MATH 1071 — Calculus for Business & Economics

Lecture Notes. This is NOT meant to be a substitute for coming to lecture or consulting the textbook directly, but rather a helpful resource to review for exams/ review what was discussed in lecture. Each section ends with a Self-Check taken from the in-class activity — try the problem first, then click the button to see the answer.

Rules of Exponents

↑ Back to top

Positive Exponents

Exponents are shorthand for repeated multiplication. Instead of writing $2\cdot2\cdot2\cdot2\cdot2$, we write $2^5$.

Definition For a positive integer $n$, \[ x^n = \underbrace{x\cdot x\cdots x}_{n \text{ times}}. \] The number $x$ being multiplied is called the base, and $n$, the number of times it is multiplied, is called the exponent (or power). We read $x^n$ as "$x$ to the $n$th power." Special cases: $x^2$ is "$x$ squared" and $x^3$ is "$x$ cubed."

Example \[ 2^3 = 2\cdot2\cdot2 = 8, \qquad 5^2 = 5\cdot 5 = 25, \qquad 10^4 = 10\cdot10\cdot10\cdot10 = 10{,}000, \qquad 7^1 = 7, \] \[ \left(\tfrac12\right)^3 = \tfrac12\cdot\tfrac12\cdot\tfrac12 = \tfrac18, \qquad x^4 = x\cdot x\cdot x\cdot x. \]

Remark (Watch the Parentheses) The exponent only applies to what it is directly attached to. \[ (-3)^2 = (-3)(-3) = 9, \qquad\text{but}\qquad -3^2 = -(3\cdot 3) = -9. \] Similarly, $2x^3 = 2\cdot x\cdot x\cdot x$, while $(2x)^3 = (2x)(2x)(2x) = 8x^3$.

Where the Rules Come From

Writing exponents out as repeated multiplication explains every rule below. You don't need to memorize them if you can rebuild them this way.

Example \[ x^2\cdot x^3 = (x\cdot x)(x\cdot x\cdot x) = x^5 \qquad\text{(count the } x\text{'s: } 2+3 = 5\text{)} \] \[ \frac{x^5}{x^2} = \frac{\cancel{x}\cdot\cancel{x}\cdot x\cdot x\cdot x}{\cancel{x}\cdot\cancel{x}} = x^3 \qquad\text{(two } x\text{'s cancel: } 5-2 = 3\text{)} \] \[ (x^2)^3 = x^2\cdot x^2\cdot x^2 = x^6 \qquad\text{(three groups of two: } 2\cdot 3 = 6\text{)} \] \[ (xy)^2 = (xy)(xy) = x\cdot x\cdot y\cdot y = x^2y^2 \]

Zero and Negative Exponents

So far an exponent counts how many times we multiply. But what should $2^0$ or $2^{-1}$ mean? You can't multiply 2 by itself "zero times" or "negative one times." Look at the pattern of powers of 2. Moving down one row means dividing by 2:

$2^5$	$2^4$	$2^3$	$2^2$	$2^1$	$2^0$	$2^{-1}$	$2^{-2}$
32	16	8	4	2	1	$\tfrac12$	$\tfrac14$

Continuing the "divide by 2" pattern past $2^1$ forces the following rules.

Zero and Negative Exponents \[ a^0 = 1 \quad (a \neq 0), \qquad\qquad a^{-b} = \frac{1}{a^b}. \]

Example

$3^{-2} = \dfrac{1}{3^2} = \dfrac19$
$5^0 = 1$
$4^2\cdot 2^{-4} = 16\cdot \dfrac1{16} = 1$

More Exponent Rules

Rules of Exponents \[ \begin{aligned} x^m\cdot x^n &= x^{m+n} &&\text{(same base: add exponents)}\\ \frac{x^m}{x^n} &= x^{m-n} &&\text{(same base, divide: subtract exponents)}\\ (x^m)^n &= x^{mn} &&\text{(power to a power: multiply)}\\ (xy)^n &= x^n y^n, \qquad \left(\frac{x}{y}\right)^n = \frac{x^n}{y^n}\\ \sqrt[n]{x} &= x^{1/n}, \qquad \sqrt[n]{x^m} = x^{m/n} &&\text{(power on top, root on bottom)} \end{aligned} \]

Remark A very common mistake: exponents do not distribute over addition. In general $(x+y)^n \neq x^n + y^n$. For example, $(1+2)^2 = 9$ but $1^2+2^2 = 5$.

Example Some quick examples of each rule: \[ 2^3\cdot 2^2 = 2^{5}, \qquad \frac{2^3}{2^2} = 2^{3-2} = 2, \qquad (2^3)^2 = 2^{6}, \qquad (1\cdot 2)^2 = 1^2\cdot 2^2, \qquad \sqrt{x^3} = x^{3/2}, \qquad \sqrt[4]{x^5} = x^{5/4}. \]

Simplifying Expressions

To simplify a messy expression, split it up so you can focus on one variable at a time, convert every radical to a fractional exponent, and then combine exponents on the same base.

Example Simplify $\dfrac{x^3\sqrt z}{y^2z^5}$ so there are no radicals and no division.

Split by variable:

\[ \frac{x^3\sqrt z}{y^2 z^5} = x^3\cdot\frac{1}{y^2}\cdot\frac{\sqrt z}{z^5}. \]

Since $\frac{1}{y^2} = y^{-2}$ and $\frac{\sqrt z}{z^5} = \frac{z^{1/2}}{z^5} = z^{1/2-5} = z^{-9/2}$, we get

\[ x^3y^{-2}z^{-9/2}. \]

Example Simplify $\dfrac{x^2\sqrt[4]{z^8}}{(xy)^2z^{-4}}$. \[ \frac{x^2 z^{8/4}}{x^2y^2z^{-4}} = x^{2}x^{-2}\,y^{-2}\,z^{2}z^{4} = y^{-2}z^{6}. \]

Self-Check

From Activity 1: Rules of Exponents

(a) Simplify so that there are only positive exponents and no fractional exponents (use radicals instead):

\[ \text{(i)}\ \ \frac{(xy)^3 z}{x^{5/2}y^5z^{1/2}} \qquad\qquad \text{(ii)}\ \ \frac{x^3(yz)^{4/5}}{y^{-2}z} \]

(b) Use exponent rules and the fact that $\sqrt[3]{64} = 4$ to show that $8^{4/3} = 16$.

Review of Functions, Part A

↑ Back to top

Goals Define functions, domain, codomain and range; represent functions with tables, graphs and formulas; evaluate functions using function notation; find the natural domain and write it in interval notation; find zeros and intercepts and use the vertical line test.

What Is a Function?

Informally, a function is a rule that gives exactly one output for every input.

Definition A function is a map that assigns to each element of the input set (called the domain) exactly one element of the target/output set (called the codomain). The set of outputs that actually get hit is called the range.

Example Suppose a function sends $1\mapsto 8$, $2\mapsto 9$, $3\mapsto 10$, and $4\mapsto 9$, with target set $\{8,9,10,11\}$. Then \[ \text{Domain} = \{1,2,3,4\}, \qquad \text{Codomain} = \{8,9,10,11\}, \qquad \text{Range} = \{8,9,10\}. \] Nothing gets sent to 11, so it is in the codomain but not the range. It is fine that two inputs share the output 9.

Representing Functions

The same function can be described several ways: as a diagram of arrows, a table, a graph, or a formula (for example $f(x)=x^2$, or a piecewise rule such as $f(x) = x^2$ for $x\ge 0$ and $5x$ for $x<0$).

Example Does each of the following represent a function? If so, give the domain and range.

$y = x^3$: yes. Domain $(-\infty,\infty)$, range $(-\infty,\infty)$.
$y = \pm\sqrt{x}$: not a function — the input $x=4$ gives two outputs, $2$ and $-2$.
The table $0\mapsto 3,\ 1\mapsto 3,\ 2\mapsto 3$: yes. Domain $\{0,1,2\}$, range $\{3\}$.
The table $0\mapsto 5,\ 1\mapsto 6,\ 2\mapsto 7,\ 1\mapsto 8$: not a function — the input 1 has two outputs.

Vertical Line Test A graph is the graph of a function if and only if every vertical line crosses it at most once.

Function Notation

Think of $f(x) = x^2+3x$ as $f(\text{input}) = (\text{input})^2 + 3(\text{input})$. Whatever goes in the parentheses replaces every $x$.

Example Let $f(x) = x^2 + 3x$. Then \[ \begin{aligned} f(1) &= 1^2+3(1) = 4, \qquad f(0) = 0, \qquad f(2) = 4+6 = 10,\\ f(x+2) &= (x+2)^2 + 3(x+2) = x^2+4x+4+3x+6 = x^2+7x+10,\\ f(a+h) &= (a+h)^2+3(a+h) = a^2+2ah+h^2+3a+3h. \end{aligned} \]

Example Let $f(x) = \sqrt{4-x}$. Then $f(1) = \sqrt3$ and $f(0) = 2$, but $f(6) = \sqrt{-2}$ is not allowed — we cannot take the square root of a negative number, so $6$ is not in the domain.

Finding the Natural Domain

Idea: the natural domain is the set of all allowable inputs. So we ask: what can't we do?

Domain Restrictions to Check

No square root (or any even root) of a negative number: set the inside $\ge 0$.
No division by zero: set the denominator $\neq 0$.
(Later, with logarithms) No log or ln of zero or a negative number: set the inside $> 0$.

Interval Notation

We use square brackets $[\ ,\ ]$ when an endpoint is included ($\le,\ \ge$) and parentheses $(\ ,\ )$ when it is not ($<,\ >$, or $\pm\infty$). The symbol $\cup$ ("union") glues intervals together.

Inequality	Interval notation
$2 < x < 10$	$(2,10)$
$x \ge 5$	$[5,\infty)$
$x \ne 2$	$(-\infty,2)\cup(2,\infty)$
$\|x\| > 6$	$(-\infty,-6)\cup(6,\infty)$

Example Find the domain of $f(x) = \dfrac{\sqrt{x-1}}{x^2+2x-24}$.

We need two things:

\[ x - 1 \ge 0 \;\Longrightarrow\; x\ge 1, \qquad\qquad x^2+2x-24 = (x-4)(x+6) \ne 0 \;\Longrightarrow\; x\ne 4,\ x\ne -6. \]

The restriction $x\neq -6$ is already handled by $x\ge 1$. Putting it together, the domain is $[1,4)\cup(4,\infty)$.

Example What is the domain of each function?

$x-1$: $(-\infty,\infty)$, all real numbers.
$\sqrt{2-x}$: need $2-x\ge 0$, so $x\le 2$. Domain $(-\infty,2]$.
$\dfrac{1}{x-1}$: need $x\ne 1$. Domain $(-\infty,1)\cup(1,\infty)$.
$\dfrac{\sqrt{2-x}}{x-1}$: need $x\le 2$ and $x\ne 1$. Domain $(-\infty,1)\cup(1,2]$.

Example Find a function whose domain is $[0,1)\cup(1,\infty)$.

We want $x\ge 0$ (a square root of $x$) and $x\ne 1$ (a factor of $x-1$ in a denominator). One answer is $f(x) = \dfrac{\sqrt x}{x-1}$.

Zeros and Intercepts

Definition The zeros (or $x$-intercepts) of $f$ are the inputs where $f(x) = 0$. The $y$-intercept is the value $f(0)$, if $0$ is in the domain.

A function can have many zeros but at most one $y$-intercept (otherwise the vertical line $x=0$ would hit the graph twice).

Self-Check

From Activity 2: Introduction to Functions

(a) Determine whether each is a function. If it is, find the domain, $x$-intercepts, and $y$-intercept. If not, explain why.

\[ \text{(i)}\ f(x) = \frac{\sqrt x - 3}{x^2+2x-8} \qquad \text{(ii)}\ f(x) = \begin{cases} x+1 & \text{if } x \text{ is even}\\ x & \text{if } x \text{ is odd}\\ 0 & \text{if } x = 0\end{cases} \qquad \text{(iii)}\ f(x) = \frac{e^x(x^2-9)}{(7-x)^{1/2}} \]

(b) The domain of $x^{1/2} = \sqrt x$ is $[0,\infty)$. What is the domain of $x^{1/3}$? What about $x^{1/n}$ for a positive integer $n$?

Review of Functions, Part B

↑ Back to top

Goals Create new functions from old by adding, subtracting, multiplying and dividing; define the composition of functions; identify where functions are increasing, decreasing, concave up and concave down.

Combining Functions

Given functions $f$ and $g$, we can add, subtract, multiply and divide them to get new functions. Things to be careful of:

Subtraction: distribute the negative, e.g. $x^2+1 - (3x+2) = x^2 - 3x - 1$.
Division: we can't divide by zero, so check the "bottom" for new domain restrictions.
Dividing fractions: multiply by the reciprocal, e.g. $\dfrac{x^2+1}{x+1}\div\dfrac{1}{x^3} = \dfrac{x^2+1}{x+1}\cdot x^3$.

Example Let $f(x) = x^2+4x-12$ and $g(x) = 4x-8$. \[ \begin{aligned} f(x)+g(x) &= x^2+8x-20 = (x-2)(x+10),\\ f(x)-g(x) &= x^2+4x-12-(4x-8) = x^2-4 = (x+2)(x-2),\\ f(x)g(x) &= (x^2+4x-12)(4x-8) = 4x^3+8x^2-80x+96,\\ \frac{f(x)}{g(x)} &= \frac{(x-2)(x+6)}{4(x-2)} = \frac{x+6}{4}, \quad x\neq 2. \end{aligned} \]

Even though the $(x-2)$ cancels, $x = 2$ is still not in the domain of $f/g$, because the original denominator $4x-8$ is zero there.

Example Let $h(x) = \dfrac{2x}{x-5}$ and $g(x) = x^2-25$. Then \[ h(x)g(x) = \frac{2x(x-5)(x+5)}{x-5} = 2x(x+5) \ (x\ne 5), \qquad \frac{g(x)}{h(x)} = (x^2-25)\cdot\frac{x-5}{2x} = \frac{(x-5)^2(x+5)}{2x}. \]

Composition of Functions

Definition The composition $f\circ g$ (read "$f$ compose $g$") is \[ (f\circ g)(x) = f(g(x)), \] and similarly $(g\circ f)(x) = g(f(x))$. To compute $f(g(x))$, plug the whole function $g(x)$ in as the input of $f$.

Example Let $f(x) = 2x^2+x$ and $g(x) = x^2-1$. Then \[ (f\circ g)(x) = f(x^2-1) = 2(x^2-1)^2 + (x^2-1), \qquad (g\circ f)(x) = g(2x^2+x) = (2x^2+x)^2 - 1. \]

These are different! In general $f\circ g \ne g\circ f$.

Example Let $h(x) = 3x^2-1$ and $k(x) = 2x$. Then \[ (h\circ k)(x) = 3(2x)^2-1, \qquad (k\circ h)(x) = 2(3x^2-1), \qquad (h\circ h)(x) = 3(3x^2-1)^2-1. \]

Increasing, Decreasing, and Concavity

Definition A function is increasing on an interval if for all points $x < y$ in the interval, $f(x)\le f(y)$. It is decreasing on an interval if for all $x<y$ in the interval, $f(x)\ge f(y)$.

Another way to say it: reading the graph left to right, a function is increasing where the outputs are getting bigger (the graph goes up) and decreasing where the outputs are getting smaller (the graph goes down).

Definition A function is concave up if it is getting steeper as you move left to right — it is shaped like a cup and "can hold water," and its secant lines lie above the graph. A function is concave down if it is getting less steep as you move left to right — it is shaped like a frown and "can't hold water," and its secant lines lie below the graph.

Remark Increasing/decreasing and concavity are separate questions. A graph can be increasing and concave up (getting steeper upward), or increasing and concave down (still rising, but leveling off). Always answer with intervals of $x$-values.

Self-Check

From Activities 3 and 4: Functions; Function Composition/Characteristics

(a) Find the domain, $x$-intercepts, and $y$-intercept of $f(x) = \dfrac{x^2-1}{x\sqrt{3x-6}}$.

(b) Let $f(x) = x^2+1$ and $g(x) = 2x$. Find $f(x)+g(x)$, $f(x)g(x)$, $(f\circ g)(x)$, and $(g\circ f)(x)$.

(c) Let $f(x) = x^4-2x^2+1$ and $g(x) = 7\sqrt x$. Find $(f\circ g)(x)$ and its domain. Then find $(g\circ f)(x)$.

(d) Below is the graph of $f(x) = \tfrac13x^3-4x$ on $[-3,3]$. Where is it increasing, decreasing, concave up, and concave down?

$f(x) = \tfrac13x^3 - 4x$ on $[-3,3]$

Lines

↑ Back to top

Goals Learn what makes a function's graph a line; find the equation of a line in point-slope form and slope-intercept form.

Slope

The key idea: lines have constant slope.

Definition The slope of the line through $(x_1,y_1)$ and $(x_2,y_2)$ is \[ m = \frac{\text{rise}}{\text{run}} = \frac{\Delta y}{\Delta x} = \frac{y_2-y_1}{x_2-x_1}. \]

If a line has slope $m$ and passes through $(a,b)$, then for any other point $(x,y)$ on the line, the slope from $(a,b)$ to $(x,y)$ is still $m$:

\[ \frac{y-b}{x-a} = m \quad\Longrightarrow\quad y-b = m(x-a). \]

Forms of a Line

Definition (Point-Slope Form) Given a point $(x_1,y_1)$ and slope $m$, the line through that point with that slope is \[ y - y_1 = m(x-x_1). \]

Definition (Slope-Intercept Form) \[ y = mx + b, \] where $m$ is the slope and $b$ is the $y$-intercept (the line passes through $(0,b)$).

Example Find the point-slope form of the line through $(1,2)$ with slope $5$. \[ y - 2 = 5(x-1). \]

Solving for $y$: $y = 5x - 5 + 2 = 5x - 3$. So the slope is $5$ and the $y$-intercept is $-3$.

Example A line passes through $(0,1)$ and $(5,5)$. The $y$-intercept is $b = 1$, and the slope is $m = \frac{5-1}{5-0} = \frac45$, so \[ y = \tfrac45 x + 1. \]

Recognizing a Line from a Table

When the $x$-values go up by a constant step, a table comes from a line exactly when the $y$-values change by a constant amount.

Example Does each table represent a line? If so, find its equation.

$x$	1	2	3	4	5
$f(x)$	2	5	8	11	14

The outputs go up by $3$ each time, so this is a line with slope 3. Using $(1,2)$: $y-2 = 3(x-1)$, so $y = 3x-1$. (Using $(5,14)$ instead gives $y-14 = 3(x-5)$, which simplifies to the same line.)

$x$	2	4	6	8
$f(x)$	2	4	8	16

The outputs go up by 2, then 4, then 8 — not constant, so this is not a line. (It is the exponential $f(x) = 2^{x/2}$; see Exponential Functions below.)

Remark Two different lines with the same slope are parallel, so they never intersect.

Self-Check

From Activities 5 and 6: Function Characteristics and Lines; Lines and Polynomials

(a) Find the equation of the line through $(1,2)$ and $(-1,4)$ in both point-slope and slope-intercept form. Then find a second line $g(x)$ that never intersects it.

(b) Find the equation of the line through $(-5,-3)$ and $(1,7)$ in both point-slope and slope-intercept form.

Polynomials, Rational & Piecewise Functions

↑ Back to top

Goals Learn what polynomials and power functions are and the basic shapes of their graphs; find zeros of polynomials by factoring and the quadratic formula; learn what rational functions are and find their domains; learn about piecewise functions and their graphs.

Polynomials

Definition A polynomial is a function of the form \[ a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0, \] where $a_0, a_1, \dots, a_n$ are real numbers and $n$ is a non-negative integer. The integer $n$ (the highest power) is called the degree.

Example Which are polynomials?

$x^2+2x+1$: yes, degree 2. $\quad 2x+1$: yes, degree 1. $\quad x^5$: yes, degree 5. $\quad 8$: yes, degree 0.
$\sqrt x = x^{1/2}$: no — fractional powers are not allowed.
$x^3 + \frac1x = x^3+x^{-1}$: no — negative powers are not allowed.
$\dfrac{2x}{x^2+1}$: no — dividing by a polynomial is not allowed (this is a rational function).

Power Functions

Definition A power function is a function of the form $f(x) = ax^b$, where $a$ and $b$ are real numbers.

Examples: $x^5$, $8$, $\sqrt x = x^{1/2}$, $\frac5x = 5x^{-1}$, $4x^3$, $3x^2$.

End Behavior

End behavior describes what $f(x)$ does as $x\to\infty$ and as $x\to-\infty$.

Function	As $x\to -\infty$	As $x\to\infty$
constant $c$	$f(x)\to c$	$f(x)\to c$
$x^2, x^4, x^6,\dots$ (even)	$f(x)\to\infty$	$f(x)\to\infty$
$x, x^3, x^5,\dots$ (odd)	$f(x)\to-\infty$	$f(x)\to\infty$
$-x^2, -x^4,\dots$	$f(x)\to-\infty$	$f(x)\to-\infty$
$-x, -x^3,\dots$	$f(x)\to\infty$	$f(x)\to-\infty$

End Behavior of Polynomials The graph of a polynomial has the same end behavior as its highest degree term.

Example Let $f(x) = x^2 - x^3 + 2x$. The highest degree term is $-x^3$, so as $x\to\infty$, $f(x)\to-\infty$, and as $x\to-\infty$, $f(x)\to\infty$. (In the middle, other terms matter and different things can happen.)

Finding Zeros

To find the zeros (roots) of a polynomial, factor it, then set each factor equal to 0.

Example Find the zeros of $x^2-x^3+2x$. \[ \begin{aligned} x^2 - x^3 + 2x &= 0\\ -x(x^2 - x - 2) &= 0\\ -x(x+1)(x-2) &= 0 \end{aligned} \]

(For $x^2-x-2$ we want two numbers that multiply to $-2$ and add to $-1$: $1$ and $-2$.) So the zeros are $x = 0,\ -1,\ 2$. Combined with the end behavior from before, the graph comes down from the upper left, crosses at $-1$, $0$, and $2$, and heads down to the lower right.

Example Factor, then solve.

$x^2+6x+8 = (x+2)(x+4) = 0$, so $x = -2, -4$.
$x^2+4x-5 = (x-1)(x+5) = 0$, so $x = 1,-5$.
$x^3-4x^2-21x = x(x^2-4x-21) = x(x+3)(x-7) = 0$, so $x = 0,-3,7$.
$3x^2+5x-2 = 0$. Split the middle term using $6x - x$ (since $3\cdot(-2) = -6 = 6\cdot(-1)$ and $6 + (-1) = 5$): \[ 3x^2+6x-x-2 = 3x(x+2) - (x+2) = (3x-1)(x+2) = 0, \] so $x = \tfrac13, -2$.
$2x-1 = x^2$. Move everything to one side: $x^2-2x+1 = (x-1)^2 = 0$, so $x = 1$.
$x^2+4x = 10$, i.e. $x^2+4x-10 = 0$. This one doesn't factor nicely — see below.

Quadratic Formula If $ax^2+bx+c = 0$, then \[ x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}. \]

Example Solve $x^2+4x = 10$. Rewrite as $x^2+4x-10 = 0$, so $a=1,\ b=4,\ c=-10$. \[ x = \frac{-4\pm\sqrt{16-4(1)(-10)}}{2(1)} = \frac{-4\pm\sqrt{56}}{2} = \frac{-4\pm2\sqrt{14}}{2} = -2\pm\sqrt{14}. \]

(We simplified $\sqrt{56} = \sqrt{4\cdot14} = 2\sqrt{14}$.)

Rational Functions

Definition A rational function is a function of the form $\dfrac{\text{polynomial}}{\text{polynomial}}$.

Examples: $\dfrac{4x^2+2x+8}{x^2}$, $\ \dfrac1x$, $\ x^2+2x$ (a polynomial over $1$), and $\dfrac{x^2+3x}{x} = x+3$ when $x\neq 0$.

When we find the domain of a rational function, each excluded $x$-value is a "gap." If the factor causing the gap cancels with the numerator, the gap is a hole. If it only appears in the denominator, the gap is a vertical asymptote.

Example Find the domain of $f(x) = \dfrac{x^2+3x+2}{x^2+5x+4}$. \[ f(x) = \frac{(x+2)(x+1)}{(x+4)(x+1)}. \]

We need $x\ne -4$ and $x\ne-1$. Even though $(x+1)$ cancels, $x=-1$ is still excluded. The domain is

\[ (-\infty,-4)\cup(-4,-1)\cup(-1,\infty). \]

The graph has a hole at $x=-1$ (the factor cancelled) and a vertical asymptote at $x = -4$.

Piecewise Functions

Sometimes a function is made up of different pieces depending on the input.

Example Let \[ f(x) = \begin{cases} 3x-1 & x > 4,\\ 2x & x\le 4.\end{cases} \]

Check which condition the input satisfies, then use that piece:

\[ f(3) = 2(3) = 6, \qquad f(5) = 3(5)-1 = 14, \qquad f(4) = 2(4) = 8. \]

The graph is the line $y=2x$ up to and including the point $(4,8)$, then jumps: the line $y = 3x-1$ starts with an open circle at $(4,11)$ and continues with slope 3. Because of the jump, this function is not continuous at $x=4$.

Example Find $b$ so that the following function is continuous (no jump): \[ f(x) = \begin{cases} x^2+1 & x\ge 2,\\ bx & x < 2.\end{cases} \]

At $x=2$ the top piece gives $f(2) = 2^2+1 = 5$. For the pieces to meet, we want the bottom piece to also equal $5$ at $x = 2$:

\[ b(2) = 5 \quad\Longrightarrow\quad b = \tfrac52 = 2.5. \]

Self-Check

From Activities 6 and 7: Lines and Polynomials; Piecewise and Rational Functions

(a) Find the domain and zeros of \[ f(x) = \frac{x^2+8x+15}{x^2+6x-16}. \] Then create a piecewise function $g(x)$ whose domain is all real numbers and with $g(x) = f(x)$ for every $x$ in the domain of $f$.

(b) Find the domain of the function below and classify each gap as a vertical asymptote or a hole. \[ f(x) = \begin{cases} \dfrac{x+3}{x^2-9} & \text{if } x<0,\\[2mm] \dfrac{1}{x-5} & \text{if } 0\le x\le 20,\\[2mm] 2x & \text{if } x>20.\end{cases} \]

Mathematical Modeling

↑ Back to top

Goals Discuss linear models for cost, revenue and profit, and for supply and demand; find and analyze break-even quantities and equilibrium points; discuss quadratic models and find the vertex of their graphs; optimize cost, revenue and profit functions.

Cost, Revenue, and Profit

Throughout, let $x$ be the number of items.

Definition (Linear Cost, Revenue, Profit)

Cost to produce $x$ items: $C(x) = px + b$, where $b$ is the fixed cost (an upfront cost) and $p$ is the cost per item, also called the variable or marginal cost.
Revenue from selling $x$ items: $R(x) = mx$, where $m$ is the selling price, also called the marginal revenue.
Profit from selling $x$ items: \[ P(x) = R(x) - C(x). \]

Remark Be careful to distribute the negative when computing profit: $mx - (px+b) = mx - px - b$.

Break-Even Analysis

Definition The break-even point is where $P(x) = 0$, or equivalently $C(x) = R(x)$. Its $x$-coordinate, the break-even quantity, tells us how much we need to sell to start making a profit.

Graphically, the break-even point is where the revenue line crosses the cost line. To the left of it, cost is above revenue (a loss); to the right, revenue is above cost (a gain).

Example A lemonade stand costs $300 to build and $1 per cup in supplies. Lemonade sells for $3 a cup. How much do we need to sell? \[ C(x) = x + 300, \qquad R(x) = 3x, \qquad P(x) = 3x - (x+300) = 2x-300. \]

Set $P(x) = 0$: $2x - 300 = 0$, so $x = 150$. We break even after selling 150 cups. The break-even point is $(150, 450)$, since $R(150) = C(150) = 450$.

Supply and Demand

Definition

The demand equation $p(x)$ gives the price needed in order to be able to sell $x$ items. Demand has negative slope: as price goes up, demand goes down.
The supply equation $p(x)$ gives the price needed in order for the producer to create (and bring to market) $x$ items. Supply has positive slope.

Remark These have $x$ = number of items as the input and price $p$ as the output, which might feel backwards. It can help to think of setting a price and then finding $x$.

Example A company will produce 200 items if they can sell at $10 and 300 items if they can sell at $15. Find a linear supply equation.

The two points are $(200,10)$ and $(300,15)$, so the slope is

\[ m = \frac{15-10}{300-200} = \frac{5}{100} = \frac1{20} = 0.05. \]

Point-slope form: $p - 10 = \tfrac1{20}(x-200)$, i.e. $p = \tfrac1{20}x$.

Definition The equilibrium point is where supply equals demand. Its $x$-coordinate is the equilibrium quantity. Above the equilibrium price there is a surplus (more supply than demand); below it there is a shortage (more demand than supply).

Example The supply equation is $p = 2x+1$ and the demand equation is $p = -x+4$. Find the equilibrium point.

Set supply equal to demand:

\[ 2x+1 = -x+4 \;\Longrightarrow\; 3x = 3 \;\Longrightarrow\; x = 1. \]

Then $p = 2(1)+1 = 3$, so the equilibrium point is $(1,3)$.

Quadratic Models

Vertex of a Parabola The graph of $ax^2+bx+c$ has its vertex at \[ x = -\frac{b}{2a}. \] If $a>0$ the parabola opens up and the vertex is a minimum; if $a<0$ it opens down and the vertex is a maximum.

Example Find the vertex of $f(x) = x^2+8x+1$. \[ x = -\frac{8}{2(1)} = -4, \qquad f(-4) = 16-32+1 = -15. \]

The vertex is $(-4,-15)$, a minimum since $a = 1 > 0$.

Example (Maximizing Profit and Minimizing Cost) A clothing store's supply equation for $x$ dresses is $p = 140-0.25x$, and the cost of producing $x$ dresses is $C(x) = 2000-10x+0.25x^2$.

1. Revenue (assuming demand meets supply).

\[ R(x) = (\text{price per item})(\text{number of items}) = (140-0.25x)x = -0.25x^2+140x. \]

2. Profit.

\[ P(x) = R(x)-C(x) = -0.25x^2+140x - 2000 + 10x - 0.25x^2 = -0.5x^2+150x-2000. \]

3. Number of dresses that maximizes profit.

$P$ is a downward parabola, so its vertex is a maximum: $x = -\dfrac{150}{2(-0.5)} = 150$ dresses.

4. Maximum profit.

\[ P(150) = -0.5(150)^2 + 150(150) - 2000 = 11{,}250 - 2000 = \$9{,}250. \]

5. Number of dresses that minimizes cost.

$C(x) = 0.25x^2 - 10x + 2000$ opens up, so its vertex is a minimum: $x = -\dfrac{-10}{2(0.25)} = \dfrac{10}{0.5} = 20$ dresses.

Self-Check

From Activities 7 and 8: Modeling; Mathematical Modeling

(a) The demand function for an item is $p_D(x) = 15-0.25x$.

What is the revenue function?
How many items do we need to sell to maximize revenue?
What price per item maximizes revenue?
What is the maximum revenue?

(b) A company has linear demand function $p(x) = 125-5x$. Find the revenue function and the number of items that maximizes revenue.

(c) It costs $7 per item to produce the first 5 items and $5 per item after that. Write a piecewise cost function $C(x)$.

Exponential Functions

↑ Back to top

Goals Identify exponential functions and the basic shape of their graphs; derive the formula for interest compounded $n$ times per year and see how it relates to the formula for continuously compounded interest.

The Exponential Function

Definition An exponential function is a function of the form $f(x) = b^x$, where $b>0$ (and $b\neq1$).

Remark Don't confuse $b^x$ (variable in the exponent: exponential) with $x^b$ (variable in the base: power function).

Example Tables of values for $f(x) = 2^x$ and $g(x) = \left(\tfrac13\right)^x$:

$x$	$-2$	$-1$	$0$	$\tfrac12$	$1$	$2$
$2^x$	$\tfrac14$	$\tfrac12$	$1$	$\sqrt2\approx1.41$	$2$	$4$
$(1/3)^x$	$9$	$3$	$1$	$\approx0.58$	$\tfrac13$	$\tfrac19$

$2^x$ increases quickly to the right and flattens toward $0$ on the left (it gets close to $0$ but never touches it). $(1/3)^x$ is the mirror image: large on the left, decaying toward $0$ on the right.

Shape of $b^x$

If $b>1$: exponential growth (e.g. $2^x$, $4^x$).
If $0<b<1$: exponential decay (e.g. $(1/2)^x$, $(1/4)^x$).
Every $b^x$ passes through $(0,1)$, since $b^0 = 1$, and $b^x > 0$ for every $x$.

Compound Interest

A key business application of exponential growth is compound interest.

Example You put $1000 into an account earning 6% interest compounded yearly. \[ \text{After 1 year: } 1000(1.06) = 1060, \qquad \text{after 2 years: } 1000(1.06)(1.06) = 1000(1.06)^2 = 1123.60, \] \[ \text{after } n \text{ years: } 1000(1.06)^n. \]

Example What if the 6% is compounded monthly? Then each month earns $\frac{6\%}{12} = 0.5\%$, and there are 12 compoundings per year: \[ \text{after 1 year: } 1000\left(1+\frac{0.06}{12}\right)^{12} \approx 1061.68, \qquad \text{after } n \text{ years: } 1000\left(1+\frac{0.06}{12}\right)^{12n}. \]

Compound Interest Formulas Let $F$ = future value, $P$ = principal (starting amount), $r$ = interest rate (as a decimal), $t$ = time in years, and $n$ = number of times compounded per year. \[ \text{Compounded } n \text{ times per year:}\qquad F = P\left(1+\frac rn\right)^{nt}. \] Letting $n\to\infty$ gives \[ \text{Compounded continuously:}\qquad F = Pe^{rt}, \] where $e\approx 2.71828\dots$

Example You put $500 into an account earning 6% compounded continuously. How much do you have after 15 years? \[ P = 500,\quad r = 0.06,\quad t = 15, \qquad F = 500e^{0.06(15)} = 500e^{0.9}\approx \$1229.80. \]

Solving Exponential Equations

Problem-Solving Strategy

To solve equations involving exponentials, try to rewrite both sides with the same base, then set the exponents equal to each other.

Example Solve $9^x = 27^{4x-10}$. Both 9 and 27 are powers of 3: \[ (3^2)^x = (3^3)^{4x-10} \;\Longrightarrow\; 3^{2x} = 3^{12x-30} \;\Longrightarrow\; 2x = 12x-30 \;\Longrightarrow\; 30 = 10x \;\Longrightarrow\; x = 3. \]

Example Solve $\dfrac{1}{2^{6x}} = 2^{9x-13}$. \[ 2^{-6x} = 2^{9x-13} \;\Longrightarrow\; -6x = 9x-13 \;\Longrightarrow\; 13 = 15x \;\Longrightarrow\; x = \tfrac{13}{15}. \]

Self-Check

Practice in the style of the lecture examples (there is no in-class activity for this section)

(a) You invest $2000 at 5% annual interest. How much do you have after 10 years if interest is compounded (i) quarterly, (ii) continuously?

(b) Solve $4^{x+1} = 8^x$.

(c) Is $f(x) = (0.8)^x$ exponential growth or decay? What is its $y$-intercept?

Logarithms

↑ Back to top

Goals Learn what logarithm functions are and how they relate to exponential functions; use logarithms to solve equations involving exponentials; learn properties of logarithms and use them to solve equations involving logarithms.

Guiding Example We have $1000 and want $1500 after 12 years. What interest rate do we need if interest is compounded continuously? Using $F = Pe^{rt}$: \[ 1500 = 1000e^{12r}. \]

The unknown $r$ is stuck in the exponent, and we can't make the bases match. We need a new tool: logarithms.

What Is a Logarithm?

Basic idea: logarithms undo exponentials. Logarithms and exponentials are inverse functions.

Definition \[ \log_b a = x \quad\text{means}\quad b^x = a. \] In words: $\log_b a$ is the power we raise $b$ to in order to get $a$.

Example \[ 10^2 = 100 \text{ so } \log_{10}100 = 2, \qquad 2^5 = 32 \text{ so } \log_2 32 = 5. \] Evaluate: \[ \log_3 9 = 2, \qquad \log_2 8 = 3, \qquad \log_{10}1000 = 3, \qquad \log_7 7^5 = 5, \qquad \log_5 125 = 3. \]

Inverse Rules \[ a^{\log_a b} = b, \qquad\qquad \log_a(a^x) = x. \]

Our Three Favorite Bases

Base 10: $\log_{10}x$ is written $\log x$ and called the common log.
Base $e$: $\log_e x$ is written $\ln x$ and called the natural log. This is the one we use most.
Base 2: $\log_2 x$ has no special name.

Example (Guiding Example, solved) Solve $1500 = 1000e^{12r}$. First isolate the exponential, then take $\ln$ of both sides to cancel the $e$: \[ 1.5 = e^{12r} \;\Longrightarrow\; \ln(1.5) = \ln\!\left(e^{12r}\right) = 12r \;\Longrightarrow\; r = \frac{\ln(1.5)}{12}\approx 0.0338. \]

We need an interest rate of about 3.4%.

Solving Exponential Equations with Logs

Isolate the exponential (get $e^{(\cdots)}$ by itself).
Take $\ln$ of both sides to cancel the $e$.
Solve for the variable.

Example Solve $5e^x - 1 = 16$. \[ 5e^x = 17 \;\Longrightarrow\; e^x = \tfrac{17}{5} \;\Longrightarrow\; x = \ln\!\left(\tfrac{17}{5}\right). \]

Example Solve $8e^{2x} = 17e^x$. Divide both sides by $e^x$ (which is never zero): \[ 8e^{x} = 17 \;\Longrightarrow\; e^x = \tfrac{17}{8} \;\Longrightarrow\; x = \ln\!\left(\tfrac{17}{8}\right). \]

Graphs and Domain of Logarithms

Since $\log_a x$ and $a^x$ are inverses, the graph of $y = \log_a x$ is the graph of $y = a^x$ flipped over the line $y = x$. The point $(0,1)$ on $a^x$ becomes the point $(1,0)$ on $\log_a x$, because $\log_a 1 = 0$.

Domain of a Logarithm $\ln(x)$ and $\log_a(x)$ are only defined for $x>0$. Their domain is $(0,\infty)$.

Example Find the domain of $\dfrac{\ln(2-x)}{\sqrt{x+1}}$.

The log needs $2-x>0$, so $x<2$. The square root needs $x+1\ge0$, and since it is in the denominator, $x+1\neq0$; together, $x>-1$. The domain is $(-1,2)$.

Log Rules

Properties of Logarithms \[ \begin{aligned} \log_a(xy) &= \log_a x + \log_a y &&\text{(multiplication inside = addition outside)}\\ \log_a\!\left(\frac xy\right) &= \log_a x - \log_a y &&\text{(division inside = subtraction outside)}\\ \log_a(x^c) &= c\log_a x &&\text{(the power "rolls off" the log, or jumps back inside)} \end{aligned} \]

Example Combine into a single logarithm: $\log x + 2\log y - 4\log z$. \[ \log x + \log(y^2) - \log(z^4) = \log\!\left(\frac{xy^2}{z^4}\right). \]

Example Combine $\log x - 3\log(y^2) + 6\log(\sqrt z)$. \[ \log x - \log\!\left((y^2)^3\right) + \log\!\left((z^{1/2})^6\right) = \log x - \log(y^6) + \log(z^3) = \log\!\left(\frac{xz^3}{y^6}\right). \]

Example Solve $\ln(x+2) + \ln(x+7) = \ln(18x)$.

Step 1. Get a single $\ln$ on each side.

\[ \ln\big((x+2)(x+7)\big) = \ln(18x) \;\Longrightarrow\; \ln(x^2+9x+14) = \ln(18x). \]

Step 2. Clear the $\ln$'s from both sides.

\[ x^2+9x+14 = 18x. \]

Step 3. Solve the quadratic by setting it equal to 0 and factoring (or using the quadratic formula).

\[ x^2-9x+14 = 0 \;\Longrightarrow\; (x-2)(x-7) = 0 \;\Longrightarrow\; x = 2,\ x = 7. \]

Step 4. Check the answers (every log must have a positive input).

\[ x=2:\ \ln4+\ln9 = \ln36 = \ln(18\cdot2)\ ✓ \qquad x=7:\ \ln9+\ln14 = \ln126 = \ln(18\cdot7)\ ✓ \]

Self-Check

From the Chapter 1 Review, plus practice in the style of the lecture examples

(a) We put $1000 into an account compounded continuously at 3%. How long does it take to have $1300 in the account?

(b) Combine into a single logarithm: $2\ln x - \ln y + 3\ln z$.

(c) Solve $3e^{2x} + 4 = 19$.

(d) Find the domain of $f(x) = \ln(x-4)$.

Inequality	Interval notation
\(2 < x < 10\)	\((2,10)\)
\(x \ge 5\)	\([5,\infty)\)
\(x \ne 2\)	\((-\infty,2)\cup(2,\infty)\)
\(\|x\| > 6\)	\((-\infty,-6)\cup(6,\infty)\)

Function	As \(x\to -\infty\)	As \(x\to\infty\)
constant \(c\)	\(f(x)\to c\)	\(f(x)\to c\)
\(x^2, x^4, x^6,\dots\) (even)	\(f(x)\to\infty\)	\(f(x)\to\infty\)
\(x, x^3, x^5,\dots\) (odd)	\(f(x)\to-\infty\)	\(f(x)\to\infty\)
\(-x^2, -x^4,\dots\)	\(f(x)\to-\infty\)	\(f(x)\to-\infty\)
\(-x, -x^3,\dots\)	\(f(x)\to\infty\)	\(f(x)\to-\infty\)

\(x\)	\(-2\)	\(-1\)	\(0\)	\(\tfrac12\)	\(1\)	\(2\)
\(2^x\)	\(\tfrac14\)	\(\tfrac12\)	\(1\)	\(\sqrt2\approx1.41\)	\(2\)	\(4\)
\((1/3)^x\)	\(9\)	\(3\)	\(1\)	\(\approx0.58\)	\(\tfrac13\)	\(\tfrac19\)

\(x\)	1	2	3	4	5
\(f(x)\)	2	5	8	11	14

\(x\)	2	4	6	8
\(f(x)\)	2	4	8	16