Int-Alg Some Basic Graphs

Section 5.3 Some Basic Graphs

In this section we study the graphs of some important basic functions. Many functions fall into families or classes of similar functions, and recognizing the appropriate family for a given situation is an important part of modeling.

We’ll need two new algebraic operations.

Subsection 5.3.1 Cube Roots

You are familiar with square roots. Every non-negative number has two square roots, defined as follows.

\begin{equation*} \blert{~s~~~\text{is a square root of}~~~n~~~\text{if}~~~ s^2=n} \end{equation*}

There are several other kinds of roots, one of which is called the cube root. We define the cube root as follows.

Definition 5.3.1. Cube Root.

$b$ is the cube root of $a$ if $b$ cubed equals $a\text{.}$ In symbols, we write

\begin{equation*} \blert{b=\sqrt[3]{a}~~~~\text{if}~~~~b^3=a} \end{equation*}

Square roots of negative numbers are not real (they are complex), but every real number has a real cube root. For example,

\begin{align*} 4=\sqrt[3]{64}~~~~~~ \amp \text{because} \amp \amp 4^3=64\\ -3=\sqrt[3]{-27}~~~~~~ \amp \text{because} \amp \amp (-3)^3=-27 \end{align*}

Simplifying radicals occupies the same position in the order of operations as computing powers: after parentheses, and before products and quotients.

Example 5.3.2.

Simplify each expression. $~\alert{\text{[TK]}}$

$\displaystyle 3\sqrt[3]{-8}$
$\displaystyle 2-\sqrt[3]{-125}$

Solution.

$\displaystyle 3\sqrt[3]{-8}=3(-2)=-6$
$\displaystyle 2-\sqrt[3]{-125}=2-(-5)=7$

Checkpoint 5.3.3. Practice 1.

Simplify each expression.

$\displaystyle 5-3\sqrt[3]{64}$
$\displaystyle \dfrac{6-\sqrt[3]{-27}}{2}$

Solution.

$\displaystyle -7$
$\displaystyle \dfrac{9}{2}$

Checkpoint 5.3.4. QuickCheck 1.

Decide whether each statement is true or false.

A negative number has a negative cube root.
A negative number has a negative square root.
A positive number has a negative square root
A positive number has a negative cube root.

Note 5.3.5.

We can use the calculator to find cube roots as follows. Press the MATH key to get a menu of options. Option 4 is labeled $\sqrt[3]{~~}\text{;}$ this is the cube root key. To find the cube root of, say, 15.625, we key in $\qquad\qquad$ MATH 4 $15.625$ ) ENTER and the calculator returns the result, 2.5. Thus, $\sqrt[3]{15.625} = 2.5\text{.}$ You can check this result by verifying that $~2.5^3=15.625\text{.}$

Subsection 5.3.2 Absolute Value

We use the absolute value to discuss problems involving distance. For example, consider the number line below. Starting at the origin, we travel in opposite directions to reach the two numbers $6$ and $-6\text{,}$ but the distance we travel in each case is the same.

The distance from a number $c$ to the origin is called the absolute value of $c\text{,}$ denoted by $\abs{c}\text{.}$ Because distance is never negative, the absolute value of a number is always positive (or zero). Thus, $\abs{6}= 6$ and $\abs{-6} = 6\text{.}$ In general, we define the absolute value of a number $x$ as follows.

Definition 5.3.6. Absolute Value.

The absolute value of $x$ is defined by

\begin{align*} \abs{x} = \begin{cases} x \amp \text{if } x\ge 0\\ -x \amp \text{if } x\lt 0 \end{cases} \end{align*}

This definition is called piecewise, because the formula has two pieces. It says that the absolute value of a positive number (or zero) is the same as the number. To find the absolute value of a negative number, we take the opposite of the number, which is then positive. For instance,

\begin{equation*} \abs{-6}=-(-6)=6 \end{equation*}

Checkpoint 5.3.7. QuickCheck 2.

If $\left| x \right| =-x \text{,}$ what can you say about $x \text{?}$

$x$ must be zero.
$x$ must be negative.
$x$ must be zero or negative.
This cannot happen for any value of $x\text{.}$

Absolute value bars act like grouping devices in the order of operations: you should complete any operations that appear inside absolute value bars before you compute the absolute value.

Example 5.3.8.

Simplify each expression. $~\alert{\text{[TK]}}$

$\displaystyle \abs{3 - 8}$
$\displaystyle \abs{3} - \abs{8}$

Solution.

We simplify the expression inside the absolute value bars first.

\begin{equation*} \abs{3 - 8} = \abs{-5} = 5 \end{equation*}
We simplify each absolute value; then subtract.

\begin{equation*} \abs{3} - \abs{8} = 3 - 8 = -5 \end{equation*}

Checkpoint 5.3.9. Practice 2.

Simplify each expression.

$\displaystyle 12 - 3 \left|-6\right|=$
$\displaystyle -7 - 3\left|2 - 9\right|=$

Solution.

$\displaystyle -6$
$\displaystyle -28$

Checkpoint 5.3.10. QuickCheck 3.

True or False: $\left| {a+b} \right| = \left|{a}\right| + \left|{b}\right|$

Hint: Try some values of $a$ and $b\text{.}$

Subsection 5.3.3 Eight Basic Graphs

Most of the graphs in this section will be new to you, but many useful graphs are variations of the eight basic functions shown below.

Consider the first pair of graphs. You have already studied the graph of $f(x)=x^2\text{,}$ the basic parabola. Compare that graph with the graph of $g(x)=x^3\text{.}$ Notice several differences in the shape of the two graphs. Once you have a good idea of the shape of a graph, up can make a quick sketch with just a few "guide points." For these two graphs, complete a short table of values to find useful guide points:

$x$	$-2$	$-1$	$0$	$1$	$2$
$f(x)$	$\qquad$	$\qquad$	$\qquad$	$\qquad$	$\qquad$
$g(x)$	$~~$	$~~$	$~~$	$~~$	$~~$

The next pair of graphs are $f(x)=\sqrt{x}$ and $g(x)=\sqrt[3]{x}\text{.}$ Once again, notice the differences in the two graphs. For example, we cannot take the square root of a negative number, but we can take its cube root. How is this reflected in the graphs?

$square root graph$

$cube root graph$

The next pair of functions, $f(x)=\dfrac{1}{x}$ and $g(x)=\dfrac{1}{x^2}\text{,}$ are both undefined at $x=0\text{,}$ so thier graphs do not include any points with $x$-coordinate zero. For very small positive values of $x\text{,}$ both $f(x)$ and $g(x)$ get very large. As $x$ gets closer to zero, the graphs approach the vertical line $x=0$ (the $y$-axis). This line is called a vertical asymptote for the graph.

Also, notice that for very large values of $x\text{,}$ both $f(x)$ and $g(x)$ get very close to zero. Their graphs approach the horizontal line $y=0$ (the $x$-axis). This line is called the horizontal asymptote for the graph.

Finally, compare the familiar graph of $f(x)=x$ with the graph of $g(x)=\abs{x}\text{.}$ The piecewise definition of $\abs{x}$ means that we graph $y=x$ in the first quadrant (where $x \ge 0$), and $y=-x$ in the first quadrant $x \lt 0$). The result is the V-shaped graph shown below.

$y=x$

$y=abs(x)$

Because they are fundamental to further study of mathematics and its applications, you should become familiar with the properties of these eight graphs, and be able to sketch them easily from memory, using their basic shapes and a few guidepoints.

Exercises 5.3.4 Problem Set 5.3

Warm Up

1.

Evaluate each function.

$\displaystyle f(x)=-2x^3-3x^2;~~f(-2)$
$\displaystyle g(x)=\dfrac{x-1}{x^2+2x};~~g(-1)$

Exercise Group.

For Problems 2 and 3, compute each cube root. Round your answers to three decimal places if necessary. Verify your answers by cubing them.

2.

$\displaystyle \sqrt[3]{512}$
$\displaystyle \sqrt[3]{-125}$
$\displaystyle \sqrt[3]{-0.064}$
$\displaystyle \sqrt[3]{1.728}$

3.

$\displaystyle \sqrt[3]{9}$
$\displaystyle \sqrt[3]{258}$
$\displaystyle \sqrt[3]{-0.02}$
$\displaystyle \sqrt[3]{-3.1}$

Exercise Group.

For Problems 4–6, simplify each by following the order of operations.

4.

$\displaystyle \dfrac{6-2\sqrt[3]{64}}{2}$
$\displaystyle 2\sqrt[3]{-125}-\sqrt[3]{6^2-3^2}$

5.

$\displaystyle \sqrt[3]{\dfrac{8-1}{64-8}}$
$\displaystyle \dfrac{4+\sqrt[3]{-216}}{8-\sqrt[3]{8}}$

6.

$\displaystyle \sqrt[3]{3^3+4^3+5^3}$
$\displaystyle \sqrt[3]{9^3+10^3-1^3}$

Exercise Group.

For problems 7–10, simplify the expression according to the order of operations.

7.

$\displaystyle -\abs{-9} $
$\displaystyle -(-9) $
$\displaystyle 2-(-6) $
$\displaystyle 2-\abs{-6} $

8.

$\displaystyle \abs{-8}-\abs{12} $
$\displaystyle \abs{-8-12} $
$\displaystyle \abs{-3}+\abs{-5} $
$\displaystyle \abs{-3+(-5)} $

9.

$\displaystyle 4-9\abs{2-8} $
$\displaystyle 2-5\abs{-6-3} $
$\displaystyle \abs{-4-5} \abs{1-3(-5)} $

10.

$\displaystyle \abs{-3+7}\abs{-2(6-10)} $
$\displaystyle \abs{ \abs{-5}-\abs{-6}} $
$\displaystyle \abs{ \abs{4}-\abs{-6}} $

Skills Practice

Exercise Group.

For Problems 11–16, sketch the graph of the function by hand, paying attention to the shape of the graph. Carefully plot at least three “guide points” to ensure accuracy. If possible, plot the points with $x$-coordinates $-1, ~0,$ and $1\text{.}$

11.

$f(x)=x^3$

$grid$

12.

$f(x)=\abs{x}$

$grid$

13.

$f(x)=\sqrt{x}$

$grid$

14.

$f(x)=\sqrt[3]{x}$

$grid$

15.

$f(x)=\dfrac{1}{x}$

$grid$

16.

$f(x)=\dfrac{1}{x^2}$

17.

Use your calculator to graph $f(x)=x^2$ and $g(x)=x^3$ on the same axes for $0 \le x \le 1\text{.}$ Which function is greater on that interval?
Use your calculator to graph $f(x)=\sqrt{x}$ and $g(x)=\sqrt[3]{x}$ on the same axes for $0 \le x \le 1\text{.}$ Which function is greater on that interval?

18.

Use your calculator to graph $f(x)=\dfrac{1}{x}$ and $g(x)=\dfrac{1}{x^2}$ on the same axes for $0 \le x \le 1\text{.}$ Which function is greater on that interval?
Use your calculator to graph $f(x)=\dfrac{1}{x}$ and $g(x)=\dfrac{1}{x^2}$ on the same axes for $1 \le x \le 4\text{.}$ Which function is greater on that interval?

Applications

19.

Use the graph of $y=\dfrac{1}{x}$ to solve the inequality $\dfrac{1}{x} \le 2\text{.}$

$graph of 1/x$

20.

Use the graph of $y=\abs{x-2}$ to solve the inequality $\abs{x-2} \gt 1\text{.}$

$graph of abs(x-2)$

Exercise Group.

For Problems 21–26, graph the functions in the same window on your calculator. Describe how the graphs in parts (b) and (c) are different from the basic graph.