \end{cases}[/latex].

Therefore, we see there are two solutions:

[latex]x=±\frac{1}{2}(y-4)+3[/latex].

The range of this function is [latex]\{y|y\ge 4\}[/latex].

For the function [latex]f(x)=|x+2|-4[/latex], find the domain and range.

Hint

[latex]|x+2|\ge 0[/latex] for all real numbers [latex]x[/latex].

Solution

Domain = [latex](-\infty ,\infty )[/latex], range = [latex]\{y|y\ge -4\}[/latex].

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 [latex]y=f(x)[/latex], the domain is considered to be the set of all real numbers [latex]x[/latex] for which the function is defined.
• When sketching the graph of a function [latex]f[/latex], each vertical line may intersect the graph, at most, once.
• A function may have any number of zeros, but it has, at most, one [latex]y[/latex]-intercept.
• To define the composition [latex]g\circ f[/latex], the range of [latex]f[/latex] must be contained in the domain of [latex]g[/latex].
• Even functions are symmetric about the [latex]y[/latex]-axis whereas odd functions are symmetric about the origin.

Key Equations

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

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

1.

[latex]x[/latex]	-3	-2	-1	0	1	2	3
[latex]f(x)[/latex]	9	4	1	0	1	4	9

 

Solution

a. Domain = [latex]\{-3,-2,-1,0,1,2,3\}[/latex], range = [latex]\{0,1,4,9\}[/latex]

b. Yes, a function

2.

[latex]x[/latex]	-3	-2	-1	0	1	2	3
[latex]f(x)[/latex]	-2	-8	-1	0	1	8	2

3.

[latex]x[/latex]	1	2	3	0	1	2	3
[latex]f(x)[/latex]	-3	-2	-1	0	1	2	3

Solution

a. Domain = [latex]\{0,1,2,3\}[/latex], range = [latex]\{-3,-2,-1,0,1,2,3\}[/latex]

b. No, not a function because the domain value is mapped to more than one range value.

4.

[latex]x[/latex]	1	2	3	4	5	6	7
[latex]y[/latex]	1	1	1	1	1	1	1

 

5.

[latex]x[/latex]	3	5	8	10	15	21	33
[latex]y[/latex]	3	2	1	0	1	2	3

Solution

a. Domain = [latex]\{3,5,8,10,15,21,33\}[/latex], range = [latex]\{0,1,2,3\}[/latex]

b. Yes, a function

6.

[latex]x[/latex]	-7	-2	-2	0	1	3	6
[latex]f(x)[/latex]	11	5	1	-1	-2	4	11

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

a. [latex]f(0)[/latex] b. [latex]f(1)[/latex] c. [latex]f(3)[/latex] d. [latex]f(-x)[/latex] e. [latex]f(a)[/latex] f. [latex]f(a+h)[/latex]

7.  [latex]f(x)=5x-2[/latex]

Solution

a. -2

b. 3

c. 13

d. [latex]-5x-2[/latex]

e. [latex]5a-2[/latex]

f. [latex]5a+5h-2[/latex]

8.  [latex]f(x)=4x^2-3x+1[/latex]

9.  [latex]f(x)=\frac{2}{x}[/latex]

Solution

a. Undefined

b. 2

c. [latex]\frac{2}{3}[/latex]

d. [latex]-\frac{2}{x}[/latex]

e. [latex]\frac{2}{a}[/latex]

f. [latex]\frac{2}{a+h}[/latex]

10.  [latex]f(x)=|x-7|+8[/latex]

11.  [latex]f(x)=\sqrt{6x+5}[/latex]

Solution

a. [latex]\sqrt{5}[/latex]

b. [latex]\sqrt{11}[/latex]

c. [latex]\sqrt{23}[/latex]

d. [latex]\sqrt{-6x+5}[/latex]

e. [latex]\sqrt{6a+5}[/latex]

f. [latex]\sqrt{6a+6h+5}[/latex]

12.  [latex]f(x)=\frac{x-2}{3x+7}[/latex]

13.  [latex]f(x)=9[/latex]

Solution

a. 9 b. 9 c. 9 d. 9 e. 9 f. 9

For the following exercises, find the domain of the functions.

14.  [latex]f(x)=\sqrt{x^2+x-6}[/latex]

15.  [latex]g(x)=\frac{\sqrt{x^2-4}}{x-3}[/latex]

Solution

Domain: [latex](-\infty,-2]\cup [2,3) \cup (3,\infty)[/latex]

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

16.  [latex]f(x)=\frac{x}{x^2-16}[/latex]

17.  [latex]g(x)=\sqrt{8x-1}[/latex]

Solution

Domain: [latex]\left[\frac{1}{8},\infty\right)[/latex]

Range: [latex][0,\infty)[/latex]

Zeros: [latex]x=\frac{1}{8}[/latex]

no [latex]y[/latex]-intercepts

18.  [latex]h(x)=\frac{3}{x^2+4}[/latex]

19.  [latex]f(x)=-1+\sqrt{x+2}[/latex]

Solution

Domain: [latex][-2,\infty)[/latex]

Range: [latex][-1,\infty)[/latex]

Zeros: [latex]x=-1[/latex]

[latex]y[/latex]-intercepts: [latex]y=-1+\sqrt{2}[/latex]

20.  [latex]f(x)=\frac{1}{\sqrt{x-9}}[/latex]

21.  [latex]g(x)=\frac{3}{x-4}[/latex]

Solution

Domain: [latex](-\infty,4)\cup(4,\infty)[/latex]

Range: [latex](-\infty,0)\cup(0,\infty)[/latex]

no [latex]x[/latex]-intercept

[latex]y[/latex]-intercept: [latex]y=-\frac{3}{4}[/latex]

22.  [latex]f(x)=4|x+5|[/latex]

23.  [latex]g(x)=\sqrt{\frac{7}{x-5}}[/latex]

Solution

Domain: [latex](5,\infty)[/latex]

Range: [latex](0,\infty)[/latex]

no intercepts

For the following exercises, set up a table of values to sketch the graph of each function using the following values: [latex]x=-3,-2,-1,0,1,2,3[/latex].

24.  [latex]f(x)=x^2+1[/latex]

25.  [latex]f(x)=3x-6[/latex]

Solution

[latex]x[/latex]	-3	-2	-1	0	1	2	3
[latex]f(x)[/latex]	-15	-12	-9	-6	-3	0	3

26.  [latex]f(x)=\frac{1}{2}x+1[/latex]

27.  [latex]f(x)=2|x|[/latex]

Solution

[latex]x[/latex]	-3	-2	-1	0	1	2	3
[latex]f(x)[/latex]	6	4	2	0	2	4	6

28.  [latex]f(x)=-x^2[/latex]

29.  [latex]f(x)=x^3[/latex]

Solution

[latex]x[/latex]	-3	-2	-1	0	1	2	3
[latex]f(x)[/latex]	-27	-8	-1	0	1	8	27

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:

1. Domain and range
2. [latex]x[/latex]-intercept, if any (estimate where necessary)
3. [latex]y[/latex]-Intercept, if any (estimate where necessary)
4. The intervals for which the function is increasing
5. The intervals for which the function is decreasing
6. The intervals for which the function is constant
7. Symmetry about any axis and/or the origin
8. Whether the function is even, odd, or neither

30.

31.

Solution

Function;

a. Domain: all real numbers, Range: [latex][0,\infty)[/latex]

b. [latex]x=±1[/latex]

c. [latex]y=1[/latex]

d. [latex]-1

32.

33.

Solution

Function;

a. Domain: all real numbers, Range: [latex][-1.5, 1.5][/latex]

b. [latex]x=0[/latex]

c. [latex]y=0[/latex]

d. All real numbers

e. None

f. Not constant

g. Origin

h. Odd

34.

35.

Solution

Function;

a. Domain: [latex](-\infty,\infty)[/latex], Range: [latex][-2, 2][/latex]

b. [latex]x=0[/latex]

c. [latex]y=0[/latex]

d. [latex]-2

36.

37.

Solution

Function;

a. Domain: [latex][-4, 4][/latex], Range: [latex][-4, 4][/latex]

b. [latex]x=1.2[/latex]

c. [latex]y=4[/latex]

d. Not increasing

e. [latex]0

For the following exercises, for each pair of functions, find a. [latex]f+g[/latex] b. [latex]f-g[/latex] c. [latex]f\cdot g[/latex] d. [latex]f/g[/latex]. Determine the domain of each of these new functions.

38.  [latex]f(x)=3x+4, \, g(x)=x-2[/latex]

39.  [latex]f(x)=x-8, \, g(x)=5x^2[/latex]

Solution

a. [latex]5x^2+x-8[/latex]; all real numbers

b. [latex]-5x^2+x-8[/latex]; all real numbers

c. [latex]5x^3-40x^2[/latex]; all real numbers

d. [latex]\frac{x-8}{5x^2}; \, (-\infty,0)\cup (0,\infty)[/latex]

40.  [latex]f(x)=3x^2+4x+1, \, g(x)=x+1[/latex]

41.  [latex]f(x)=9-x^2, \, g(x)=x^2-2x-3[/latex]

Solution

a. [latex]-2x+6[/latex]; all real numbers

b. [latex]-2x^2+2x+12[/latex]; all real numbers

c. [latex]-x^4+2x^3+12x^2-18x-27[/latex]; all real numbers

d. [latex]-\frac{x+3}{x+1}; \, (-\infty,-1)\cup(-1,3)\cup(3,\infty)[/latex]

42.  [latex]f(x)=\sqrt{x}, \, g(x)=x-2[/latex]

43.  [latex]f(x)=6+\frac{1}{x},g(x)=\frac{1}{x}[/latex]

Solution

a. [latex]6+\frac{2}{x}; \, (-\infty,0)\cup (0,\infty)[/latex]

b. 6; [latex](-\infty,0)\cup (0,\infty)[/latex]

c. [latex]\frac{6}{x}+\frac{1}{x^2}; \, (-\infty,0)\cup (0,\infty)[/latex]

d. [latex]6x+1; \, (-\infty,0)\cup (0,\infty)[/latex]

For the following exercises, for each pair of functions, find a. [latex](f\circ g)(x)[/latex] and b. [latex](g\circ f)(x)[/latex] Simplify the results. Find the domain of each of the results.

44.  [latex]f(x)=3x, \, g(x)=x+5[/latex]

45.  [latex]f(x)=x+4, \, g(x)=4x-1[/latex]

Solution

a. [latex]4x+3[/latex]; all real numbers

b. [latex]4x+15[/latex]; all real numbers

46.  [latex]f(x)=2x+4, \, g(x)=x^2-2[/latex]

47.  [latex]f(x)=x^2+7, \, g(x)=x^2-3[/latex]

Solution

a. [latex]x^4-6x^2+16[/latex]; all real numbers

b. [latex]x^4+14x^2+46[/latex]; all real numbers

48.  [latex]f(x)=\sqrt{x}, \, g(x)=x+9[/latex]

49.  [latex]f(x)=\frac{3}{2x+1}, \, g(x)=\frac{2}{x}[/latex]

Solution

a. [latex]\frac{3x}{4+x}; \, (-\infty,-4)\cup (-4,0)\cup (0,\infty)[/latex]

b. [latex]\frac{4x+2}{3}; \, \left (-\infty,-\frac{1}{2}\right)\cup \left(-\frac{1}{2},\infty\right)[/latex]

50.  [latex]f(x)=|x+1|, \, g(x)=x^2+x-4[/latex]

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

Year	Winner
2001	LA Lakers
2002	LA Lakers
2003	San Antonio Spurs
2004	Detroit Pistons
2005	San Antonio Spurs
2006	Miami Heat
2007	San Antonio Spurs
2008	Boston Celtics
2009	LA Lakers
2010	LA Lakers
2011	Dallas Mavericks
2012	Miami Heat

1. 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.
2. 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.

Solution

a. Yes, because there is only one winner for each year.

b. No, because there are three teams that won more than once during the years 2001 to 2012.

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

1. Write a function [latex]A(s)[/latex] for the area of a square.
2. Find and interpret [latex]A(6.5)[/latex].
3. Find the exact and the two-significant-digit approximation to the length of the sides of a square with area 56 square units.

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

1. Write a function [latex]V(s)[/latex] for the area of a square.
2. Find and interpret [latex]V(11.8)[/latex].

Solution

a. [latex]V(s)=s^3[/latex]

b. [latex]V(11.8)\approx 1643[/latex]; a cube of side length 11.8 each has a volume of approximately 1643 cubic units.

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

1. Write the formula for the function that models this situation.
2. Find the total cost to rent a car for 2 days and 7 hours.
3. Determine how long the car was rented if the bill is $432.73.

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

1. Write a formula that models this situation.
2. 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.
3. Determine the domain and range of the function.
4. Determine how many times the driver had to stop for gas if she has driven a total of 578 mi.

Solution

a. [latex]N(x)=15x[/latex]

b. i. [latex]N(20)=15(20)=300[/latex]; therefore, the vehicle can travel 300 mi on a full tank of gas. ii. [latex]N(15)=225[/latex]; therefore, the vehicle can travel 225 mi on 3/4 of a tank of gas.

c. Domain: [latex]0\le x\le 20[/latex]; Range: [latex][0,300][/latex] d. The driver had to stop at least once, given that it takes approximately 39 gal of gas to drive a total of 578 mi.

56. [T] The volume [latex]V[/latex] of a sphere depends on the length of its radius as [latex]V=(4/3)\pi r^3[/latex]. 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 [latex]6.371×10^6[/latex] m.

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

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

Solution

a. [latex]A(t)=A(r(t))=\pi \cdot(6-\frac{5}{t^2+1})^2[/latex]

b. Exact: [latex]\frac{121\pi }{4}[/latex]; approximately 95 cm ²

c. [latex]C(t)=C(r(t))=2\pi (6-\frac{5}{t^2+1})[/latex]

d. Exact: [latex]11\pi[/latex]; approximately 35 cm

58. [T] An American tourist visits Paris and must convert U.S. dollars to Euros, which can be done using the function [latex]E(x)=0.79x[/latex], where [latex]x[/latex] is the number of U.S. dollars and [latex]E(x)[/latex] 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 [latex]D(x)=1.245x[/latex], where [latex]x[/latex] is the number of Euros and [latex]D(x)[/latex] is the equivalent number of U.S. dollars.

1. 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?
2. 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.

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

1. Write a function [latex]y=S(x)[/latex] that models a worker’s monthly salary based on the number of skateboards [latex]x[/latex] he or she sells.
2. Find the approximate monthly salary when a worker sells 25, 40, or 55 skateboards.
3. 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 [latex]y=1400[/latex].)

Solution

a. [latex]S(x)=8.5x+750[/latex]

b. $962.50, $1090, $1217.50

c. 77 skateboards

60. [T] Use a graphing calculator to graph the half-circle [latex]y=\sqrt{25-(x-4)^2}[/latex]. Then, use the INTERCEPT feature to find the value of both the [latex]x[/latex]– and [latex]y[/latex]-intercepts.

Glossary

absolute value function

[latex]f(x) = |x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases}[/latex]

composite function

given two functions [latex]f[/latex] and [latex]g[/latex], a new function, denoted [latex]g\circ f[/latex], such that [latex](g\circ f)(x)=g(f(x))[/latex]

decreasing on the interval [latex]I[/latex]

a function decreasing on the interval [latex]I[/latex] if, for all [latex]x_1, \, x_2\in I, \, f(x_1)\ge f(x_2)[/latex] if [latex]x_1

dependent variable

the output variable for a function

domain

the set of inputs for a function

even function

a function is even if [latex]f(-x)=f(x)[/latex] for all [latex]x[/latex] in the domain of [latex]f[/latex]

function

a set of inputs, a set of outputs, and a rule for mapping each input to exactly one output

graph of a function

the set of points [latex](x,y)[/latex] such that [latex]x[/latex] is in the domain of [latex]f[/latex] and [latex]y=f(x)[/latex]

increasing on the interval [latex]I[/latex]

a function increasing on the interval [latex]I[/latex] if for all [latex]x_1, \, x_2\in I, \, f(x_1)\le f(x_2)[/latex] if [latex]x_1

independent variable

the input variable for a function

odd function

a function is odd if [latex]f(-x)=-f(x)[/latex] for all [latex]x[/latex] in the domain of [latex]f[/latex]

range

the set of outputs for a function

symmetry about the origin

the graph of a function [latex]f[/latex] is symmetric about the origin if [latex](-x,-y)[/latex] is on the graph of [latex]f[/latex] whenever [latex](x,y)[/latex] is on the graph

symmetry about the [latex]y[/latex]-axis

the graph of a function [latex]f[/latex] is symmetric about the [latex]y[/latex]-axis if [latex](-x,y)[/latex] is on the graph of [latex]f[/latex] whenever [latex](x,y)[/latex] is on the graph

table of values

a table containing a list of inputs and their corresponding outputs

vertical line test

given the graph of a function, every vertical line intersects the graph, at most, once

zeros of a function

when a real number [latex]x[/latex] is a zero of a function [latex]f[/latex], [latex]f(x)=0[/latex]