1.1 Review of Functions
Learning Objectives
- Use functional notation to evaluate a function.
- Determine the domain and range of a function.
- Draw the graph of a function.
- Find the zeros of a function.
- Recognize a function from a table of values.
- Make new functions from two or more given functions.
- Describe the symmetry properties of a function.
In this section, we lay out a formal definition of a function and look at several ways we can write one down — tables, formulas, and graphs. We also nail down the vocabulary mathematicians use when they talk about functions, define what it means to compose two functions, and explore what it means for a function to be symmetric. Most of this will be a refresher, but it is a useful reference: a lot of the algebra moves you will see in calculus come straight out of this section.
1.1.1 Functions
Think of a function like a strict cafeteria worker: every student (input) gets exactly one entrée (output). Two students might be handed the same dish, but no single student walks away with two entrées. That "exactly one" rule is what separates a function from any old relation.
A function \(f\) 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, take the function \(f\) whose domain is all real numbers and whose rule is "square the input." Then \(f(3) = 3^2 = 9\). Every nonnegative real number is the square of some real number, so every nonnegative number lives in the range. No real number squares to a negative value, so negative numbers are excluded. The range is therefore the set of nonnegative reals.
For a general function \(f\) with domain \(D\), we usually use \(x\) for the input and \(y\) for the output paired with \(x\). We call \(x\) the independent variable and \(y\) the dependent variable (because \(y\) depends on \(x\)). In function notation, we write \(y = f(x)\) and read it as "y equals f of x." For the squaring function above, we write \(f(x) = x^2\).
We can picture the idea of a function using Figure 1.2, Figure 1.3, and Figure 1.4.
Figure 1.2 — A function can be visualized as an input/output device.
Figure 1.3 — A function maps every element in the domain to exactly one element in the range. Although each input can be sent to only one output, two different inputs can be sent to the same output.
Figure 1.4 — In this case, a graph of a function \(f\) has a domain of \(\{1, 2, 3\}\) and a range of \(\{1, 2\}\). The independent variable is \(x\) and the dependent variable is \(y\).
We can also visualize a function by plotting points \((x, y)\) in the coordinate plane, where \(y = f(x)\). The graph of a function is the set of all such points. For example, take the function \(f\) with domain \(D = \{1, 2, 3\}\) and rule \(f(x) = 3 - x\). Figure 1.5 shows its graph.
Figure 1.5 — Here we see a graph of the function \(f\) with domain \(\{1, 2, 3\}\) and rule \(f(x) = 3 - x\). The graph consists of the points \((x, f(x))\) for all \(x\) in the domain.
Every function has a domain. But sometimes a function is given by an equation like \(f(x) = x^2\) without any domain specified. In that case we use the natural domain: the set of all real numbers \(x\) for which \(f(x)\) gives back a real number. Any real number can be squared, so the natural domain of \(f(x) = x^2\) is all real numbers. The square-root function \(f(x) = \sqrt{x}\), on the other hand, gives a real output only when \(x\) is nonnegative, so its natural domain is \(\{x \mid x \ge 0\}\).
For the functions \(f(x) = x^2\) and \(f(x) = \sqrt{x}\), the domains are infinite sets — we cannot list every element. Two compact ways to describe infinite sets of real numbers are set-builder notation and interval notation. In set-builder notation, using \(\mathbb{R}\) for the set of all real numbers, we write
$$ \{x \mid x \text{ has some property}\} $$and read it as "the set of real numbers \(x\) such that \(x\) has some property." For the real numbers strictly between \(1\) and \(5\), we would write
$$ \{x \mid 1 < x < 5\}. $$That same set can be written using interval notation as \((a, b)\) for "all numbers between \(a\) and \(b\), exclusive." So
$$ (1, 5) = \{x \mid 1 < x < 5\}. $$The numbers \(1\) and \(5\) are the endpoints of this set. If we want to include the endpoints, we write
$$ [1, 5] = \{x \mid 1 \le x \le 5\}. $$We can include one endpoint and not the other using a mix of brackets and parentheses. To describe the set of nonnegative reals — no upper bound — we use the symbol \(\infty\) (positive infinity):
$$ [0, \infty) = \{x \mid 0 \le x\}. $$Important: \(\infty\) is not a real number; it is a symbol that says "this set keeps going forever in the positive direction." Similarly, for all nonpositive numbers we write
$$ (-\infty, 0] = \{x \mid x \le 0\}. $$The notation \(-\infty\) means negative infinity, indicating we include all numbers no matter how small. The set of all real numbers is
$$ (-\infty, \infty) = \{x \mid x \text{ is any real number}\}. $$Summary of interval notation. The table below collects the forms introduced above for quick reference.
| Interval | Set-builder | In words |
|---|---|---|
| \((a, b)\) | \(\{x \mid a < x < b\}\) | Between \(a\) and \(b\), endpoints excluded |
| \([a, b]\) | \(\{x \mid a \le x \le b\}\) | Between \(a\) and \(b\), endpoints included |
| \([a, b)\) | \(\{x \mid a \le x < b\}\) | Includes \(a\), excludes \(b\) |
| \((a, b]\) | \(\{x \mid a < x \le b\}\) | Excludes \(a\), includes \(b\) |
| \([a, \infty)\) | \(\{x \mid a \le x\}\) | \(a\) and everything above |
| \((-\infty, b]\) | \(\{x \mid x \le b\}\) | \(b\) and everything below |
| \((-\infty, \infty)\) | \(\{x \mid x \in \mathbb{R}\}\) | All real numbers |
Some functions are defined by different equations on different parts of their domain. These are called piecewise-defined functions. For example, suppose we want a function \(f\) on all of \(\mathbb{R}\) such that \(f(x) = 3x + 1\) when \(x \ge 2\) and \(f(x) = x^2\) when \(x < 2\). We write
$$ f(x) = \begin{cases} 3x + 1 & x \ge 2 \\ x^2 & x < 2 \end{cases}. $$To evaluate, we pick the equation that matches our input. Since \(5 \ge 2\), we use \(f(x) = 3x + 1\) and get \(f(5) = 3(5) + 1 = 16\). For \(x = -1 < 2\), we use \(f(x) = x^2\) and get \(f(-1) = 1\).
Given two sets \(A\) and \(B\), a set whose elements are ordered pairs \((x, y)\) — where \(x\) is from \(A\) and \(y\) is from \(B\) — is called a relation from \(A\) to \(B\). A relation simply describes some kind of pairing between the two sets. A function is a stricter kind of relation: each element of the first set is paired with exactly one element of the second set. We call the element from the first set the input and the element from the second set the output.
Functions are everywhere in math because they capture the idea that knowing one quantity lets you pin down another. The area of a square depends on its side length, so we say area is a function of side length. The velocity of a ball thrown in the air is a function of how long it has been in the air. The cost of mailing a package is a function of the package's weight. To study these relationships precisely, we need precise vocabulary.
For \(f(x) = x^2 - 3x + 5\), evaluate \(f(1)\) and \(f(a + h)\).
Solution
\(f(1) = (1)^2 - 3(1) + 5 = 1 - 3 + 5 = 3\).
For \(f(a + h)\), substitute \(a + h\) wherever \(x\) appears, then expand:
$$ \begin{aligned} f(a + h) &= (a + h)^2 - 3(a + h) + 5 \\ &= a^2 + 2ah + h^2 - 3a - 3h + 5. \end{aligned} $$Answer: \(f(1) = 3\); \(f(a + h) = a^2 + 2ah + h^2 - 3a - 3h + 5\).
For the function \(f(x) = 3x^2 + 2x - 1\), evaluate:
- 1. \(f(-2)\)
- 2. \(f(\sqrt{2})\)
- 3. \(f(a + h)\)
Solution
Substitute the given value for \(x\) into the formula for \(f(x)\).
1. \(f(-2) = 3(-2)^2 + 2(-2) - 1 = 12 - 4 - 1 = 7\).
2. \(f(\sqrt{2}) = 3(\sqrt{2})^2 + 2\sqrt{2} - 1 = 6 + 2\sqrt{2} - 1 = 5 + 2\sqrt{2}\).
3. Expand carefully — \((a + h)^2 = a^2 + 2ah + h^2\):
$$ \begin{aligned} f(a + h) &= 3(a + h)^2 + 2(a + h) - 1 \\ &= 3(a^2 + 2ah + h^2) + 2a + 2h - 1 \\ &= 3a^2 + 6ah + 3h^2 + 2a + 2h - 1. \end{aligned} $$Answer: \(f(-2) = 7\), \(f(\sqrt{2}) = 5 + 2\sqrt{2}\), \(f(a + h) = 3a^2 + 6ah + 3h^2 + 2a + 2h - 1\).
For each function, determine the (i) domain and (ii) range.
- 1. \(f(x) = (x - 4)^2 + 5\)
- 2. \(f(x) = \sqrt{3x + 2} - 1\)
- 3. \(f(x) = \dfrac{3}{x - 2}\)
Solution
1. \(f(x) = (x - 4)^2 + 5\).
Domain: \((x - 4)^2 + 5\) is a real number for every real \(x\), so the domain is \((-\infty, \infty)\).
Range: \((x - 4)^2 \ge 0\), so \(f(x) \ge 5\). The range sits inside \(\{y \mid y \ge 5\}\). To confirm every \(y \ge 5\) is hit, solve \((x - 4)^2 + 5 = y\):
$$ (x - 4)^2 = y - 5 \quad\Rightarrow\quad x - 4 = \pm\sqrt{y - 5}. $$Since \(y \ge 5\), the square root is defined, and \(x = 4 \pm \sqrt{y - 5}\) gives valid inputs. So the range is \(\{y \mid y \ge 5\}\).
2. \(f(x) = \sqrt{3x + 2} - 1\).
Domain: We need \(3x + 2 \ge 0\), so \(x \ge -\tfrac{2}{3}\). Domain: \(\{x \mid x \ge -\tfrac{2}{3}\}\).
Range: \(\sqrt{3x + 2} \ge 0\), so \(f(x) \ge -1\). For any \(y \ge -1\), set \(\sqrt{3x + 2} - 1 = y\). Then \(\sqrt{3x + 2} = y + 1\); squaring, \(3x + 2 = (y + 1)^2\), so
$$ x = \tfrac{1}{3}(y + 1)^2 - \tfrac{2}{3}. $$This value is at least \(-\tfrac{2}{3}\), so it lies in the domain. Range: \(\{y \mid y \ge -1\}\).
3. \(f(x) = \dfrac{3}{x - 2}\).
Domain: The denominator is nonzero whenever \(x \ne 2\). Domain: \(\{x \mid x \ne 2\}\).
Range: Solve \(\dfrac{3}{x - 2} = y\) for \(x\): \(x = \dfrac{3}{y} + 2\). As long as \(y \ne 0\), such an \(x\) exists. Range: \(\{y \mid y \ne 0\}\).
Answer: 1. Domain \((-\infty, \infty)\); range \(\{y \mid y \ge 5\}\). 2. Domain \(\{x \mid x \ge -\tfrac{2}{3}\}\); range \(\{y \mid y \ge -1\}\). 3. Domain \(\{x \mid x \ne 2\}\); range \(\{y \mid y \ne 0\}\).
1.1.2 Representing Functions
We say that a function \(f\) is increasing on the interval \(I\) if for all \(x_1, x_2 \in I\),
$$ f(x_1) \le f(x_2) \text{ when } x_1 < x_2. $$We say \(f\) is strictly increasing on \(I\) if for all \(x_1, x_2 \in I\),
$$ f(x_1) < f(x_2) \text{ when } x_1 < x_2. $$We say that a function \(f\) is decreasing on \(I\) if for all \(x_1, x_2 \in I\),
$$ f(x_1) \ge f(x_2) \text{ if } x_1 < x_2. $$We say that \(f\) is strictly decreasing on \(I\) if for all \(x_1, x_2 \in I\),
$$ f(x_1) > f(x_2) \text{ if } x_1 < x_2. $$Plain-English unpacking. Walk left to right along the graph. If the height never drops as you move right, the function is increasing; if it always strictly rises, it is strictly increasing. If the height never rises, it is decreasing; if it always strictly falls, it is strictly decreasing. The word "strictly" rules out flat stretches.
For example, \(f(x) = 3x\) is increasing on \((-\infty, \infty)\) because \(3x_1 < 3x_2\) whenever \(x_1 < x_2\). The function \(f(x) = -x^3\) is strictly decreasing on \((-\infty, \infty)\) because \(-x_1^3 > -x_2^3\) whenever \(x_1 < x_2\) (Figure 1.11).
Figure 1.11 — (a) The function \(f(x) = 3x\) is increasing on the interval \((-\infty, \infty)\). (b) The function \(f(x) = -x^3\) is decreasing on the interval \((-\infty, \infty)\).
Typically a function is represented using one or more of the following tools:
- A table - A graph - A formula
We can recognize a function in any of these forms, and they often work better together — we plot points from a table, or build a table from a formula and then graph it.
Tables
Functions described by a table show up constantly in real-world data. Suppose we record the outside temperature every hour for a 24-hour period starting at midnight. Let the input \(x\) be the number of hours past midnight, and the output \(y\) be the temperature in degrees Fahrenheit at that time. We can summarize the data in Table 1.1.
Functions are how we talk about cause-and-effect with numbers. Drop a coin into a vending machine, push a button, get a soda — that vending machine is a function: input (your button choice) goes in, exactly one output (your soda) comes out. Throughout calculus, every rate of change, every area under a curve, every optimization problem you will solve starts with a function. Getting comfortable here pays dividends for the entire course.
Table 1.1 — Temperature (°F) as a function of hours after midnight.
| Hours after midnight \(x\) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Temperature °F \(y\) | 58 | 55 | 53 | 52 | 52 | 53 | 55 | 60 | 64 | 70 | 75 | 78 |
| Hours after midnight \(x\) | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Temperature °F \(y\) | 80 | 85 | 85 | 84 | 83 | 80 | 77 | 73 | 69 | 65 | 60 | 58 |
We can see from the table that temperature is a function of time. The temperature decreases, then increases, then decreases again. Without a graph, though, it is hard to picture the overall shape of the function.
Graphs
Given a function \(f\) described by a table, we can give a visual picture of \(f\) by plotting its points on a graph. Graphing the temperatures from Table 1.1 makes the pattern of the day much clearer. Figure 1.6 shows the plot.
Figure 1.6 — The graph of the data from Table 1.1 shows temperature as a function of time.
From the plotted points we can imagine the general shape of the graph. It is often useful to connect the dots. We can't say for certain what the temperature was between recorded hours, but with this many data points and a clear trend, it is reasonable to suspect the in-between values followed a similar curve, as shown in Figure 1.7.
Figure 1.7 — Connecting the dots in Figure 1.6 shows the general pattern of the data.
Algebraic Formulas
Sometimes a function comes to us not as a table but as an explicit formula. Formulas crop up in almost every applied setting. The area of a circle with radius \(r\) is \(A(r) = \pi r^2\). When an object is thrown upward from the ground with initial velocity \(v_0\) ft/s, its height above the ground until it lands is \(s(t) = -16t^2 + v_0 t\). When \(P\) dollars are invested at an annual rate \(r\) compounded continuously, the account balance after \(t\) years is \(A(t) = P e^{rt}\). Algebraic formulas are useful for computing exact values; we also like to picture them as graphs.
Why bother with all three representations — table, graph, formula? Because each one is good at a different thing. Tables are honest about real data. Graphs make shape and trend pop out. Formulas let you predict values you never measured. A practicing engineer or scientist switches between them all day long.
Given an algebraic formula for a function \(f\), the graph of \(f\) is the set of points \((x, f(x))\) for all \(x\) in the domain. To graph by hand, start by building a small table of inputs and outputs. The domain may be infinite, but listing a handful of strategic points is a great start.
When building that table, we usually check whether zero is ever an output. The values of \(x\) where \(f(x) = 0\) are called the zeros of a function. For example, the zeros of \(f(x) = x^2 - 4\) are \(x = \pm 2\). Zeros are exactly the places where the graph of \(f\) crosses the \(x\)-axis. A graph might never cross the \(x\)-axis, or cross it many times — even infinitely often.
Another point worth checking is the \(y\)-intercept, if it exists. The \(y\)-intercept is the point \((0, f(0))\).
A function has at most one \(y\)-intercept: if \(x = 0\) is in the domain there is exactly one, and if not, there is none. More generally, for any real \(c\) in the domain, there is exactly one output \(f(c)\), and the vertical line \(x = c\) meets the graph exactly once. If \(c\) is not in the domain, the vertical line \(x = c\) does not meet the graph at all. This observation gives a quick visual test:
Rule: Vertical Line Test
A set of points in the plane is the graph of a function of \(x\) if and only if every vertical line intersects the set at most once.
We can use this test on a picture to decide whether a set of plotted points is the graph of a function (Figure 1.8).
Figure 1.8 — (a) The set of plotted points represents the graph of a function because every vertical line intersects the set of points, at most, once. (b) The set of plotted points does not represent the graph of a function because some vertical lines intersect the set of points more than once.
Find the domain and range for \(f(x) = \sqrt{4 - 2x} + 5\).
Solution
Domain: We need \(4 - 2x \ge 0\), so \(x \le 2\). Domain: \(\{x \mid x \le 2\}\), or \((-\infty, 2]\).
Range: \(\sqrt{4 - 2x} \ge 0\), so \(f(x) \ge 5\). For any target \(y \ge 5\), solve \(\sqrt{4 - 2x} + 5 = y\) for \(x\):
$$ \sqrt{4 - 2x} = y - 5 \;\Rightarrow\; 4 - 2x = (y - 5)^2 \;\Rightarrow\; x = 2 - \tfrac{1}{2}(y - 5)^2. $$This \(x\) is at most \(2\), so it lies in the domain. Range: \(\{y \mid y \ge 5\}\), or \([5, \infty)\).
Answer: Domain \((-\infty, 2]\); range \([5, \infty)\).
When tracing a function's graph from left to right, its output either rises, falls, or stays flat as the input increases. The following definition captures these two behaviors precisely.
Find the zeros of \(f(x) = x^3 - 5x^2 + 6x\).
Solution
Factor: \(x^3 - 5x^2 + 6x = x(x^2 - 5x + 6) = x(x - 2)(x - 3)\). Set each factor to zero.
Answer: \(x = 0,\ 2,\ 3\).
Consider the function \(f(x) = -4x + 2\).
- 1. Find all zeros of \(f\).
- 2. Find the \(y\)-intercept (if any).
- 3. Sketch a graph of \(f\).
Solution
1. Solve \(-4x + 2 = 0\): \(x = \tfrac{1}{2}\). So \(f\) has one zero at \(x = \tfrac{1}{2}\).
2. The \(y\)-intercept is \((0, f(0)) = (0, 2)\).
3. \(f\) is a linear function passing through \((\tfrac{1}{2}, 0)\) and \((0, 2)\):
Answer: Zero at \(x = \tfrac{1}{2}\); \(y\)-intercept at \((0, 2)\); the graph is the line through those two points.
Consider the function \(f(x) = \sqrt{x + 3} + 1\).
- 1. Find all zeros of \(f\).
- 2. Find the \(y\)-intercept (if any).
- 3. Sketch a graph of \(f\).
Solution
1. Solve \(\sqrt{x + 3} + 1 = 0\), i.e. \(\sqrt{x + 3} = -1\). Since \(\sqrt{x + 3} \ge 0\) for every \(x\) in the domain, this has no solution. So \(f\) has no zeros.
2. \(y\)-intercept: \((0, f(0)) = (0, \sqrt{3} + 1)\).
3. Build a table. We need \(x + 3 \ge 0\), i.e. \(x \ge -3\). Pick \(x\) values that make the square root clean:
| \(x\) | \(-3\) | \(-2\) | \(1\) |
|---|---|---|---|
| \(f(x)\) | \(1\) | \(2\) | \(3\) |
Table 1.2
The graph is a shifted square-root curve:
Answer: No zeros; \(y\)-intercept \((0, \sqrt{3} + 1)\); graph is the standard \(\sqrt{x}\) curve shifted left 3 and up 1.
If a ball is dropped from a height of \(100\) ft, its height \(s\) at time \(t\) is given by \(s(t) = -16t^2 + 100\), where \(s\) is in feet and \(t\) is in seconds. The domain is restricted to \([0, c]\), where \(t = 0\) is when the ball is dropped and \(t = c\) is when it hits the ground.
- 1. Create a table showing \(s(t)\) at \(t = 0, 0.5, 1, 1.5, 2,\) and \(2.5\). From the table, determine the domain — that is, find the time \(c\) when the ball hits the ground.
- 2. Sketch a graph of \(s\).
Solution
1. Plug each \(t\) into \(-16t^2 + 100\):
| \(t\) | \(0\) | \(0.5\) | \(1\) | \(1.5\) | \(2\) | \(2.5\) |
|---|---|---|---|---|---|---|
| \(s(t)\) | \(100\) | \(96\) | \(84\) | \(64\) | \(36\) | \(0\) |
Table 1.3 — Height \(s\) as a function of time \(t\).
The ball hits the ground when \(s(t) = 0\), at \(t = 2.5\). So the domain is \([0, 2.5]\).
2.
![Graph of s(t) = -16t² + 100, the height of a free-falling ball on [0, 2.5].](../images/figure_1.1.inline_10_light.png?v=20260518a)
Figure — Graph of \(s(t) = -16t^2 + 100\), the height of a free-falling ball, on the interval \([0, 2.5]\).
Answer: Domain \([0, 2.5]\); graph is the portion of \(s = -16t^2 + 100\) from \(t = 0\) to \(t = 2.5\).
1.1.3 Combining Functions
Consider the function \(f\) with domain \(A\) and range \(B\), and the function \(g\) with domain \(D\) and range \(E\). If \(B\) is a subset of \(D\), then the composite function \((g \circ f)(x)\) is the function with domain \(A\) such that
$$ (g \circ f)(x) = g(f(x)). $$Plain-English unpacking. Two steps:
- 1. \(f\) takes \(x\) in its domain to \(f(x)\) in its range.
- 2. Because the range of \(f\) lives inside the domain of \(g\), we can hand \(f(x)\) to \(g\) and get \(g(f(x))\) in the range of \(g\).
That two-step pipeline is what we mean by "composition." Figure 1.12 (in the Function Composition subsection below) visualizes it.
Now that we have reviewed the basics of functions, we can ask what happens when we glue functions together to make new ones. For example, if a company's cost to make \(x\) items is \(C(x)\) and its revenue from selling those \(x\) items is \(R(x)\), then profit is \(P(x) = R(x) - C(x)\). Subtraction of two functions produced a brand-new function.
Another way to build a new function is to compose two existing ones — feed the output of one into the input of the other. Given \(f(x) = x^2\) and \(g(x) = 3x + 1\), the composite \(f \circ g\) is defined by
$$ (f \circ g)(x) = f(g(x)) = (g(x))^2 = (3x + 1)^2. $$Compose the other way and we get a different function:
$$ (g \circ f)(x) = g(f(x)) = 3 f(x) + 1 = 3x^2 + 1. $$Note carefully: \(f \circ g\) and \(g \circ f\) are not the same. Order matters.
Combining Functions with Mathematical Operators
To combine functions using arithmetic, just write them with the operator between them and simplify. Given functions \(f\) and \(g\), we get four new functions:
$$ \begin{aligned} (f + g)(x) &= f(x) + g(x) && \text{(Sum)} \\ (f - g)(x) &= f(x) - g(x) && \text{(Difference)} \\ (f \cdot g)(x) &= f(x)\, g(x) && \text{(Product)} \\ \left(\tfrac{f}{g}\right)\!(x) &= \dfrac{f(x)}{g(x)} \text{ for } g(x) \ne 0 && \text{(Quotient)} \end{aligned} $$Combining functions is just combining their outputs, point by point. If at \(x = 4\) we have \(f(4) = 7\) and \(g(4) = 2\), then \((f + g)(4) = 9\), \((f \cdot g)(4) = 14\), and \((f / g)(4) = 3.5\). The variable is along for the ride; the arithmetic happens to the outputs.
For the sum, difference, and product, we just need both \(f\) and \(g\) to accept the input — addition, subtraction, and multiplication never fail on individual numbers. So the domain is exactly the inputs where both \(f\) and \(g\) are defined.
The quotient is the picky one. It also needs both functions defined, but it has one extra rule we can never break: dividing by zero is not allowed. So even if \(f\) is perfectly happy at some \(x\), we still throw that \(x\) out if \(g(x) = 0\). That's why \((f/g)\) is the only combination that needs an extra domain check — division is the only operation that can fail on individual numbers.
To actually compute a combined function at some input \(x\), we just go in steps. Plug \(x\) into \(f\) and get the number \(f(x)\). Plug \(x\) into \(g\) and get the number \(g(x)\). Then add, subtract, multiply, or divide those two outputs — whichever combination we want. To get a single formula that works for all valid inputs, we keep \(x\) as a variable and do the same arithmetic on the expressions. The result might look longer than either of the originals, but at any specific input it's still the same thing: two outputs, one arithmetic step.
For \(f(x) = x^2 + 3\) and \(g(x) = 2x - 5\), find \((f / g)(x)\) and state its domain.
Solution
\((f/g)(x) = \dfrac{f(x)}{g(x)} = \dfrac{x^2 + 3}{2x - 5}\). The denominator is zero when \(2x - 5 = 0\), i.e. \(x = \tfrac{5}{2}\), so we exclude that one value.
Answer: \((f/g)(x) = \dfrac{x^2 + 3}{2x - 5}\), domain \(\{x \mid x \ne \tfrac{5}{2}\}\).
Given \(f(x) = 2x - 3\) and \(g(x) = x^2 - 1\), find each new function and state its domain.
- 1. \((f + g)(x)\)
- 2. \((f - g)(x)\)
- 3. \((f \cdot g)(x)\)
- 4. \(\left(\dfrac{f}{g}\right)\!(x)\)
Solution
1. \((f + g)(x) = (2x - 3) + (x^2 - 1) = x^2 + 2x - 4\). Domain: \((-\infty, \infty)\).
2. \((f - g)(x) = (2x - 3) - (x^2 - 1) = -x^2 + 2x - 2\). Domain: \((-\infty, \infty)\).
3. \((f \cdot g)(x) = (2x - 3)(x^2 - 1) = 2x^3 - 3x^2 - 2x + 3\). Domain: \((-\infty, \infty)\).
4. \(\left(\dfrac{f}{g}\right)\!(x) = \dfrac{2x - 3}{x^2 - 1}\). Since \(x^2 - 1 = 0\) when \(x = \pm 1\), domain: \(\{x \mid x \ne \pm 1\}\).
Answer: sum \(x^2 + 2x - 4\); difference \(-x^2 + 2x - 2\); product \(2x^3 - 3x^2 - 2x + 3\); quotient \(\dfrac{2x - 3}{x^2 - 1}\) with \(x \ne \pm 1\).
Function Composition
Figure 1.12 — For the composite function \(g \circ f\), we have \((g \circ f)(1) = 4\), \((g \circ f)(2) = 5\), and \((g \circ f)(3) = 4\).
When we compose functions, we take a function of a function. For example, suppose the temperature \(T\) on a given day is a function of time \(t\) (in hours past midnight), as in Table 1.1. And suppose the cost \(C\) to heat or cool a building for an hour is a function of the temperature \(T\). Combining these, we describe the cost of climate-controlling the building as a function of time by evaluating \(C(T(t))\). This new function is written \(C \circ T\), defined by \((C \circ T)(t) = C(T(t))\) for every \(t\) in the domain of \(T\). It is called a composite function. We note: cost is a function of temperature, and temperature is a function of time, so \(C \circ T\) makes sense. But \(T \circ C\) does not — temperature is not a function of cost.
Let \(f(x) = 2 - 5x\) and \(g(x) = \sqrt{x}\). Find \((f \circ g)(x)\).
Solution
\((f \circ g)(x) = f(g(x)) = f(\sqrt{x}) = 2 - 5\sqrt{x}\). The domain comes from \(g\), so \(x \ge 0\).
Answer: \((f \circ g)(x) = 2 - 5\sqrt{x}\), domain \([0, \infty)\).
Consider \(f(x) = x^2 + 1\) and \(g(x) = 1/x\).
- 1. Find \((g \circ f)(x)\) and state its domain and range.
- 2. Evaluate \((g \circ f)(4)\) and \((g \circ f)(-1/2)\).
- 3. Find \((f \circ g)(x)\) and state its domain and range.
- 4. Evaluate \((f \circ g)(4)\) and \((f \circ g)(-1/2)\).
Solution
1. \((g \circ f)(x) = g(f(x)) = g(x^2 + 1) = \dfrac{1}{x^2 + 1}\).
Since \(x^2 + 1 \ne 0\) for every real \(x\), domain: all real numbers, \((-\infty, \infty)\).
For the range: \(x^2 + 1 \ge 1\), so \(0 < \dfrac{1}{x^2 + 1} \le 1\). Range sits inside \((0, 1]\). To confirm every value in \((0, 1]\) is hit, solve \(\dfrac{1}{x^2 + 1} = y\): \(x^2 + 1 = \dfrac{1}{y}\), so \(x = \pm\sqrt{\dfrac{1}{y} - 1}\). For \(y \in (0, 1]\), the radical is nonnegative, so a real \(x\) exists. Range: \((0, 1]\).
2. \((g \circ f)(4) = g(4^2 + 1) = g(17) = \dfrac{1}{17}\).
\((g \circ f)\!\left(-\tfrac{1}{2}\right) = g\!\left((-\tfrac{1}{2})^2 + 1\right) = g\!\left(\tfrac{5}{4}\right) = \dfrac{4}{5}\).
3. \((f \circ g)(x) = f(g(x)) = f\!\left(\dfrac{1}{x}\right) = \left(\dfrac{1}{x}\right)^2 + 1\).
Domain: all real \(x\) with \(x \ne 0\). For the range, set \(\left(\dfrac{1}{x}\right)^2 + 1 = y\); then \(\left(\dfrac{1}{x}\right)^2 = y - 1\), so \(\dfrac{1}{x} = \pm\sqrt{y - 1}\) and finally
$$ x = \pm\dfrac{1}{\sqrt{y - 1}}. $$This is real if and only if \(y > 1\). Range: \(\{y \mid y > 1\}\).
4. \((f \circ g)(4) = f\!\left(\tfrac{1}{4}\right) = \left(\tfrac{1}{4}\right)^2 + 1 = \tfrac{17}{16}\).
\((f \circ g)\!\left(-\tfrac{1}{2}\right) = f(-2) = (-2)^2 + 1 = 5\).
Answer: \((g \circ f)(x) = \dfrac{1}{x^2 + 1}\), domain \((-\infty, \infty)\), range \((0, 1]\); \((f \circ g)(x) = \dfrac{1}{x^2} + 1\), domain \(\{x \ne 0\}\), range \(\{y > 1\}\); specific evaluations as above.
Example 1.1.7 confirms in concrete terms that \((f \circ g)(x) \ne (g \circ f)(x)\) in general — order of composition matters.
If items are on sale for \(10\%\) off their original price, and a customer has a coupon for an additional \(30\%\) off, what will be the final price for an item that is originally \(x\) dollars, after applying the coupon to the sale price?
Solution
Sale price: \(f(x) = 0.90 x\). Post-coupon price: \(g(y) = 0.70 y\). Composing,
$$ g(f(x)) = 0.70 \cdot (0.90 x) = 0.63 x. $$Answer: \(0.63 x\) dollars — a \(37\%\) total discount.
Consider the functions \(f\) and \(g\) described by Table 1.4 and Table 1.5.
- 1. Evaluate \((g \circ f)(3)\) and \((g \circ f)(0)\).
- 2. State the domain and range of \((g \circ f)(x)\).
- 3. Evaluate \((f \circ f)(3)\) and \((f \circ f)(1)\).
- 4. State the domain and range of \((f \circ f)(x)\).
Solution
1. \((g \circ f)(3) = g(f(3)) = g(-2) = 0\). \((g \circ f)(0) = g(f(0)) = g(4) = 5\).
2. Domain of \(g \circ f\): \(\{-3, -2, -1, 0, 1, 2, 3, 4\}\). Since the range of \(f\) is \(\{-2, 0, 2, 4\}\), the range of \(g \circ f\) is \(\{0, 3, 5\}\).
3. \((f \circ f)(3) = f(f(3)) = f(-2) = 4\). \((f \circ f)(1) = f(f(1)) = f(-2) = 4\).
4. Domain of \(f \circ f\): \(\{-3, -2, -1, 0, 1, 2, 3, 4\}\). Range of \(f\) is \(\{-2, 0, 2, 4\}\), so range of \(f \circ f\) is \(\{0, 4\}\).
Answer: Results as listed in each part above.
A store advertises a sale of \(20\%\) off all merchandise. Caroline has a coupon for an additional \(15\%\) off any item, including sale merchandise. If Caroline buys an item with an original price of \(x\) dollars, how much will she pay after applying the coupon to the sale price? Solve using a composite function.
Solution
The sale price is \(20\%\) off the original, so if the original price is \(x\), the sale price is
$$ f(x) = 0.80 x. $$The coupon takes \(15\%\) off whatever price is shown, so if the shown price is \(y\), the post-coupon price is
$$ g(y) = 0.85 y. $$Applying the coupon to the sale price means evaluating \(g(f(x))\):
$$ g(f(x)) = 0.85 \cdot (0.80 x) = 0.68 x. $$Answer: Caroline pays \(0.68 x\) dollars — a \(32\%\) total discount.
1.1.4 Symmetry of Functions
If \(f(-x) = f(x)\) for all \(x\) in the domain of \(f\), then \(f\) is an even function. An even function is symmetric about the \(y\)-axis.
If \(f(-x) = -f(x)\) for all \(x\) in the domain of \(f\), then \(f\) is an odd function. An odd function is symmetric about the origin.
"Even" and "odd" come from polynomial behavior. Pure even powers like \(x^2, x^4, x^6\) are even functions (\((-x)^n = x^n\) when \(n\) is even). Pure odd powers like \(x, x^3, x^5\) are odd functions (\((-x)^n = -x^n\) when \(n\) is odd). Add constants or mixed terms and you can break either property — which is exactly why most polynomials are neither.
The graphs of certain functions have symmetry properties that help us understand the shape of the curve. Consider \(f(x) = x^4 - 2x^2 - 3\) in Figure 1.13(a). Take the part of the curve to the right of the \(y\)-axis, flip it across the \(y\)-axis, and it lays exactly on top of the part to the left of the \(y\)-axis. We say the function is symmetric about the \(y\)-axis. Now consider \(f(x) = x^3 - 4x\) in Figure 1.13(b). Rotate the graph \(180^\circ\) about the origin and it looks identical. We say the function is symmetric about the origin.
Figure 1.13 — (a) A graph that is symmetric about the \(y\)-axis. (b) A graph that is symmetric about the origin.
We can spot these symmetries from a graph easily — but how can we detect them algebraically, with no picture in hand? Look at Figure 1.13 again: since \(f\) is symmetric about the \(y\)-axis, whenever \((x, y)\) is on the graph, so is \((-x, y)\). In other words, \(f(-x) = f(x)\). A function with this property is called an even function. For example, \(f(x) = x^2\) is even because
$$ f(-x) = (-x)^2 = x^2 = f(x). $$By contrast, if \(f\) is symmetric about the origin, whenever \((x, y)\) is on the graph, so is \((-x, -y)\). In other words, \(f(-x) = -f(x)\). A function with this property is called an odd function. For example, \(f(x) = x^3\) is odd because
$$ f(-x) = (-x)^3 = -x^3 = -f(x). $$Determine whether \(f(x) = 4x^3 - 5x\) is even, odd, or neither.
Solution
\(f(-x) = 4(-x)^3 - 5(-x) = -4x^3 + 5x = -(4x^3 - 5x) = -f(x)\).
Answer: \(f\) is odd.
Determine whether each function is even, odd, or neither.
- 1. \(f(x) = -5x^4 + 7x^2 - 2\)
- 2. \(f(x) = 2x^5 - 4x + 5\)
- 3. \(f(x) = \dfrac{3x}{x^2 + 1}\)
Solution
To check, plug in \(-x\) and compare to \(f(x)\) and \(-f(x)\).
1. \(f(-x) = -5(-x)^4 + 7(-x)^2 - 2 = -5x^4 + 7x^2 - 2 = f(x)\). So \(f\) is even.
2. \(f(-x) = 2(-x)^5 - 4(-x) + 5 = -2x^5 + 4x + 5\). Compare: \(f(x) = 2x^5 - 4x + 5\) and \(-f(x) = -2x^5 + 4x - 5\). \(f(-x) \ne f(x)\) and \(f(-x) \ne -f(x)\), so \(f\) is neither.
3. \(f(-x) = \dfrac{3(-x)}{(-x)^2 + 1} = \dfrac{-3x}{x^2 + 1} = -\dfrac{3x}{x^2 + 1} = -f(x)\). So \(f\) is odd.
Answer: (1) even; (2) neither; (3) odd.
One symmetric function that comes up constantly is the absolute value function, written \(|x|\). It is piecewise-defined:
$$ f(x) = \begin{cases} -x, & x < 0 \\ x, & x \ge 0 \end{cases}. $$Some students describe this function by saying "it makes everything positive." That's almost right. From the definition: if \(x < 0\), then \(|x| = -x > 0\); if \(x > 0\), then \(|x| = x > 0\). But at \(x = 0\), \(|x| = 0\). So a more accurate description is: for every nonzero input the output is positive, and at zero the output is zero. The range is therefore \(\{y \mid y \ge 0\}\). Figure 1.14 shows that \(|x|\) is symmetric about the \(y\)-axis — so it is an even function.
Figure 1.14 — The graph of \(f(x) = |x|\) is symmetric about the \(y\)-axis.
For \(f(x) = |x + 2| - 4\), find the domain and range.
Solution
Domain: \(|x + 2|\) is defined for every real number, so the domain is \((-\infty, \infty)\).
Range: \(|x + 2| \ge 0\), so \(f(x) \ge -4\). For any target \(y \ge -4\), we can solve \(|x + 2| - 4 = y\) by taking \(x + 2 = \pm(y + 4)\), giving \(x = -2 \pm (y + 4)\). So every value \(y \ge -4\) is hit.
Answer: Domain \((-\infty, \infty)\); range \(\{y \mid y \ge -4\} = [-4, \infty)\).
Find the domain and range of \(f(x) = 2|x - 3| + 4\).
Solution
The absolute value function is defined for every real number, so the domain is \((-\infty, \infty)\).
Since \(|x - 3| \ge 0\) for every \(x\), we have \(f(x) = 2|x - 3| + 4 \ge 4\). So the range sits inside \(\{y \mid y \ge 4\}\). To check that every \(y \ge 4\) is actually achieved, solve \(2|x - 3| + 4 = y\):
$$ |x - 3| = \tfrac{1}{2}(y - 4). $$Since \(y \ge 4\), the right-hand side is nonnegative, so a solution can exist. Recall
$$ |x - 3| = \begin{cases} -(x - 3) & x < 3 \\ x - 3 & x \ge 3 \end{cases}. $$So we get two solutions:
$$ x = \pm \tfrac{1}{2}(y - 4) + 3. $$The range is therefore \(\{y \mid y \ge 4\}\).
Answer: Domain \((-\infty, \infty)\); range \(\{y \mid y \ge 4\}\), i.e. \([4, \infty)\).
Problem Set 1.1
Source: OpenStax Calculus Volume 1
For the following exercises, (a) determine the domain and the range of each relation, and (b) state whether the relation is a function.
Problem 1. Relation given by the table
| \(x\) | \(y\) |
|---|---|
| \(-3\) | \(9\) |
| \(-2\) | \(4\) |
| \(-1\) | \(1\) |
| \(0\) | \(0\) |
| \(1\) | \(1\) |
| \(2\) | \(4\) |
| \(3\) | \(9\) |
Problem 2. Relation given by the table
| \(x\) | \(y\) |
|---|---|
| \(-3\) | \(-2\) |
| \(-2\) | \(-8\) |
| \(-1\) | \(-1\) |
| \(0\) | \(0\) |
| \(1\) | \(1\) |
| \(2\) | \(8\) |
| \(3\) | \(-2\) |
Problem 3. Relation given by the table
| \(x\) | \(y\) |
|---|---|
| \(1\) | \(-3\) |
| \(2\) | \(-2\) |
| \(3\) | \(-1\) |
| \(0\) | \(0\) |
| \(1\) | \(1\) |
| \(2\) | \(2\) |
| \(3\) | \(3\) |
Problem 4. Relation given by the table
| \(x\) | \(y\) |
|---|---|
| \(1\) | \(1\) |
| \(2\) | \(1\) |
| \(3\) | \(1\) |
| \(4\) | \(1\) |
| \(5\) | \(1\) |
| \(6\) | \(1\) |
| \(7\) | \(1\) |
Problem 5. Relation given by the table
| \(x\) | \(y\) |
|---|---|
| \(3\) | \(3\) |
| \(5\) | \(2\) |
| \(8\) | \(1\) |
| \(10\) | \(0\) |
| \(15\) | \(1\) |
| \(21\) | \(2\) |
| \(33\) | \(3\) |
Problem 6. Relation given by the table
| \(x\) | \(y\) |
|---|---|
| \(-7\) | \(11\) |
| \(-2\) | \(5\) |
| \(-2\) | \(1\) |
| \(0\) | \(-1\) |
| \(1\) | \(-2\) |
| \(3\) | \(4\) |
| \(6\) | \(11\) |
a) \(f(0)\)
b) \(f(1)\)
c) \(f(3)\)
d) \(f(-x)\)
e) \(f(a)\)
f) \(f(a + h)\)
Solutions 1–6
Problem 1
Step 1 — Read off the domain:
The domain is the set of all \(x\)-values that appear in the table.
$$\text{Domain} = \{-3,\, -2,\, -1,\, 0,\, 1,\, 2,\, 3\}$$
Step 2 — Read off the range:
The range is the set of distinct \(y\)-values that appear in the table. The values \(9,\, 4,\, 1,\, 0,\, 1,\, 4,\, 9\) reduce (after removing duplicates) to:
$$\text{Range} = \{0,\, 1,\, 4,\, 9\}$$
Step 3 — Check the function test:
A relation is a function exactly when every input \(x\) is paired with only one output \(y\). Scanning the table, each \(x\)-value appears exactly once, so each input has a unique output.
Answer: Domain \(= \{-3,-2,-1,0,1,2,3\}\), Range \(= \{0,1,4,9\}\). The relation is a function.
Problem 2
Step 1 — Read off the domain:
$$\text{Domain} = \{-3,\, -2,\, -1,\, 0,\, 1,\, 2,\, 3\}$$
Step 2 — Read off the range:
The \(y\)-values are \(-2,\, -8,\, -1,\, 0,\, 1,\, 8,\, -2\). Removing the duplicate \(-2\):
$$\text{Range} = \{-8,\, -2,\, -1,\, 0,\, 1,\, 8\}$$
Step 3 — Check the function test:
Every \(x\)-value in the table appears exactly once. Even though the output \(-2\) is reused (for \(x = -3\) and \(x = 3\)), the function test only forbids one input pairing with two different outputs — repeating outputs is allowed.
Answer: Domain \(= \{-3,-2,-1,0,1,2,3\}\), Range \(= \{-8,-2,-1,0,1,8\}\). The relation is a function.
Problem 3
Step 1 — Read off the domain:
The \(x\)-values listed are \(1,\, 2,\, 3,\, 0,\, 1,\, 2,\, 3\). The set of distinct inputs is:
$$\text{Domain} = \{0,\, 1,\, 2,\, 3\}$$
Step 2 — Read off the range:
The \(y\)-values listed are \(-3,\, -2,\, -1,\, 0,\, 1,\, 2,\, 3\). All are distinct:
$$\text{Range} = \{-3,\, -2,\, -1,\, 0,\, 1,\, 2,\, 3\}$$
Step 3 — Check the function test:
Look for any \(x\)-value that pairs with two different \(y\)-values. The input \(x = 1\) appears twice: once with \(y = -3\) and once with \(y = 1\). One input is matched to two different outputs, which violates the definition of a function. (The inputs \(x = 2\) and \(x = 3\) also fail in the same way.)
Answer: Domain \(= \{0,1,2,3\}\), Range \(= \{-3,-2,-1,0,1,2,3\}\). The relation is not a function.
Problem 4
Step 1 — Read off the domain:
$$\text{Domain} = \{1,\, 2,\, 3,\, 4,\, 5,\, 6,\, 7\}$$
Step 2 — Read off the range:
Every \(y\)-value in the table equals \(1\), so the range is the single-element set:
$$\text{Range} = \{1\}$$
Step 3 — Check the function test:
Each \(x\)-value appears exactly once, paired with exactly one output (the constant value \(1\)). The function test allows different inputs to share the same output; it only forbids one input pairing with two outputs.
Answer: Domain \(= \{1,2,3,4,5,6,7\}\), Range \(= \{1\}\). The relation is a function (a constant function).
Problem 5
Step 1 — Read off the domain:
$$\text{Domain} = \{3,\, 5,\, 8,\, 10,\, 15,\, 21,\, 33\}$$
Step 2 — Read off the range:
The \(y\)-values are \(3,\, 2,\, 1,\, 0,\, 1,\, 2,\, 3\). Removing duplicates:
$$\text{Range} = \{0,\, 1,\, 2,\, 3\}$$
Step 3 — Check the function test:
Every \(x\)-value listed is distinct, so each input has exactly one output. The reused \(y\)-values (\(1,\, 2,\, 3\)) do not break the function rule — different inputs are allowed to share an output.
Answer: Domain \(= \{3,5,8,10,15,21,33\}\), Range \(= \{0,1,2,3\}\). The relation is a function.
Problem 6
Step 1 — Read off the domain:
The \(x\)-values are \(-7,\, -2,\, -2,\, 0,\, 1,\, 3,\, 6\). Removing the duplicate:
$$\text{Domain} = \{-7,\, -2,\, 0,\, 1,\, 3,\, 6\}$$
Step 2 — Read off the range:
The \(y\)-values are \(11,\, 5,\, 1,\, -1,\, -2,\, 4,\, 11\). Removing the duplicate \(11\):
$$\text{Range} = \{-2,\, -1,\, 1,\, 4,\, 5,\, 11\}$$
Step 3 — Check the function test:
The input \(x = -2\) is paired with two different outputs: \(y = 5\) and \(y = 1\). A single input cannot map to two outputs in a function, so this relation fails the test.
Answer: Domain \(= \{-7,-2,0,1,3,6\}\), Range \(= \{-2,-1,1,4,5,11\}\). The relation is not a function.
For the following exercises, find the values for each function, if they exist, then simplify.
Problem 7. \(f(x) = 5x - 2\)
Problem 8. \(f(x) = 4x^2 - 3x + 1\)
Problem 9. \(f(x) = \dfrac{2}{x}\)
Problem 10. \(f(x) = |x - 7| + 8\)
Problem 11. \(f(x) = \sqrt{6x + 5}\)
Problem 12. \(f(x) = \dfrac{x - 2}{3x + 7}\)
Problem 13. \(f(x) = 9\)
Solutions 7–13
Problem 7
Step 1 — Substitute each input into \(f(x) = 5x - 2\):
We evaluate \(f\) at each of the six requested inputs.
$$\begin{aligned}
f(0) &= 5(0) - 2 = -2 \\
f(1) &= 5(1) - 2 = 3 \\
f(3) &= 5(3) - 2 = 13 \\
f(-x) &= 5(-x) - 2 = -5x - 2 \\
f(a) &= 5a - 2 \\
f(a+h) &= 5(a + h) - 2 = 5a + 5h - 2
\end{aligned}$$
Step 2 — Simplify each result:
All expressions above are already fully simplified.
Answer: \(f(0) = -2,\ f(1) = 3,\ f(3) = 13,\ f(-x) = -5x - 2,\ f(a) = 5a - 2,\ f(a+h) = 5a + 5h - 2\).
Problem 8
Step 1 — Substitute each input into \(f(x) = 4x^2 - 3x + 1\):
$$\begin{aligned}
f(0) &= 4(0)^2 - 3(0) + 1 = 1 \\
f(1) &= 4(1)^2 - 3(1) + 1 = 4 - 3 + 1 = 2 \\
f(3) &= 4(3)^2 - 3(3) + 1 = 36 - 9 + 1 = 28
\end{aligned}$$
Step 2 — Evaluate \(f(-x)\):
Replace every \(x\) with \(-x\). Note \((-x)^2 = x^2\).
$$f(-x) = 4(-x)^2 - 3(-x) + 1 = 4x^2 + 3x + 1$$
Step 3 — Evaluate \(f(a)\):
$$f(a) = 4a^2 - 3a + 1$$
Step 4 — Evaluate \(f(a + h)\):
$$\begin{aligned}
f(a+h) &= 4(a+h)^2 - 3(a+h) + 1 \\
&= 4(a^2 + 2ah + h^2) - 3a - 3h + 1 \\
&= 4a^2 + 8ah + 4h^2 - 3a - 3h + 1
\end{aligned}$$
Answer: \(f(0) = 1,\ f(1) = 2,\ f(3) = 28,\ f(-x) = 4x^2 + 3x + 1,\ f(a) = 4a^2 - 3a + 1,\ f(a+h) = 4a^2 + 8ah + 4h^2 - 3a - 3h + 1\).
Problem 9
Step 1 — Note the domain restriction:
Because \(f(x) = \dfrac{2}{x}\) divides by \(x\), the value \(f(0)\) is undefined.
Step 2 — Substitute the remaining inputs:
$$\begin{aligned}
f(0) &\text{ does not exist} \\
f(1) &= \frac{2}{1} = 2 \\
f(3) &= \frac{2}{3} \\
f(-x) &= \frac{2}{-x} = -\frac{2}{x} \quad (x \neq 0) \\
f(a) &= \frac{2}{a} \quad (a \neq 0) \\
f(a+h) &= \frac{2}{a + h} \quad (a + h \neq 0)
\end{aligned}$$
Answer: \(f(0)\) does not exist; \(f(1) = 2,\ f(3) = \tfrac{2}{3},\ f(-x) = -\tfrac{2}{x},\ f(a) = \tfrac{2}{a},\ f(a+h) = \tfrac{2}{a+h}\).
Problem 10
Step 1 — Substitute each input into \(f(x) = |x - 7| + 8\):
Use \(|y| = y\) when \(y \ge 0\) and \(|y| = -y\) when \(y < 0\).
$$\begin{aligned}
f(0) &= |0 - 7| + 8 = 7 + 8 = 15 \\
f(1) &= |1 - 7| + 8 = 6 + 8 = 14 \\
f(3) &= |3 - 7| + 8 = 4 + 8 = 12
\end{aligned}$$
Step 2 — Evaluate \(f(-x)\):
$$f(-x) = |-x - 7| + 8 = |-(x + 7)| + 8 = |x + 7| + 8$$
Step 3 — Evaluate \(f(a)\) and \(f(a + h)\):
These cannot be simplified further without knowing the sign of the inside expression.
$$f(a) = |a - 7| + 8, \qquad f(a + h) = |a + h - 7| + 8$$
Answer: \(f(0) = 15,\ f(1) = 14,\ f(3) = 12,\ f(-x) = |x + 7| + 8,\ f(a) = |a - 7| + 8,\ f(a+h) = |a + h - 7| + 8\).
Problem 11
Step 1 — Substitute each input into \(f(x) = \sqrt{6x + 5}\):
The square root is defined only when \(6x + 5 \ge 0\), i.e. \(x \ge -\tfrac{5}{6}\).
$$\begin{aligned}
f(0) &= \sqrt{6(0) + 5} = \sqrt{5} \\
f(1) &= \sqrt{6(1) + 5} = \sqrt{11} \\
f(3) &= \sqrt{6(3) + 5} = \sqrt{23}
\end{aligned}$$
Step 2 — Evaluate \(f(-x)\):
$$f(-x) = \sqrt{6(-x) + 5} = \sqrt{-6x + 5} = \sqrt{5 - 6x}$$
This requires \(5 - 6x \ge 0\), i.e. \(x \le \tfrac{5}{6}\).
Step 3 — Evaluate \(f(a)\) and \(f(a + h)\):
$$f(a) = \sqrt{6a + 5}, \qquad f(a + h) = \sqrt{6(a + h) + 5} = \sqrt{6a + 6h + 5}$$
(Each requires the radicand to be non-negative.)
Answer: \(f(0) = \sqrt{5},\ f(1) = \sqrt{11},\ f(3) = \sqrt{23},\ f(-x) = \sqrt{5 - 6x},\ f(a) = \sqrt{6a + 5},\ f(a+h) = \sqrt{6a + 6h + 5}\).
Problem 12
Step 1 — Substitute each input into \(f(x) = \dfrac{x - 2}{3x + 7}\):
The denominator vanishes when \(3x + 7 = 0\), i.e. \(x = -\tfrac{7}{3}\), so that input is excluded.
$$\begin{aligned}
f(0) &= \frac{0 - 2}{3(0) + 7} = -\frac{2}{7} \\
f(1) &= \frac{1 - 2}{3(1) + 7} = \frac{-1}{10} = -\frac{1}{10} \\
f(3) &= \frac{3 - 2}{3(3) + 7} = \frac{1}{16}
\end{aligned}$$
Step 2 — Evaluate \(f(-x)\):
$$f(-x) = \frac{-x - 2}{3(-x) + 7} = \frac{-(x + 2)}{-3x + 7} = \frac{-(x+2)}{7 - 3x}$$
Step 3 — Evaluate \(f(a)\) and \(f(a + h)\):
$$f(a) = \frac{a - 2}{3a + 7}, \qquad f(a + h) = \frac{(a + h) - 2}{3(a + h) + 7} = \frac{a + h - 2}{3a + 3h + 7}$$
(In each case the denominator must be non-zero.)
Answer: \(f(0) = -\tfrac{2}{7},\ f(1) = -\tfrac{1}{10},\ f(3) = \tfrac{1}{16},\ f(-x) = \dfrac{-(x+2)}{7 - 3x},\ f(a) = \dfrac{a - 2}{3a + 7},\ f(a+h) = \dfrac{a + h - 2}{3a + 3h + 7}\).
Problem 13
Step 1 — Recognize the constant function:
\(f(x) = 9\) returns \(9\) regardless of the input.
Step 2 — Evaluate at every requested input:
$$f(0) = f(1) = f(3) = f(-x) = f(a) = f(a + h) = 9$$
Answer: All six values equal \(9\).
For the following exercises, find the domain, range, and all zeros/intercepts, if any, of the functions.
Problem 14. \(f(x) = \dfrac{x}{x^2 - 16}\)
Problem 15. \(g(x) = \sqrt{8x - 1}\)
Problem 16. \(h(x) = \dfrac{3}{x^2 + 4}\)
Problem 17. \(f(x) = -1 + \sqrt{x + 2}\)
Problem 18. \(f(x) = \dfrac{1}{\sqrt{x - 9}}\)
Problem 19. \(g(x) = \dfrac{3}{x - 4}\)
Problem 20. \(f(x) = 4|x + 5|\)
Problem 21. \(g(x) = \sqrt{\dfrac{7}{x - 5}}\)
Solutions 14–21
Problem 14
Step 1 — Find the domain of \(f(x) = \dfrac{x}{x^2 - 16}\):
The function is undefined where the denominator is zero. Solve \(x^2 - 16 = 0\):
$$x^2 = 16 \implies x = \pm 4$$
So we must exclude \(x = 4\) and \(x = -4\).
$$\text{Domain} = (-\infty,\, -4) \cup (-4,\, 4) \cup (4,\, \infty)$$
Step 2 — Find the zeros:
Set the numerator to zero (and confirm the denominator is non-zero there):
$$x = 0, \quad \text{denominator} = 0^2 - 16 = -16 \neq 0$$
So \(x = 0\) is a zero, giving \(x\)-intercept \((0, 0)\). Because \(f(0) = 0\), the \(y\)-intercept is also \((0, 0)\).
Step 3 — Determine the range:
Solve \(y = \dfrac{x}{x^2 - 16}\) for \(x\). Multiplying through, \(y(x^2 - 16) = x\), i.e.
$$y x^2 - x - 16 y = 0.$$
If \(y = 0\), then \(x = 0\), which lies in the domain — so \(y = 0\) is attainable. Otherwise treat it as a quadratic in \(x\) with discriminant \(1 + 64 y^2 \ge 1 > 0\); the discriminant is always positive, so real \(x\) exists for every non-zero \(y\) as well. Therefore every real number is in the range.
$$\text{Range} = (-\infty,\, \infty)$$
Answer: Domain \(= (-\infty, -4) \cup (-4, 4) \cup (4, \infty)\); Range \(= (-\infty, \infty)\); only intercept is the origin \((0, 0)\).
Problem 15
Step 1 — Find the domain of \(g(x) = \sqrt{8x - 1}\):
The radicand must be non-negative.
$$8x - 1 \ge 0 \implies x \ge \tfrac{1}{8}$$
$$\text{Domain} = \left[\tfrac{1}{8},\, \infty\right)$$
Step 2 — Find the range:
A principal square root is always \(\ge 0\), and as \(x \to \infty\) the expression grows without bound.
$$\text{Range} = [0,\, \infty)$$
Step 3 — Find the zeros and intercepts:
Set \(g(x) = 0\): \(\sqrt{8x - 1} = 0 \implies 8x - 1 = 0 \implies x = \tfrac{1}{8}\). So the \(x\)-intercept is \(\left(\tfrac{1}{8},\, 0\right)\).
For the \(y\)-intercept, evaluate \(g(0)\). But \(x = 0\) is not in the domain, so there is no \(y\)-intercept.
Answer: Domain \(= \left[\tfrac{1}{8}, \infty\right)\); Range \(= [0, \infty)\); \(x\)-intercept \(\left(\tfrac{1}{8}, 0\right)\); no \(y\)-intercept.
Problem 16
Step 1 — Find the domain of \(h(x) = \dfrac{3}{x^2 + 4}\):
The denominator \(x^2 + 4\) is always positive (\(\ge 4\)), so the function is defined for every real \(x\).
$$\text{Domain} = (-\infty,\, \infty)$$
Step 2 — Find the zeros:
\(h(x) = 0\) would require \(3 = 0\), which is impossible. There are no zeros, so there is no \(x\)-intercept.
Step 3 — Find the \(y\)-intercept:
$$h(0) = \frac{3}{0 + 4} = \frac{3}{4}$$
So the \(y\)-intercept is \(\left(0,\, \tfrac{3}{4}\right)\).
Step 4 — Find the range:
\(h(x)\) is positive everywhere. It is maximized when the denominator is smallest, at \(x = 0\), giving \(h(0) = \tfrac{3}{4}\). As \(|x| \to \infty\), \(h(x) \to 0^+\) but never reaches \(0\).
$$\text{Range} = \left(0,\, \tfrac{3}{4}\right]$$
Answer: Domain \(= (-\infty, \infty)\); Range \(= \left(0, \tfrac{3}{4}\right]\); no \(x\)-intercept; \(y\)-intercept \(\left(0, \tfrac{3}{4}\right)\).
Problem 17
Step 1 — Find the domain of \(f(x) = -1 + \sqrt{x + 2}\):
Require \(x + 2 \ge 0\):
$$x \ge -2 \implies \text{Domain} = [-2,\, \infty)$$
Step 2 — Find the range:
The square root \(\sqrt{x + 2}\) ranges over \([0, \infty)\). Subtracting \(1\) shifts the range down by \(1\):
$$\text{Range} = [-1,\, \infty)$$
Step 3 — Find the \(x\)-intercept (zeros):
Set \(f(x) = 0\):
$$\sqrt{x + 2} = 1 \implies x + 2 = 1 \implies x = -1$$
So the \(x\)-intercept is \((-1, 0)\).
Step 4 — Find the \(y\)-intercept:
$$f(0) = -1 + \sqrt{2}$$
So the \(y\)-intercept is \(\left(0,\, -1 + \sqrt{2}\right) \approx (0,\, 0.414)\).
Answer: Domain \(= [-2, \infty)\); Range \(= [-1, \infty)\); \(x\)-intercept \((-1, 0)\); \(y\)-intercept \(\left(0,\, \sqrt{2} - 1\right)\).
Problem 18
Step 1 — Find the domain of \(f(x) = \dfrac{1}{\sqrt{x - 9}}\):
The radicand must be strictly positive (the square root sits in a denominator, so it cannot be zero either).
$$x - 9 > 0 \implies x > 9 \implies \text{Domain} = (9,\, \infty)$$
Step 2 — Find the range:
For \(x > 9\), \(\sqrt{x - 9}\) takes every positive value, so its reciprocal also takes every positive value.
$$\text{Range} = (0,\, \infty)$$
Step 3 — Find the zeros and intercepts:
\(f(x) = 0\) would require the numerator to be zero, but the numerator is the constant \(1\). No zeros, so no \(x\)-intercept.
Since \(0\) is not in the domain, there is no \(y\)-intercept.
Answer: Domain \(= (9, \infty)\); Range \(= (0, \infty)\); no \(x\)- or \(y\)-intercepts.
Problem 19
Step 1 — Find the domain of \(g(x) = \dfrac{3}{x - 4}\):
Exclude where the denominator is zero: \(x - 4 = 0 \implies x = 4\).
$$\text{Domain} = (-\infty,\, 4) \cup (4,\, \infty)$$
Step 2 — Find the range:
Solving \(y = \dfrac{3}{x - 4}\) for \(x\) gives \(x = 4 + \dfrac{3}{y}\), which is defined for every \(y \neq 0\). So every non-zero \(y\) is attainable.
$$\text{Range} = (-\infty,\, 0) \cup (0,\, \infty)$$
Step 3 — Find the zeros and intercepts:
\(g(x) = 0\) would require \(3 = 0\), impossible — no \(x\)-intercept.
For the \(y\)-intercept, \(g(0) = \dfrac{3}{0 - 4} = -\dfrac{3}{4}\), so the \(y\)-intercept is \(\left(0,\, -\tfrac{3}{4}\right)\).
Answer: Domain \(= (-\infty, 4) \cup (4, \infty)\); Range \(= (-\infty, 0) \cup (0, \infty)\); no \(x\)-intercept; \(y\)-intercept \(\left(0,\, -\tfrac{3}{4}\right)\).
Problem 20
Step 1 — Find the domain of \(f(x) = 4|x + 5|\):
Absolute value is defined for every real input.
$$\text{Domain} = (-\infty,\, \infty)$$
Step 2 — Find the range:
\(|x + 5| \ge 0\), so \(4|x + 5| \ge 0\). The minimum value \(0\) occurs at \(x = -5\), and the expression grows without bound as \(|x + 5| \to \infty\).
$$\text{Range} = [0,\, \infty)$$
Step 3 — Find the zeros:
Set \(4|x + 5| = 0\): \(|x + 5| = 0 \implies x = -5\). So the \(x\)-intercept is \((-5, 0)\).
Step 4 — Find the \(y\)-intercept:
$$f(0) = 4|0 + 5| = 4(5) = 20$$
\(y\)-intercept: \((0, 20)\).
Answer: Domain \(= (-\infty, \infty)\); Range \(= [0, \infty)\); \(x\)-intercept \((-5, 0)\); \(y\)-intercept \((0, 20)\).
Problem 21
Step 1 — Find the domain of \(g(x) = \sqrt{\dfrac{7}{x - 5}}\):
The expression inside the square root must be \(\ge 0\), and the denominator \(x - 5\) cannot be zero.
Because \(7 > 0\), the sign of \(\dfrac{7}{x - 5}\) matches the sign of \(x - 5\). So we need \(x - 5 > 0\) (we cannot include \(x = 5\) — it makes the denominator zero, and the expression would not be \(\ge 0\) below \(5\) either).
$$x > 5 \implies \text{Domain} = (5,\, \infty)$$
Step 2 — Find the range:
As \(x \to 5^+\), \(\dfrac{7}{x - 5} \to +\infty\), so \(g(x) \to \infty\). As \(x \to \infty\), \(\dfrac{7}{x - 5} \to 0^+\), so \(g(x) \to 0^+\). The function is continuous and strictly decreasing on \((5, \infty)\), taking every value in between.
$$\text{Range} = (0,\, \infty)$$
Step 3 — Find the zeros and intercepts:
\(g(x) = 0\) would need \(\dfrac{7}{x - 5} = 0\), which is impossible. So no \(x\)-intercept. Since \(0 \notin\) domain, no \(y\)-intercept.
Answer: Domain \(= (5, \infty)\); Range \(= (0, \infty)\); no \(x\)- or \(y\)-intercepts.
For the following exercises, sketch the graph with the aid of the tables given.
Problem 22. \(f(x) = x^2 + 1\)
| \(x\) | \(y\) |
|---|---|
| \(-3\) | \(10\) |
| \(-2\) | \(5\) |
| \(-1\) | \(2\) |
| \(0\) | \(1\) |
| \(1\) | \(2\) |
| \(2\) | \(5\) |
| \(3\) | \(10\) |
Problem 23. \(f(x) = 3x - 6\)
| \(x\) | \(y\) |
|---|---|
| \(-3\) | \(-15\) |
| \(-2\) | \(-12\) |
| \(-1\) | \(-9\) |
| \(0\) | \(-6\) |
| \(1\) | \(-3\) |
| \(2\) | \(0\) |
| \(3\) | \(3\) |
Problem 24. \(f(x) = \dfrac{1}{2}x + 1\)
| \(x\) | \(y\) |
|---|---|
| \(-3\) | \(-\tfrac{1}{2}\) |
| \(-2\) | \(0\) |
| \(-1\) | \(\tfrac{1}{2}\) |
| \(0\) | \(1\) |
| \(1\) | \(\tfrac{3}{2}\) |
| \(2\) | \(2\) |
| \(3\) | \(\tfrac{5}{2}\) |
Problem 25. \(f(x) = 2|x|\)
| \(x\) | \(y\) |
|---|---|
| \(-3\) | \(6\) |
| \(-2\) | \(4\) |
| \(-1\) | \(2\) |
| \(0\) | \(0\) |
| \(1\) | \(2\) |
| \(2\) | \(4\) |
| \(3\) | \(6\) |
Problem 26. \(f(x) = -x^2\)
| \(x\) | \(y\) |
|---|---|
| \(-3\) | \(-9\) |
| \(-2\) | \(-4\) |
| \(-1\) | \(-1\) |
| \(0\) | \(0\) |
| \(1\) | \(-1\) |
| \(2\) | \(-4\) |
| \(3\) | \(-9\) |
Problem 27. \(f(x) = x^3\)
| \(x\) | \(y\) |
|---|---|
| \(-3\) | \(-27\) |
| \(-2\) | \(-8\) |
| \(-1\) | \(-1\) |
| \(0\) | \(0\) |
| \(1\) | \(1\) |
| \(2\) | \(8\) |
| \(3\) | \(27\) |
Solutions 22–27
Problem 22
Step 1 — Plot the seven listed points:
From the table, plot \((-3, 10),\ (-2, 5),\ (-1, 2),\ (0, 1),\ (1, 2),\ (2, 5),\ (3, 10)\).
Step 2 — Identify the shape:
\(f(x) = x^2 + 1\) is a parabola. Adding \(1\) to \(x^2\) shifts the standard parabola up by one unit, so the vertex is at \((0, 1)\) and the curve opens upward.
Step 3 — Connect the points with a smooth curve:
Draw a smooth, symmetric U-shape through the plotted points. The graph is symmetric about the \(y\)-axis because \(f(-x) = (-x)^2 + 1 = x^2 + 1 = f(x)\) (the function is even).
Step 4 — Describe key features:
- Vertex (minimum) at \((0, 1)\). - \(y\)-intercept \((0, 1)\). - No \(x\)-intercepts (since \(x^2 + 1 \ge 1 > 0\)). - Symmetric about the \(y\)-axis.
Answer: The graph is an upward-opening parabola with vertex \((0, 1)\), passing through the seven tabulated points and symmetric about the \(y\)-axis.
Problem 23
Step 1 — Plot the seven points from the table:
Plot \((-3, -15),\ (-2, -12),\ (-1, -9),\ (0, -6),\ (1, -3),\ (2, 0),\ (3, 3)\).
Note: the table appears to have used the rule \(y = 3x - 6\) with an unrelated tabulation error at \(x = -3, -2, -1, 0\); however the problem text fixes the rule as \(f(x) = 3x - 6\). Using the rule directly: \(f(-3) = -15,\ f(-2) = -12,\ f(-1) = -9,\ f(0) = -6,\ f(1) = -3,\ f(2) = 0,\ f(3) = 3\). These match the table (the table is consistent with the rule \(f(x) = 3x - 6\)).
Step 2 — Identify the shape:
The function is linear with slope \(3\) and \(y\)-intercept \(-6\).
Step 3 — Connect the points with a straight line:
Draw a straight line through the plotted points, extending in both directions.
Step 4 — Describe key features:
- Slope \(= 3\), \(y\)-intercept \((0, -6)\). - \(x\)-intercept: set \(3x - 6 = 0 \implies x = 2\), so \((2, 0)\). - The line is increasing everywhere; no symmetry about either axis.
Answer: The graph is a straight line through \((0, -6)\) with slope \(3\), passing through every point in the table.
Problem 24
Step 1 — Plot the seven points from the table:
Plot \(\left(-3, -\tfrac{1}{2}\right),\ (-2, 0),\ \left(-1, \tfrac{1}{2}\right),\ (0, 1),\ \left(1, \tfrac{3}{2}\right),\ (2, 2),\ \left(3, \tfrac{5}{2}\right)\).
Step 2 — Identify the shape:
\(f(x) = \tfrac{1}{2}x + 1\) is linear with slope \(\tfrac{1}{2}\) and \(y\)-intercept \(1\).
Step 3 — Connect the points with a straight line:
The line rises by \(\tfrac{1}{2}\) for every increase of \(1\) in \(x\); each plotted \(y\)-value matches \(\tfrac{x}{2} + 1\). Draw the line through the points and extend in both directions.
Step 4 — Describe key features:
- Slope \(\tfrac{1}{2}\), \(y\)-intercept \((0, 1)\). - \(x\)-intercept: \(\tfrac{1}{2}x + 1 = 0 \implies x = -2\), so \((-2, 0)\). - Strictly increasing; no symmetry.
Answer: The graph is a straight line through \((0, 1)\) with slope \(\tfrac{1}{2}\), passing through each tabulated point.
Problem 25
Step 1 — Plot the seven points from the table:
Plot \((-3, 6),\ (-2, 4),\ (-1, 2),\ (0, 0),\ (1, 2),\ (2, 4),\ (3, 6)\).
Step 2 — Identify the shape:
\(f(x) = 2|x|\) is a V-shaped absolute-value graph. The factor \(2\) makes the V twice as steep as \(y = |x|\). For \(x \ge 0\) the graph is the line \(y = 2x\); for \(x < 0\) it is the line \(y = -2x\).
Step 3 — Connect the points:
Draw a V whose vertex is at the origin, with both arms having slope \(\pm 2\).
Step 4 — Describe key features:
- Vertex (minimum) at \((0, 0)\). - \(x\)- and \(y\)-intercepts both at the origin. - Symmetric about the \(y\)-axis (since \(f(-x) = 2|-x| = 2|x| = f(x)\)); the function is even. - Decreasing on \((-\infty, 0]\), increasing on \([0, \infty)\).
Answer: The graph is a V opening upward with vertex \((0, 0)\) and arms of slope \(\pm 2\), passing through every tabulated point.
Problem 26
Step 1 — Plot the seven points from the table:
Plot \((-3, -9),\ (-2, -4),\ (-1, -1),\ (0, 0),\ (1, -1),\ (2, -4),\ (3, -9)\).
Step 2 — Identify the shape:
\(f(x) = -x^2\) is a downward-opening parabola — the reflection of \(y = x^2\) across the \(x\)-axis.
Step 3 — Connect the points with a smooth curve:
Draw a smooth, symmetric arch through the plotted points, opening downward.
Step 4 — Describe key features:
- Vertex (maximum) at \((0, 0)\). - \(x\)- and \(y\)-intercepts both at the origin. - Symmetric about the \(y\)-axis; the function is even because \(f(-x) = -(-x)^2 = -x^2 = f(x)\). - Increasing on \((-\infty, 0]\), decreasing on \([0, \infty)\).
Answer: The graph is a downward-opening parabola with vertex at the origin, passing through every tabulated point.
Problem 27
Step 1 — Plot the seven points from the table:
Plot \((-3, -27),\ (-2, -8),\ (-1, -1),\ (0, 0),\ (1, 1),\ (2, 8),\ (3, 27)\).
Step 2 — Identify the shape:
\(f(x) = x^3\) is the standard cubic curve — passing through the origin, increasing everywhere, with an inflection point at \((0, 0)\).
Step 3 — Connect the points with a smooth curve:
Draw a smooth S-shaped curve that flattens near the origin and steepens away from it.
Step 4 — Describe key features:
- \(x\)- and \(y\)-intercepts at the origin. - Symmetric about the origin (rotational symmetry): \(f(-x) = (-x)^3 = -x^3 = -f(x)\), so the function is odd. - Strictly increasing on all of \((-\infty, \infty)\). - Inflection point at \((0, 0)\).
Answer: The graph is the standard cubic — a smooth, strictly increasing S-curve through the origin, passing through every tabulated point and symmetric about the origin.
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: (a) Domain and range; (b) \(x\)-intercept, if any (estimate where necessary); (c) \(y\)-intercept, if any (estimate where necessary); (d) The intervals for which the function is increasing; (e) The intervals for which the function is decreasing; (f) The intervals for which the function is constant; (g) Symmetry about any axis and/or the origin; (h) Whether the function is even, odd, or neither.
Problem 28. Use the vertical line test on graph 28
Problem 29. Use the vertical line test on graph 29
Problem 30. Use the vertical line test on graph 30
Problem 31. Use the vertical line test on graph 31
Problem 32. Use the vertical line test on graph 32
Problem 33. Use the vertical line test on graph 33
Problem 34. Use the vertical line test on graph 34
Problem 35. Use the vertical line test on graph 35
Solutions 28–35
Problem 28
Note on missing graph: The problem refers to "graph 28," which is a textbook figure not reproduced in the markdown source available here. The full eight-part analysis (domain/range, intercepts, intervals of monotonicity, symmetry, even/odd) cannot be completed without the image. The following solution lays out the procedure the student should follow with the actual graph.
Step 1 — Apply the vertical line test:
Slide a vertical line across the graph from left to right. If every vertical line meets the graph in at most one point, the relation is a function; otherwise it is not. If it fails the vertical line test, stop — none of the remaining items (b)–(h) apply.
Step 2 — (a) Domain and range:
- Domain: the set of \(x\)-values for which the graph has any point. Read this off the horizontal extent (and follow the instruction to assume the graph continues beyond the grid when it appears to extend to the edges). - Range: the corresponding set of \(y\)-values.
Step 3 — (b)–(c) Intercepts:
- \(x\)-intercepts: points where the curve crosses the \(x\)-axis. - \(y\)-intercept: the point where the curve crosses the \(y\)-axis (a function has at most one).
Step 4 — (d)–(f) Monotonicity intervals:
- Increasing: \(x\)-intervals where the graph rises left-to-right. - Decreasing: \(x\)-intervals where it falls. - Constant: \(x\)-intervals where it is horizontal.
Step 5 — (g) Symmetry; (h) even/odd classification:
- If the graph is mirrored across the \(y\)-axis, the function is even (\(f(-x) = f(x)\)). - If the graph has \(180^\circ\) rotational symmetry about the origin, the function is odd (\(f(-x) = -f(x)\)). - Otherwise, neither.
Answer: Solution procedure stated above; numerical values for graph 28 cannot be supplied without the figure. (Stub solution — figure missing from source.)
Problem 29
Note on missing graph: As with problem 1.1.28, "graph 29" is a textbook figure not included in the markdown source. Without the image we cannot enumerate domain, range, intercepts, monotonicity, or symmetry numerically. The procedure below is identical to 1.1.28.
Step 1 — Apply the vertical line test:
If every vertical line crosses the graph at most once, the relation is a function; otherwise it is not, and the remaining items do not apply.
Step 2 — (a) Read domain and range:
Domain: the projection of the graph onto the \(x\)-axis; range: the projection onto the \(y\)-axis. Treat segments that run off the grid as continuing.
Step 3 — (b)–(c) Intercepts:
Estimate \(x\)-intercepts (where the curve meets the \(x\)-axis) and the \(y\)-intercept (where it meets the \(y\)-axis).
Step 4 — (d)–(f) Intervals of monotonicity:
Identify the \(x\)-intervals on which the curve is rising (increasing), falling (decreasing), or horizontal (constant).
Step 5 — (g)–(h) Symmetry and parity:
- Mirror symmetry about the \(y\)-axis \(\to\) even. - Rotational symmetry about the origin \(\to\) odd. - No such symmetry \(\to\) neither.
Answer: Procedure outlined above; specific values for graph 29 cannot be provided without the figure. (Stub solution — figure missing from source.)
Problem 30
Step 1 — Recognize what is needed: This problem refers to "graph 30," a figure that is not embedded in the source text provided to me, so I cannot read the graph directly. The solution below is a stub that explains exactly which procedure to follow once the graph is visible.
Step 2 — Apply the vertical line test: Mentally (or with a ruler) sweep a vertical line from left to right across the graph. If every vertical line crosses the curve in at most one point, the relation is a function of \(x\); otherwise it is not.
Step 3 — If the relation is a function, extract (a)–(h):
- (a) Domain and range. Read the smallest and largest \(x\)-values where the curve exists; do the same for \(y\). Use interval notation, e.g. \([-5, 5]\) or \((-\infty, \infty)\). - (b) \(x\)-intercept(s). Points where the curve crosses the \(x\)-axis (i.e. \(y = 0\)); estimate to the nearest grid mark if needed. - (c) \(y\)-intercept. The single point where the curve crosses the \(y\)-axis (\(x = 0\)). - (d) Increasing intervals. Intervals where, as \(x\) moves right, \(y\) moves up. - (e) Decreasing intervals. Intervals where, as \(x\) moves right, \(y\) moves down. - (f) Constant intervals. Intervals where \(y\) does not change (flat segment). - (g) Symmetry. Check whether the picture is unchanged by reflection across the \(y\)-axis (\(y\)-axis symmetry), across the \(x\)-axis (not allowed for a function unless it is the zero function), or through the origin (\(180^\circ\) rotation). - (h) Even / odd / neither. Even iff symmetric about the \(y\)-axis; odd iff symmetric about the origin; otherwise neither.
Answer: A worked answer requires the actual figure for graph 30. Once the figure is available, run steps 2–3 above to fill in items (a)–(h).
Problem 31
Step 1 — Recognize what is needed: The figure for "graph 31" is not embedded in the supplied source text, so a numeric answer cannot be produced. Follow the procedure below once the figure is available.
Step 2 — Apply the vertical line test: If any vertical line meets the graph at two or more points, the relation is not a function and items (a)–(h) do not apply. Otherwise continue.
Step 3 — Extract (a)–(h): Use the same reading rules listed in problem 1.1.30 above: read domain/range from the horizontal/vertical extent of the curve, locate intercepts on the axes, identify rising/falling/flat intervals, and test for \(y\)-axis or origin symmetry to classify the function as even, odd, or neither.
Answer: A complete numeric answer requires the actual figure for graph 31; the procedure above produces items (a)–(h) once the graph is visible.
Problem 32
Step 1 — Recognize what is needed: The figure for "graph 32" is not embedded in the supplied source text. The solution below explains the procedure rather than reading values from a missing image.
Step 2 — Apply the vertical line test: Sweep a vertical line across the picture; if it ever intersects the curve in two or more points, the relation fails the vertical line test and is not a function.
Step 3 — Extract (a)–(h): If the test is passed, read the domain (horizontal extent of the curve), range (vertical extent), \(x\)- and \(y\)-intercepts, increasing/decreasing/constant intervals, and check for \(y\)-axis or origin symmetry to label the function even, odd, or neither.
Answer: Final answers require the figure for graph 32; the steps above show how to obtain (a)–(h) once it is visible.
Problem 33
Step 1 — Recognize what is needed: The image for "graph 33" is not embedded in the source provided here. The solution gives the procedure to follow when the figure is available.
Step 2 — Apply the vertical line test: A vertical line that intersects the curve more than once means the relation is not a function. Otherwise proceed.
Step 3 — Extract (a)–(h): Read domain and range from the horizontal/vertical extent, locate the intercepts on the axes, identify the intervals on which the graph rises, falls, or is flat, and test for symmetry across the \(y\)-axis (even) or through the origin (odd).
Answer: A definitive answer requires the figure for graph 33; apply the procedure above once the image is in hand.
Problem 34
Step 1 — Recognize what is needed: The image for "graph 34" is missing from the supplied source. The solution outlines the procedure.
Step 2 — Apply the vertical line test: Determine whether each vertical line meets the curve at most once. If so, the relation is a function; otherwise it is not, and (a)–(h) are not applicable.
Step 3 — Extract (a)–(h): Reading from the graph, record domain (horizontal extent), range (vertical extent), intercepts, increasing/decreasing/constant intervals, and check whether \(f(-x) = f(x)\) (even, \(y\)-axis symmetry) or \(f(-x) = -f(x)\) (odd, origin symmetry), or neither.
Answer: A complete numeric answer requires the actual figure for graph 34; the procedure above produces items (a)–(h) once it is visible.
Problem 35
Step 1 — Recognize what is needed: The image for "graph 35" is not embedded in the source given to me. The solution below provides the reading procedure to use with the figure.
Step 2 — Apply the vertical line test: If every vertical line meets the curve in at most one point, the relation is a function and items (a)–(h) follow. Otherwise the relation is not a function and (a)–(h) do not apply.
Step 3 — Extract (a)–(h): Read the domain and range from the curve's horizontal and vertical extent, locate any \(x\)- and \(y\)-intercepts, find intervals of increase/decrease/constancy, and check for \(y\)-axis or origin symmetry to classify the function as even, odd, or neither.
Answer: A worked numeric answer for graph 35 requires the image; the procedure above produces items (a)–(h) once the figure is available.
For the following exercises, for each pair of functions, find (a) \(f + g\); (b) \(f - g\); (c) \(f \cdot g\); (d) \(f / g\). Determine the domain of each new function.
Problem 36. \(f(x) = 3x + 4,\ g(x) = x - 2\)
Problem 37. \(f(x) = x - 8,\ g(x) = 5x^2\)
Problem 38. \(f(x) = 3x^2 + 4x + 1,\ g(x) = x + 1\)
Problem 39. \(f(x) = 9 - x^2,\ g(x) = x^2 - 2x - 3\)
Problem 40. \(f(x) = \sqrt{x},\ g(x) = x - 2\)
Problem 41. \(f(x) = 6 + \dfrac{1}{x},\ g(x) = \dfrac{1}{x}\)
Solutions 36–41
Problem 36
We are given \(f(x) = 3x + 4\) and \(g(x) = x - 2.\) Both \(f\) and \(g\) are polynomials with domain \((-\infty, \infty),\) so the domains of \(f+g,\) \(f-g,\) and \(fg\) are all \((-\infty, \infty),\) and the domain of \(f/g\) excludes the zero of \(g.\)
Step 1 — Sum \(f + g\):
$$(f + g)(x) = (3x + 4) + (x - 2) = 4x + 2.$$
Domain: \((-\infty, \infty).\)
Step 2 — Difference \(f - g\):
$$(f - g)(x) = (3x + 4) - (x - 2) = 2x + 6.$$
Domain: \((-\infty, \infty).\)
Step 3 — Product \(f \cdot g\):
$$(f \cdot g)(x) = (3x + 4)(x - 2) = 3x^2 - 6x + 4x - 8 = 3x^2 - 2x - 8.$$
Domain: \((-\infty, \infty).\)
Step 4 — Quotient \(f / g\):
$$\left(\frac{f}{g}\right)(x) = \frac{3x + 4}{x - 2}.$$
Exclude \(x\) where \(g(x) = 0,\) i.e. \(x = 2.\) Domain: \((-\infty, 2) \cup (2, \infty).\)
Answer: (a) \(4x + 2,\) domain \((-\infty, \infty);\) (b) \(2x + 6,\) domain \((-\infty, \infty);\) (c) \(3x^2 - 2x - 8,\) domain \((-\infty, \infty);\) (d) \(\dfrac{3x + 4}{x - 2},\) domain \((-\infty, 2) \cup (2, \infty).\)
Problem 37
Given \(f(x) = x - 8\) and \(g(x) = 5x^2.\) Both are polynomials with domain \((-\infty, \infty),\) so \(f \pm g\) and \(fg\) have domain \((-\infty, \infty),\) and \(f/g\) excludes the zero of \(g.\)
Step 1 — Sum:
$$(f + g)(x) = (x - 8) + 5x^2 = 5x^2 + x - 8.$$
Domain: \((-\infty, \infty).\)
Step 2 — Difference:
$$(f - g)(x) = (x - 8) - 5x^2 = -5x^2 + x - 8.$$
Domain: \((-\infty, \infty).\)
Step 3 — Product:
$$(f \cdot g)(x) = (x - 8)(5x^2) = 5x^3 - 40x^2.$$
Domain: \((-\infty, \infty).\)
Step 4 — Quotient:
$$\left(\frac{f}{g}\right)(x) = \frac{x - 8}{5x^2}.$$
\(g(x) = 5x^2 = 0\) iff \(x = 0,\) so exclude \(x = 0.\) Domain: \((-\infty, 0) \cup (0, \infty).\)
Answer: (a) \(5x^2 + x - 8;\) (b) \(-5x^2 + x - 8;\) (c) \(5x^3 - 40x^2;\) (d) \(\dfrac{x - 8}{5x^2},\) domain \((-\infty, 0) \cup (0, \infty).\) All others have domain \((-\infty, \infty).\)
Problem 38
Given \(f(x) = 3x^2 + 4x + 1\) and \(g(x) = x + 1.\) Both are polynomials, so \(f \pm g\) and \(fg\) have domain \((-\infty, \infty);\) for \(f/g\) we must exclude \(x = -1.\)
Step 1 — Sum:
$$(f + g)(x) = (3x^2 + 4x + 1) + (x + 1) = 3x^2 + 5x + 2.$$
Step 2 — Difference:
$$(f - g)(x) = (3x^2 + 4x + 1) - (x + 1) = 3x^2 + 3x.$$
Step 3 — Product:
$$(f \cdot g)(x) = (3x^2 + 4x + 1)(x + 1).$$
Expand:
$$= 3x^3 + 3x^2 + 4x^2 + 4x + x + 1 = 3x^3 + 7x^2 + 5x + 1.$$
Step 4 — Quotient:
Notice \(3x^2 + 4x + 1 = (3x + 1)(x + 1),\) so
$$\frac{f(x)}{g(x)} = \frac{(3x + 1)(x + 1)}{x + 1} = 3x + 1, \quad x \neq -1.$$
Even though the simplified form is the polynomial \(3x + 1,\) the domain still excludes \(x = -1\) because that point makes the original denominator zero.
Answer: (a) \(3x^2 + 5x + 2;\) (b) \(3x^2 + 3x;\) (c) \(3x^3 + 7x^2 + 5x + 1;\) (d) \(3x + 1,\ x \neq -1,\) domain \((-\infty, -1) \cup (-1, \infty).\) All others have domain \((-\infty, \infty).\)
Problem 39
Given \(f(x) = 9 - x^2\) and \(g(x) = x^2 - 2x - 3.\) Both are polynomials; the quotient excludes the zeros of \(g.\)
Step 1 — Sum:
$$(f + g)(x) = (9 - x^2) + (x^2 - 2x - 3) = -2x + 6.$$
Step 2 — Difference:
$$(f - g)(x) = (9 - x^2) - (x^2 - 2x - 3) = 9 - x^2 - x^2 + 2x + 3 = -2x^2 + 2x + 12.$$
Step 3 — Product:
$$(f \cdot g)(x) = (9 - x^2)(x^2 - 2x - 3).$$
Expand:
$$= 9x^2 - 18x - 27 - x^4 + 2x^3 + 3x^2 = -x^4 + 2x^3 + 12x^2 - 18x - 27.$$
Step 4 — Quotient:
Factor: \(9 - x^2 = (3-x)(3+x)\) and \(x^2 - 2x - 3 = (x-3)(x+1).\) Then
$$\frac{f(x)}{g(x)} = \frac{(3-x)(3+x)}{(x-3)(x+1)} = \frac{-(x-3)(x+3)}{(x-3)(x+1)} = \frac{-(x+3)}{x+1}, \quad x \neq 3.$$
The denominator \(g(x) = 0\) when \(x = 3\) or \(x = -1,\) so both must be excluded from the domain. Domain: \((-\infty, -1) \cup (-1, 3) \cup (3, \infty).\)
Answer: (a) \(-2x + 6,\) domain \((-\infty, \infty);\) (b) \(-2x^2 + 2x + 12,\) domain \((-\infty, \infty);\) (c) \(-x^4 + 2x^3 + 12x^2 - 18x - 27,\) domain \((-\infty, \infty);\) (d) \(\dfrac{-(x+3)}{x+1},\) domain \((-\infty, -1) \cup (-1, 3) \cup (3, \infty).\)
Problem 40
Given \(f(x) = \sqrt{x}\) (domain \([0, \infty)\)) and \(g(x) = x - 2\) (domain \((-\infty, \infty)\)). The domain of each combination is the intersection of the individual domains, further restricted by removing zeros of the denominator for the quotient.
Step 1 — Intersection of domains:
\([0, \infty) \cap (-\infty, \infty) = [0, \infty).\)
Step 2 — Sum, difference, product:
$$(f + g)(x) = \sqrt{x} + x - 2,$$
$$(f - g)(x) = \sqrt{x} - x + 2,$$
$$(f \cdot g)(x) = \sqrt{x}\,(x - 2) = x\sqrt{x} - 2\sqrt{x}.$$
All three have domain \([0, \infty).\)
Step 3 — Quotient:
$$\left(\frac{f}{g}\right)(x) = \frac{\sqrt{x}}{x - 2}.$$
We must remove \(x = 2\) (denominator zero). Domain: \([0, 2) \cup (2, \infty).\)
Answer: (a) \(\sqrt{x} + x - 2;\) (b) \(\sqrt{x} - x + 2;\) (c) \(x\sqrt{x} - 2\sqrt{x};\) each with domain \([0, \infty);\) (d) \(\dfrac{\sqrt{x}}{x - 2},\) domain \([0, 2) \cup (2, \infty).\)
Problem 41
Given \(f(x) = 6 + \dfrac{1}{x}\) (domain \(x \neq 0\)) and \(g(x) = \dfrac{1}{x}\) (domain \(x \neq 0\)). The intersection of domains is \(\{x : x \neq 0\},\) i.e. \((-\infty, 0) \cup (0, \infty).\)
Step 1 — Sum:
$$(f + g)(x) = 6 + \frac{1}{x} + \frac{1}{x} = 6 + \frac{2}{x}.$$
Step 2 — Difference:
$$(f - g)(x) = 6 + \frac{1}{x} - \frac{1}{x} = 6.$$
Step 3 — Product:
$$(f \cdot g)(x) = \left(6 + \frac{1}{x}\right)\cdot\frac{1}{x} = \frac{6}{x} + \frac{1}{x^2}.$$
Step 4 — Quotient:
$$\left(\frac{f}{g}\right)(x) = \frac{6 + \dfrac{1}{x}}{\dfrac{1}{x}} = \left(6 + \frac{1}{x}\right) \cdot x = 6x + 1, \quad x \neq 0.$$
The simplified result is a polynomial, but the original quotient was undefined at \(x = 0,\) so the domain still excludes \(0.\)
Answer: (a) \(6 + \dfrac{2}{x};\) (b) \(6;\) (c) \(\dfrac{6}{x} + \dfrac{1}{x^2};\) (d) \(6x + 1,\ x \neq 0.\) All four have domain \((-\infty, 0) \cup (0, \infty).\)
For the following exercises, for each pair of functions, find (a) \((f \circ g)(x)\) and (b) \((g \circ f)(x)\). Simplify the results. Find the domain of each of the results.
Problem 42. \(f(x) = 3x,\ g(x) = x + 5\)
Problem 43. \(f(x) = x + 4,\ g(x) = 4x - 1\)
Problem 44. \(f(x) = 2x + 4,\ g(x) = x^2 - 2\)
Problem 45. \(f(x) = x^2 + 7,\ g(x) = x^2 - 3\)
Problem 46. \(f(x) = \sqrt{x},\ g(x) = x + 9\)
Problem 47. \(f(x) = \dfrac{3}{2x + 1},\ g(x) = \dfrac{2}{x}\)
Problem 48. \(f(x) = |x + 1|,\ g(x) = x^2 + x - 4\)
Problem 49. 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 |
a) 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.
b) 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.
Problem 50. [T] The area \(A\) of a square depends on the length of the side \(s\).
a) Write a function \(A(s)\) for the area of a square.
b) Find and interpret \(A(6.5)\).
c) Find the exact and the two-significant-digit approximation to the length of the sides of a square with area 56 square units.
Problem 51. [T] The volume of a cube depends on the length of the sides \(s\).
a) Write a function \(V(s)\) for the volume of a cube.
b) Find and interpret \(V(11.8)\).
Problem 52. [T] A rental car company rents cars for a flat fee of $20 and an hourly charge of $10.25. Therefore, the total cost \(C\) to rent a car is a function of the hours \(t\) the car is rented plus the flat fee.
a) Write the formula for the function that models this situation.
b) Find the total cost to rent a car for 2 days and 7 hours.
c) Determine how long the car was rented if the bill is $430.
Problem 53. [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.
a) Write a formula that models this situation.
b) 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.
c) Determine the domain and range of the function.
d) Determine how many times the driver had to stop for gas if she has driven a total of 578 mi.
Problem 54. [T] The volume \(V\) of a sphere depends on the length of its radius as \(V = \tfrac{4}{3}\pi r^3\). 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 \(6.371 \times 10^6\) m.
Problem 55. [T] A certain bacterium grows in culture in a circular region. The radius of the circle, measured in centimeters, is given by \(r(t) = 6 - \dfrac{5}{t^2 + 1}\), where \(t\) is time measured in hours since a circle of a 1-cm radius of the bacterium was put into the culture.
a) Express the area of the bacteria as a function of time.
b) Find the exact and approximate area of the bacterial culture in 3 hours.
c) Express the circumference of the bacteria as a function of time.
d) Find the exact and approximate circumference of the bacteria in 3 hours.
Problem 56. [T] An American tourist visits Paris and must convert U.S. dollars to Euros, which can be done using the function \(E(x) = 0.79 x\), where \(x\) is the number of U.S. dollars and \(E(x)\) 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 \(D(x) = 1.245 x\), where \(x\) is the number of Euros and \(D(x)\) is the equivalent number of U.S. dollars.
a) 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?
b) 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.
Problem 57. [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.
a) Write a function \(y = S(x)\) that models a worker's monthly salary based on the number of skateboards \(x\) he or she sells.
b) Find the monthly salary when a worker sells 25, 40, or 55 skateboards.
c) 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 $1,400. (Hint: Find the intersection of the function and the line \(y = 1400\).)
Problem 58. [T] Use a graphing calculator to graph the half-circle \(y = \sqrt{25 - (x - 4)^2}\).
Then, use the INTERCEPT feature to find the value of both the \(x\)- and \(y\)-intercepts.
Solutions 42–58
Problem 42
Given \(f(x) = 3x\) and \(g(x) = x + 5.\) Both are polynomials with domain \((-\infty, \infty).\)
Step 1 — Compose \((f \circ g)(x) = f(g(x)):\)
$$f(g(x)) = 3 \cdot g(x) = 3(x + 5) = 3x + 15.$$
Since \(g\) is defined for all real \(x\) and \(f\) accepts every real input, the domain is \((-\infty, \infty).\)
Step 2 — Compose \((g \circ f)(x) = g(f(x)):\)
$$g(f(x)) = f(x) + 5 = 3x + 5.$$
Same reasoning: domain \((-\infty, \infty).\)
Answer: (a) \((f \circ g)(x) = 3x + 15,\) domain \((-\infty, \infty);\) (b) \((g \circ f)(x) = 3x + 5,\) domain \((-\infty, \infty).\)
Problem 43
Given \(f(x) = x + 4\) and \(g(x) = 4x - 1.\) Both are polynomials with domain \((-\infty, \infty).\)
Step 1 — Compute \((f \circ g)(x):\)
$$f(g(x)) = g(x) + 4 = (4x - 1) + 4 = 4x + 3.$$
Domain: \((-\infty, \infty).\)
Step 2 — Compute \((g \circ f)(x):\)
$$g(f(x)) = 4 f(x) - 1 = 4(x + 4) - 1 = 4x + 16 - 1 = 4x + 15.$$
Domain: \((-\infty, \infty).\)
Answer: (a) \((f \circ g)(x) = 4x + 3;\) (b) \((g \circ f)(x) = 4x + 15;\) each has domain \((-\infty, \infty).\)
Problem 44
Given \(f(x) = 2x + 4\) and \(g(x) = x^2 - 2.\) Both are polynomials with domain \((-\infty, \infty).\)
Step 1 — Compute \((f \circ g)(x):\)
$$f(g(x)) = 2 g(x) + 4 = 2(x^2 - 2) + 4 = 2x^2 - 4 + 4 = 2x^2.$$
Domain: \((-\infty, \infty).\)
Step 2 — Compute \((g \circ f)(x):\)
$$g(f(x)) = (f(x))^2 - 2 = (2x + 4)^2 - 2.$$
Expand \((2x + 4)^2 = 4x^2 + 16x + 16,\) so
$$g(f(x)) = 4x^2 + 16x + 16 - 2 = 4x^2 + 16x + 14.$$
Domain: \((-\infty, \infty).\)
Answer: (a) \((f \circ g)(x) = 2x^2;\) (b) \((g \circ f)(x) = 4x^2 + 16x + 14;\) each has domain \((-\infty, \infty).\)
Problem 45
Given \(f(x) = x^2 + 7\) and \(g(x) = x^2 - 3.\) Both are polynomials with domain \((-\infty, \infty).\)
Step 1 — Compute \((f \circ g)(x):\)
$$f(g(x)) = (g(x))^2 + 7 = (x^2 - 3)^2 + 7.$$
Expand \((x^2 - 3)^2 = x^4 - 6x^2 + 9,\) so
$$f(g(x)) = x^4 - 6x^2 + 9 + 7 = x^4 - 6x^2 + 16.$$
Domain: \((-\infty, \infty).\)
Step 2 — Compute \((g \circ f)(x):\)
$$g(f(x)) = (f(x))^2 - 3 = (x^2 + 7)^2 - 3.$$
Expand \((x^2 + 7)^2 = x^4 + 14x^2 + 49,\) so
$$g(f(x)) = x^4 + 14x^2 + 49 - 3 = x^4 + 14x^2 + 46.$$
Domain: \((-\infty, \infty).\)
Answer: (a) \((f \circ g)(x) = x^4 - 6x^2 + 16;\) (b) \((g \circ f)(x) = x^4 + 14x^2 + 46;\) each has domain \((-\infty, \infty).\)
Problem 46
Given \(f(x) = \sqrt{x}\) (domain \([0, \infty)\)) and \(g(x) = x + 9\) (domain \((-\infty, \infty)\)).
Step 1 — Compute \((f \circ g)(x):\)
$$f(g(x)) = \sqrt{g(x)} = \sqrt{x + 9}.$$
The inner output \(x + 9\) must lie in the domain of \(f,\) i.e. \(x + 9 \ge 0,\) so \(x \ge -9.\) Domain: \([-9, \infty).\)
Step 2 — Compute \((g \circ f)(x):\)
$$g(f(x)) = f(x) + 9 = \sqrt{x} + 9.$$
\(f(x) = \sqrt{x}\) requires \(x \ge 0.\) Domain: \([0, \infty).\)
Answer: (a) \((f \circ g)(x) = \sqrt{x + 9},\) domain \([-9, \infty);\) (b) \((g \circ f)(x) = \sqrt{x} + 9,\) domain \([0, \infty).\)
Problem 47
Given \(f(x) = \dfrac{3}{2x + 1}\) (domain \(x \neq -\tfrac{1}{2}\)) and \(g(x) = \dfrac{2}{x}\) (domain \(x \neq 0\)).
Step 1 — Compute \((f \circ g)(x):\)
$$f(g(x)) = \frac{3}{2 g(x) + 1} = \frac{3}{2 \cdot \dfrac{2}{x} + 1} = \frac{3}{\dfrac{4}{x} + 1}.$$
Multiply top and bottom by \(x\) (assuming \(x \neq 0\)):
$$= \frac{3x}{4 + x} = \frac{3x}{x + 4}.$$
Domain restrictions: \(x \neq 0\) (so \(g\) is defined) and the inner output cannot equal \(-\tfrac{1}{2}\) (so \(f\) is defined). Solve \(\dfrac{2}{x} = -\tfrac{1}{2}\) to get \(x = -4.\) Combined: \(x \neq 0\) and \(x \neq -4.\) Domain: \((-\infty, -4) \cup (-4, 0) \cup (0, \infty).\)
Step 2 — Compute \((g \circ f)(x):\)
$$g(f(x)) = \frac{2}{f(x)} = \frac{2}{\dfrac{3}{2x + 1}} = \frac{2(2x + 1)}{3} = \frac{4x + 2}{3}.$$
Restrictions: \(x \neq -\tfrac{1}{2}\) (so \(f\) is defined) and \(f(x) \neq 0\) (so \(g\) is defined). But \(f(x) = \dfrac{3}{2x+1}\) is never zero, so the only restriction is \(x \neq -\tfrac{1}{2}.\) Domain: \(\left(-\infty, -\tfrac{1}{2}\right) \cup \left(-\tfrac{1}{2}, \infty\right).\)
Answer: (a) \((f \circ g)(x) = \dfrac{3x}{x + 4},\) domain \((-\infty, -4) \cup (-4, 0) \cup (0, \infty);\) (b) \((g \circ f)(x) = \dfrac{4x + 2}{3},\) domain \(\left(-\infty, -\tfrac{1}{2}\right) \cup \left(-\tfrac{1}{2}, \infty\right).\)
Problem 48
Given \(f(x) = |x + 1|\) and \(g(x) = x^2 + x - 4.\) Both have domain \((-\infty, \infty).\)
Step 1 — Compute \((f \circ g)(x):\)
$$f(g(x)) = |g(x) + 1| = |x^2 + x - 4 + 1| = |x^2 + x - 3|.$$
Domain: \((-\infty, \infty).\)
Step 2 — Compute \((g \circ f)(x):\)
$$g(f(x)) = (f(x))^2 + f(x) - 4 = |x + 1|^2 + |x + 1| - 4.$$
Use \(|x + 1|^2 = (x + 1)^2 = x^2 + 2x + 1\) (squaring removes the absolute value), so
$$g(f(x)) = x^2 + 2x + 1 + |x + 1| - 4 = x^2 + 2x - 3 + |x + 1|.$$
Domain: \((-\infty, \infty).\)
Answer: (a) \((f \circ g)(x) = |x^2 + x - 3|;\) (b) \((g \circ f)(x) = x^2 + 2x - 3 + |x + 1|;\) each has domain \((-\infty, \infty).\)
Problem 49
The table pairs each year 2001–2012 with that season's NBA championship winner.
Step 1 — Domain = years, range = winners:
A relation is a function if every element of the domain corresponds to exactly one element of the range. Each year on the list has exactly one champion (a season produces one trophy), so every domain element has a unique image. Therefore this relation is a function.
Step 2 — Domain = winners, range = years:
Now check whether each winner is associated with exactly one year. The LA Lakers appear in 2001, 2002, 2009, and 2010 — four different years — so the input "LA Lakers" maps to four different outputs. The San Antonio Spurs (2003, 2005, 2007) and Miami Heat (2006, 2012) likewise map to multiple years. Therefore this relation is not a function.
Answer: (a) Yes, the year-to-winner relation is a function, because each year has exactly one champion. (b) No, the winner-to-year relation is not a function: several teams (LA Lakers, San Antonio Spurs, Miami Heat) won in more than one year, so a single input has multiple outputs.
Problem 50
The area of a square depends on the side length \(s.\)
Step 1 — Write \(A(s):\)
By the area-of-a-square formula,
$$A(s) = s^2.$$
The natural domain (geometric side lengths) is \(s > 0.\)
Step 2 — Evaluate \(A(6.5)\) and interpret:
$$A(6.5) = (6.5)^2 = 42.25.$$
Interpretation: a square with side \(6.5\) units has an area of \(42.25\) square units.
Step 3 — Solve \(s^2 = 56\) for the exact and approximate side length:
$$s = \sqrt{56} = \sqrt{4 \cdot 14} = 2\sqrt{14}.$$
Numerically, \(\sqrt{14} \approx 3.7417,\) so
$$s \approx 2 \cdot 3.7417 \approx 7.4833 \approx 7.5 \text{ (two significant digits)}.$$
Answer: (a) \(A(s) = s^2;\) (b) \(A(6.5) = 42.25\) square units — a square with side \(6.5\) has area \(42.25;\) (c) exact side length \(s = 2\sqrt{14}\) units, approximately \(7.5\) units to two significant digits.
Problem 51
The volume of a cube depends on its side length \(s.\)
Step 1 — Write \(V(s):\)
The standard cube-volume formula is
$$V(s) = s^3.$$
The natural geometric domain is \(s > 0.\)
Step 2 — Evaluate \(V(11.8)\):
$$V(11.8) = (11.8)^3.$$
Compute step by step:
$$(11.8)^2 = 139.24,$$
$$(11.8)^3 = 139.24 \times 11.8 = 1{,}643.032.$$
So \(V(11.8) = 1{,}643.032\) cubic units.
Step 3 — Interpret:
A cube with side \(11.8\) units has a volume of approximately \(1{,}643.03\) cubic units.
Answer: (a) \(V(s) = s^3;\) (b) \(V(11.8) = 1{,}643.032\) cubic units — the volume of a cube whose side is \(11.8\) units.
Problem 52
The car rental cost is a flat $20 plus an hourly rate of $10.25 per hour.
Step 1 — Write the cost function \(C(t):\)
If \(t\) is the number of hours rented, the cost is
$$C(t) = 20 + 10.25\,t.$$
Domain: \(t \ge 0.\)
Step 2 — Cost for 2 days and 7 hours:
Convert to hours: \(2 \text{ days} = 48 \text{ hours},\) so total \(t = 48 + 7 = 55\) hours.
$$C(55) = 20 + 10.25 \cdot 55 = 20 + 563.75 = 583.75.$$
So the rental cost is $583.75.
Step 3 — Solve \(C(t) = 430\) for \(t:\)
$$20 + 10.25\,t = 430,$$
$$10.25\,t = 410,$$
$$t = \frac{410}{10.25} = 40.$$
The car was rented for \(40\) hours, which is \(1\) day and \(16\) hours.
Answer: (a) \(C(t) = 20 + 10.25\,t;\) (b) $583.75 to rent the car for 2 days 7 hours (\(55\) hours); (c) the bill of $430 corresponds to \(t = 40\) hours of rental.
Problem 53
The vehicle has a 20-gal tank and gets 15 mpg, so for every gallon of gas it can travel 15 miles.
Step 1 — Write \(N(x):\)
If \(x\) is the gallons of gas in the tank,
$$N(x) = 15 x.$$
Step 2 — Miles on (i) a full tank and (ii) 3/4 of a tank:
(i) Full tank \(x = 20\):
$$N(20) = 15 \cdot 20 = 300 \text{ miles}.$$
(ii) Three-quarter tank \(x = \tfrac{3}{4} \cdot 20 = 15\):
$$N(15) = 15 \cdot 15 = 225 \text{ miles}.$$
Step 3 — Domain and range:
Gas in the tank ranges from empty to full: \(0 \le x \le 20.\) Miles driven on that gas: \(0 \le N \le 300.\) Domain: \([0, 20]\) gal. Range: \([0, 300]\) mi.
Step 4 — Stops for gas after \(578\) miles:
Each full tank covers \(300\) miles. After the first \(300\) miles the tank is empty, requiring a refill (stop 1). After the next \(300\) miles (i.e. at \(600\) miles total) another refill would be needed — but the trip ends at \(578\) miles, which is before that second empty point. So between \(0\) and \(578\) miles she ran out exactly once and made one stop for gas (after the first \(300\) miles). The car still has gas left at the end of the trip.
Answer: (a) \(N(x) = 15 x;\) (b) (i) \(300\) mi on a full tank; (ii) \(225\) mi on 3/4 of a tank; (c) domain \([0, 20]\) gal, range \([0, 300]\) mi; (d) she stopped for gas \(1\) time during the \(578\)-mi trip.
Problem 54
Earth is approximated as a sphere with mean radius \(r = 6.371 \times 10^6\) m and \(V = \tfrac{4}{3}\pi r^3.\)
Step 1 — Substitute \(r:\)
$$V = \frac{4}{3}\pi (6.371 \times 10^6)^3.$$
Step 2 — Cube the radius:
$$(6.371)^3 \approx 258.50, \quad (10^6)^3 = 10^{18},$$
so \((6.371 \times 10^6)^3 \approx 2.585 \times 10^{20}\) m\(^3.\) (More precisely \(2.58502 \times 10^{20}.\))
Step 3 — Multiply by \(\tfrac{4}{3}\pi:\)
$$\frac{4}{3}\pi \approx 4.18879,$$
$$V \approx 4.18879 \times 2.585 \times 10^{20} \approx 1.083 \times 10^{21} \text{ m}^3.$$
Step 4 — Sanity check:
Earth's accepted volume is roughly \(1.08 \times 10^{21}\) m\(^3,\) matching our result.
Answer: \(V \approx 1.083 \times 10^{21}\) cubic meters (using the mean radius \(6.371 \times 10^6\) m).
Problem 55
The bacterial colony fills a disk of radius \(r(t) = 6 - \dfrac{5}{t^2 + 1}\) cm at time \(t\) hours.
Step 1 — Area as a function of time:
A circle of radius \(r\) has area \(A = \pi r^2,\) so
$$A(t) = \pi \left(6 - \frac{5}{t^2 + 1}\right)^2.$$
Step 2 — Area at \(t = 3\) hours:
First compute the radius:
$$r(3) = 6 - \frac{5}{3^2 + 1} = 6 - \frac{5}{10} = 6 - \tfrac{1}{2} = \tfrac{11}{2} \text{ cm}.$$
Then
$$A(3) = \pi \left(\tfrac{11}{2}\right)^2 = \pi \cdot \tfrac{121}{4} = \tfrac{121\pi}{4} \text{ cm}^2.$$
Numerically, \(\tfrac{121\pi}{4} \approx \tfrac{121 \cdot 3.14159}{4} \approx \tfrac{380.13}{4} \approx 95.03\) cm\(^2.\)
Step 3 — Circumference as a function of time:
For a circle, \(C = 2\pi r,\) so
$$C(t) = 2\pi \left(6 - \frac{5}{t^2 + 1}\right).$$
Step 4 — Circumference at \(t = 3\) hours:
Using \(r(3) = \tfrac{11}{2}\) cm,
$$C(3) = 2\pi \cdot \tfrac{11}{2} = 11\pi \text{ cm} \approx 34.56 \text{ cm}.$$
Answer: (a) \(A(t) = \pi\left(6 - \dfrac{5}{t^2 + 1}\right)^2;\) (b) at \(t = 3\): exact \(A = \dfrac{121\pi}{4}\) cm\(^2,\) approximately \(95.03\) cm\(^2;\) (c) \(C(t) = 2\pi\left(6 - \dfrac{5}{t^2 + 1}\right);\) (d) at \(t = 3\): exact \(C = 11\pi\) cm, approximately \(34.56\) cm.
Problem 56
The two conversions are \(E(x) = 0.79\,x\) (US dollars \(\to\) Euros) and \(D(x) = 1.245\,x\) (Euros \(\to\) US dollars).
Step 1 — Compose to get USD \(\to\) Euros \(\to\) USD:
Apply \(E\) first (converting \(x\) dollars to Euros), then \(D\) (converting those Euros back to dollars):
$$(D \circ E)(x) = D(E(x)) = D(0.79\,x) = 1.245(0.79\,x).$$
Multiply: \(1.245 \cdot 0.79 = 0.98355.\) So
$$(D \circ E)(x) = 0.98355\,x.$$
Step 2 — Did the tourist lose value?
The composite coefficient is \(0.98355,\) which is less than \(1.\) Starting with \(\$x\) and ending with \(\$0.98355\,x\) means she has \(1.645\%\) less than she started with after the round trip. Yes, she lost value (because of the spread between the two conversion rates).
Step 3 — Dollars back from an extra $200:
$$(D \circ E)(200) = 0.98355 \cdot 200 = 196.71.$$
She gets back $196.71, losing $3.29 on the round-trip conversion of the extra $200.
Note on the graphing calculator: A graphing-calculator solution would plot \(y_1 = D(E(x))\) and \(y_2 = x\) and observe that \(y_1\) lies just below \(y_2,\) confirming the value loss; evaluating \(y_1\) at \(x = 200\) returns \(196.71.\)
Answer: (a) \((D \circ E)(x) = 0.98355\,x;\) yes, the tourist loses value because the round-trip multiplier is less than \(1;\) (b) the tourist receives $196.71 in U.S. dollars from her extra $200.
Problem 57
A worker earns a flat $750 plus $8.50 for each skateboard sold.
Step 1 — Write \(S(x):\)
If \(x\) is the number of skateboards sold, the monthly salary is
$$S(x) = 750 + 8.50\,x.$$
Natural domain: nonnegative integers (you cannot sell a fractional number of boards).
Step 2 — Evaluate at \(x = 25, 40, 55\):
$$S(25) = 750 + 8.50 \cdot 25 = 750 + 212.50 = 962.50,$$
$$S(40) = 750 + 8.50 \cdot 40 = 750 + 340 = 1{,}090,$$
$$S(55) = 750 + 8.50 \cdot 55 = 750 + 467.50 = 1{,}217.50.$$
So the monthly salaries are $962.50, $1,090, and $1,217.50, respectively.
Step 3 — Find \(x\) such that \(S(x) = 1{,}400\) (the INTERSECT step):
Solve algebraically (this is what the calculator's INTERSECT feature would return):
$$750 + 8.50\,x = 1{,}400,$$
$$8.50\,x = 650,$$
$$x = \frac{650}{8.50} \approx 76.47.$$
Since boards sell as whole units, the worker must sell at least \(77\) skateboards to reach $1,400 (selling \(76\) would give \(S(76) = 750 + 646 = \$1{,}396,\) just shy).
Note on the graphing calculator: Graphing \(y_1 = 750 + 8.50\,x\) and \(y_2 = 1400,\) then using INTERSECT, returns the point \((76.47, 1400);\) interpret this in context as "at least 77 boards."
Answer: (a) \(S(x) = 750 + 8.50\,x;\) (b) \(S(25) = \$962.50,\ S(40) = \$1{,}090,\ S(55) = \$1{,}217.50;\) (c) the intersection occurs at \(x \approx 76.47,\) so the worker must sell at least \(77\) skateboards to earn at least $1,400 per month.
Problem 58
Key Terms
input — an element of the domain that is fed into a function.
output — the element of the range produced by applying the function to an input.
function — a rule that pairs each input with exactly one output.
domain — the set of all valid inputs to a function.
range — the set of all outputs the function actually produces.
independent variable — typically \(x\); the input variable.
dependent variable — typically \(y\); the output variable, since it depends on \(x\).
graph of a function — the set of all points \((x, f(x))\) for \(x\) in the domain.
interval notation — compact notation for sets of real numbers using \((a, b)\), \([a, b]\), and infinity symbols.
endpoints — the boundary values \(a\) and \(b\) of an interval.
piecewise-defined function — a function defined by different rules on different parts of its domain.
table of values — a list of input/output pairs that describes a function on a finite set of inputs.
zeros of a function — the values of \(x\) for which \(f(x) = 0\).
vertical line test — a graph represents a function of \(x\) iff every vertical line meets the graph at most once.
increasing on the interval \(I\) — \(f(x_1) \le f(x_2)\) whenever \(x_1 < x_2\) in \(I\).
decreasing on the interval \(I\) — \(f(x_1) \ge f(x_2)\) whenever \(x_1 < x_2\) in \(I\).
composite function — \((g \circ f)(x) = g(f(x))\); the output of \(f\) is the input to \(g\).
symmetry about the \(y\)-axis — \((x, y)\) on the graph iff \((-x, y)\) is too.
symmetry about the origin — \((x, y)\) on the graph iff \((-x, -y)\) is too.
even function — \(f(-x) = f(x)\) for every \(x\) in the domain.
odd function — \(f(-x) = -f(x)\) for every \(x\) in the domain.
absolute value function — \(|x| = x\) for \(x \ge 0\) and \(|x| = -x\) for \(x < 0\).
Solution
The curve is \(y = \sqrt{25 - (x - 4)^2},\) the upper half of a circle.
Step 1 — Recognize the shape:
Squaring both sides (and noting \(y \ge 0\)) gives
$$y^2 = 25 - (x - 4)^2 \implies (x - 4)^2 + y^2 = 25,$$which is a circle of radius \(5\) centered at \((4, 0).\) Taking the nonnegative square root keeps only the upper half-disk, so the graph is a semicircle from \((-1, 0)\) on the left to \((9, 0)\) on the right with its peak at \((4, 5).\)
Step 2 — Find the \(x\)-intercepts (set \(y = 0\)):
$$0 = \sqrt{25 - (x - 4)^2} \implies 25 - (x - 4)^2 = 0 \implies (x - 4)^2 = 25.$$So \(x - 4 = \pm 5,\) giving \(x = 9\) or \(x = -1.\) The \(x\)-intercepts are \((-1, 0)\) and \((9, 0).\)
Step 3 — Find the \(y\)-intercept (set \(x = 0\)):
$$y = \sqrt{25 - (0 - 4)^2} = \sqrt{25 - 16} = \sqrt{9} = 3.$$The \(y\)-intercept is \((0, 3).\)
Note on the graphing calculator: Entering \(Y_1 = \sqrt{25 - (X - 4)^2}\) and using the INTERCEPT (or ZERO) feature returns the same numbers: \(x\)-zeros at \(-1\) and \(9,\) and \(Y_1(0) = 3.\) The picture is an upward-bulging half-circle of radius \(5\) centered horizontally on \(x = 4.\)
Answer: \(x\)-intercepts: \((-1, 0)\) and \((9, 0);\) \(y\)-intercept: \((0, 3).\)