Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences

Subsection 1.1.1 Sets and Number Systems

A set can be thought of as any collection of distinct objects considered as a whole. Typically, sets are represented using set-builder notation and are surrounded by braces. Recall that \((,)\) are called parentheses or round brackets; \([,]\) are called square brackets; and \(\{,\}\) are called braces or curly brackets.

Let \(S\) be any set. We use the notation \(x\in S\) to mean that \(x\) is an element inside of the set \(S\text{,}\) and the notation \(x\not\in S\) to mean that \(x\) is not an element of the set \(S\text{.}\)

The intersection between two sets \(S\) and \(T\) is denoted by \(S\cap T\) and is the collection of all elements that belong to both \(S\) and \(T\text{.}\) The union between two sets \(S\) and \(T\) is denoted by \(S\cup T\) and is the collection of all elements that belong to either \(S\) or \(T\) (or both).

Numbers can be classified into sets called number systems.

In the table, the set of rational numbers is written using set-builder notation. The colon, \(:\text{,}\) used in this manner means such that. Often times, a vertical bar \(|\) may also be used to mean such that. The expression \(\left\{\frac{p}{q}\,:\,p,q\in\mathbb{Z},q\not=0\right\}\) can be read out loud as "the set of all fractions \(p\) over \(q\) such that \(p\) and \(q\) are both integers and \(q\) is not equal to zero".

All rational numbers are real numbers with the property that their decimal expansion either terminates after a finite number of digits or begins to repeat the same finite sequence of digits over and over. Real numbers that are not rational are called irrational.

Let \(S\) and \(T\) be two sets. If every element of \(S\) is also an element of \(T\text{,}\) then we say \(S\) is a subset of \(T\) and write \(S\subseteq T\text{.}\) Furthermore, if \(S\) is a subset of \(T\) but not equal to \(T\text{,}\) we often write \(S\subset T\text{.}\) The five sets of numbers in the table give an increasing sequence of sets:

That is, all natural numbers are also integers, all integers are also rational numbers, all rational numbers are also real numbers, and all real numbers are also complex numbers.

Subsection 1.1.2 Law of Exponents

The Law of Exponents is a set of rules for simplifying expressions that governs the combination of exponents (powers). Recall that \(\sqrt[n]{~}\) denotes the \(n\)-th root. For example \(\sqrt[3]{8}=2\) represents that the cube root of \(8\) is equal to \(2\text{.}\)

In the next example, the word simplify means to make simpler or to write the expression more compactly.

Subsection 1.1.3 The Quadratic Formula and Completing the Square

The technique of completing the square allows us to solve quadratic equations and also to determine the center of a circle/ellipse or the vertex of a parabola.

The main idea behind completing the square is to turn:

into

One way to complete the square is to use the following formula:

But this formula is a bit complicated, so some students prefer following the steps outlined in the next example.

Suppose we want to solve for \(x\) in the quadratic equation \(ax^2+bx+c=0\text{,}\) where \(a\neq 0\text{.}\) The solution(s) to this equation are given by the quadratic formula.

Subsection 1.1.4 Inequality Notation

Recall that we use the symbols \(\lt ,>,\leq,\geq\) when writing an inequality. In particular,

• \(a\lt b\) means \(a\) is to the left of \(b\) (that is, \(a\) is strictly less than \(b\)),

• \(a\leq b\) means \(a\) is to the left of or the same as \(b\) (that is, \(a\) is less than or equal to \(b\)),

• \(a>b\) means \(a\) is to the right of \(b\) (that is, \(a\) is strictly greater than \(b\)),

• \(a\geq b\) means \(a\) is to the right of or the same as \(b\) (that is, \(a\) is greater than or equal to \(b\)).

To keep track of the difference between the symbols, some students use the following mnemonic.

The real numbers are ordered and are often illustrated using the real number line:

Subsection 1.1.5 Intervals

Assume \(a,b\) are real numbers with \(a\lt b\) (i.e., \(a\) is strictly less than \(b\)). An interval is a set of every real number between two indicated numbers and may or may not contain the two numbers themselves. When describing intervals we use both round brackets and square brackets.

(1) Use of round brackets in intervals: \((~,~)\text{.}\) The notation (a,b) is what we call the open interval from a to b and consists of all the numbers between \(a\) and \(b\text{,}\) but does not include \(a\) or \(b\text{.}\) Using set-builder notation we write this as:

We read \(\{x\in\mathbb{R}\,:\,a\lt x\lt b\}\) as “the set of real numbers \(x\) such that \(x\) is greater than \(a\) and less than \(b\)” On the real number line we represent this with the following diagram:

Note that the circles on \(a\) and \(b\) are not shaded in, we call these open circles and use them to denote that \(a\text{,}\) \(b\) are omitted from the set.

(2) Use of square brackets in intervals: \([~,~]\text{.}\) The notation [a,b] is what we call the closed interval from a to b and consists of all the numbers between \(a\) and \(b\) and including \(a\) and \(b\text{.}\) Using set-builder notation we write this as

On the real number line we represent this with the following diagram:

Note that the circles on \(a\) and \(b\) are shaded in, we call these closed circles and use them to denote that \(a\) and \(b\) are included in the set.

To keep track of when to shade a circle in, you may find the following mnemonic useful:

Note: Any set which is bound at positive and/or negative infinity is an open interval.

Taking combinations of round and square brackets, we can write different possible types of intervals (we assume \(a\lt b\)):

\((a,b)=\{x\in\mathbb{R}\,:\,a\lt x\lt b\} \)	\([a,b]=\{x\in\mathbb{R}\,:\,a\leq x\leq b\} \)	\([a,b)=\{x\in\mathbb{R}\,:\,a\,\leq x\lt b\}\)
\((a,b]=\{x\in\mathbb{R}\,:\,a\lt x\leq b\} \)	\((a,\infty)=\{x\in\mathbb{R}\,:\,x>a\} \)	\([a,\infty)=\{x\in\mathbb{R}\,:\,x\geq a\} \)
\((-\infty,b)=\{x\in\mathbb{R}\,:\,x\lt b\} \)	\((-\infty,b]=\{x\in\mathbb{R}\,:\,x\leq b\} \)	\((-\infty,\infty)=\mathbb{R}=\mbox{all real numbers} \)

Interactive Demonstration. Investigate interval notation by sliding the points below:

Subsection 1.1.6 Inequality Rules

Before solving inequalities, we start with the properties and rules of inequalities.

Similar rules hold for each of \(\leq\text{,}\) \(>\) and \(\geq\text{.}\)

Subsection 1.1.7 Solving Basic Inequalities

We can use the inequality rules to solve some simple inequalities.

Example 1.10. Basic Inequality.

Find all values of \(x\) satisfying

\begin{equation*} 3x+1>2x-3\text{.} \end{equation*}

Write your answer in both interval and set-builder notation. Finally, draw a number line indicating your solution set.

Solution

Subtracting \(2x\) from both sides gives \(x+1>-3\text{.}\) Subtracting \(1\) from both sides gives \(x>-4\text{.}\) Therefore, the solution is the interval \((-4,\infty)\text{.}\) In set-builder notation the solution may be written as \(\{x\in\mathbb{R}\,:\,x>-4\}\text{.}\) We illustrate the solution on the number line as follows:

Sometimes we need to split our inequality into two cases as the next example demonstrates.

We next present another method to solve more complicated looking inequalities. In the next example we will solve a rational inequality by using a number line and test points. We follow the guideline below.

Example 1.13. Rational Inequality.

Write the solution to the following inequality using interval notation:

\begin{equation*} \frac{2-x}{2+x}\geq 1\text{.} \end{equation*}

Solution

One method to solve this inequality is to multiply both sides by \(2+x\text{,}\) but because we do not know if \(2+x\) is positive or negative we must split it into two cases (Case 1: \(2+x>0\) and Case 2: \(2+x\lt 0\)).

Instead we follow the guideline for solving rational inequalities:

\begin{equation*} \begin{array}{rcl} \mbox{Start with original problem:} \amp ~ \amp \displaystyle{\frac{2-x}{2+x}\geq 1}\\ \\ \mbox{Move everything to one side:} \amp ~ \amp \displaystyle{\frac{2-x}{2+x}-1 \geq 0}\\ \\ \mbox{Find a common denominator:} \amp ~ \amp \displaystyle{\frac{2-x}{2+x}-\frac{2+x}{2+x} \geq 0}\\ \\ \mbox{Combine fractions:} \amp ~ \amp \displaystyle{\frac{(2-x)-(2+x)}{2+x} \geq 0}\\ \\ \mbox{Expand numerator:} \amp ~ \amp \displaystyle{\frac{2-x-2-x}{2+x} \geq 0}\\ \\ \mbox{Simplify numerator:} \amp ~ \amp \displaystyle{\frac{-2x}{2+x} \geq 0(*)} \end{array} \end{equation*}

Now we have the numerator and denominator in fully factored form. The split points are \(x=0\) (makes the numerator \(0\)) and \(x=-2\) (makes the denominator \(0\)). Let us draw a number line with the split points indicated on it:

The point \(x=0\) is included since if we sub \(x=0\) into (*) we get \(0\geq 0\) which is true. The point \(x=-2\) is not included since we cannot divide by zero. We indicate this with open/closed circles on the number line (remember that open means omit):

Now choosing a test point from each of the three subintervals we can determine if the expression \(\frac{-2x}{2+x}\) is positive or negative. When \(x=-3\text{,}\) it is negative. When \(x=-1\text{,}\) it is positive. When \(x=1\text{,}\) it is negative. Indicating this on the number line gives:

Since we wish to solve \(\frac{-2x}{2+x}\geq 0\text{,}\) we look at where the \(+\) signs are and shade that area on the number line:

Since there is a closed circle at \(0\text{,}\) we include it. Therefore, the solution is \((-2,0]\text{.}\)

Example 1.14. Rational Inequality.

Write the solution to the following inequality using interval notation:

\begin{equation*} \frac{2}{x+2}>{3}{x+3}\text{.} \end{equation*}

Solution

We provide a brief outline of the solution. By subtracting \((3x+3)\) from both sides and using a common denominator of \(x+2\text{,}\) we can collect like terms and simplify to get:

\begin{equation*} \frac{-(3x^2+9x+4)}{x+2}>0\text{.} \end{equation*}

The denominator is zero when \(x=-2\text{.}\) Using the quadratic formula, the numerator is zero when \(x=\frac{-9\pm\sqrt{33}}{6}\) (these two numbers are approximately \(-2.46\) and \(-0.54\)). Since the inequality uses “\(>\)” and \(0>0\) is false, we do not include any of the split points in our solution. After choosing suitable test points and determining the sign of \(\frac{-(3x^2+9x+4)}{x+2}\) we have

Looking where the \(+\) symbols are located gives the solution:

\begin{equation*} \left(-\infty,\frac{-9-\sqrt{33}}{6}\right)\bigcup\left(-2,\frac{-9+\sqrt{33}}{6}\right)\text{.} \end{equation*}

When writing the final answer we use exact expressions for numbers in mathematics, not approximations (unless stated otherwise).

Subsection 1.1.8 The Absolute Value

The absolute value of a number \(x\) is written as \(|x|\) and represents the distance \(x\) is from zero. Mathematically, we define it as follows:

Thus, if \(x\) is a negative real number, then \(-x\) is a positive real number. The absolute value does not just turn minuses into pluses. That is, \(|2x-1|\neq 2x+1\text{.}\) You should be familiar with the following properties.

When solving inequalities with absolute values, the following are helpful.

• Case 1: \(a>0\)

  • \(|x|=a\) has solutions \(x=\pm a\text{.}\)

  • \(|x|\leq a\) means \(x\geq -a\) and \(x\leq a\) (that is, \(-a\leq x\leq a\)).

  • \(|x|\lt a\) means \(x\lt -a\) and \(x\lt a\) (that is, \(-a\lt x\lt a\)).

  • \(|x|\geq a\) means \(x\leq -a\) or \(x\geq a\text{.}\)

  • \(|x|> a\) means \(x\lt -a\) or \(x> a\text{.}\)

• Case 2: \(a\lt 0\)

  • \(|x|=a\) has no solutions.

  • Both \(|x|\leq a\) and \(|x|\lt a\) have no solutions.

  • Both \(|x|\geq a\) and \(|x|>a\) have solution set \(\{x|x\in\R\}\text{.}\)

• Case 3: \(a=0\)

  • \(|x|=0\) has solution \(x=0\text{.}\)

  • \(|x|\lt 0\) has no solutions.

  • \(|x|\leq 0\) has solution \(x=0\text{.}\)

  • \(|x|>0\) has solution set \(\{x\in\R|x\neq 0\}\text{.}\)

  • \(|x|\geq 0\) has solution set \(\{x|x\in\R\}\text{.}\)

Subsection 1.1.9 Solving Inequalities that Contain Absolute Values

We start by solving an equality that contains an absolute value. To do so, we recall that if \(a\geq 0\) then the solution to \(|x|=a\) is \(x=\pm a\text{.}\) In cases where we are not sure if the right side is positive or negative, we must perform a check at the end.

We next look at absolute values and inequalities.

In some questions you must be careful when multiplying by a negative number as in the next problem.

In the previous two examples the only difference is that one had \(\lt\) in the question and the other had \(\geq\text{.}\) Combining the two solutions gives the entire real number line!