Lecture 1 · Barnett 1-1 to 1-2 · about 40 min

Functions and Their Graphs

What a function is, how to find its domain, evaluate it and build the difference quotient, the six basic graphs and how to move them, piecewise functions with open and closed dots, and the cost, revenue and profit functions of business problems.

By the end you can

Decide whether a rule or a graph defines a function, using the vertical line test
Find the domain of a function with denominators and square roots, and read the range from a graph
Evaluate f(a + h) and simplify the difference quotient
Sketch the basic functions and their shifts, reflections and stretches
Evaluate and graph piecewise functions with correct open and closed dots
Build cost, revenue and profit functions and invert a demand function

Key idea · Exam recipe (memorise this)

Domain: start with all reals, then remove: $x$ that make a denominator 0; $x$ that make an even root negative (inside $\ge 0$ ; inside $\gt 0$ if the root is in a denominator). Write the answer in interval notation.
Evaluate $f(a + h)$ : replace every $x$ by $(a + h)$ in brackets, then expand.
Difference quotient $\frac{f(x+h) - f(x)}{h}$ : expand $f(x + h)$ , subtract $f(x)$ (terms without $h$ cancel), factor out $h$ , cancel it.
Piecewise: first find which inequality $x$ satisfies, then use only that formula.
Graph $y = A\,f(x - h) + k$ : right $h$ , up $k$ , flip if $A \lt 0$ , stretch if $|A| \gt 1$ .
Range from a graph: lowest to highest $y$ the graph reaches; open dots are excluded.
Vertical line test: any vertical line hits the graph at most once.

How the exam asks it: three core exam templates

In midterm exams, functions and graph questions test three canonical structures: domain restrictions, difference quotients, and piecewise or economic models. Master these three worked solutions, then follow the cheat-sheet and 4-step method below.

Exam template 1: Finding domain with radical and denominator restrictions

Question: Consider the functions $f(x) = \dfrac{\sqrt{x + 3}}{x - 2}$ and average cost $\bar{C}(x) = \dfrac{2000}{x} + 8$ .
a) State the algebraic condition required for the square root $\sqrt{x + 3}$ .
b) State the algebraic condition required for the denominator $x - 2$ .
c) Write the domain of $f(x)$ in interval notation.
d) If $\bar{C}(x)$ is the daily average cost of a plant with maximum capacity of 500 units, determine its economic domain in interval notation.

Exam template 2: Expanding and simplifying the difference quotient

Question: Let $f(x) = x^2 - 3x + 1$ .
a) Find the expanded expression for $f(x + h)$ .
b) Compute and simplify the numerator $f(x + h) - f(x)$ .
c) Find and simplify the difference quotient $\dfrac{f(x + h) - f(x)}{h}$ for $h \ne 0$ .
d) Evaluate the difference quotient when $x = 2$ and $h = 0.1$ .

Exam template 3: Piecewise pricing schedules and demand inversion

Question: A courier charges delivery fee $C(x)$ based on distance $x$ (in km): $C(x) = \begin{cases} 20 & \text{if } 0 \lt x \le 5 \\ 20 + 3(x - 5) & \text{if } x \gt 5 \end{cases}$ Separately, market demand for an item is given by $p = D(x) = 200 - 4x$ .
a) Find the courier cost for a delivery of $x = 4$ km.
b) Find the courier cost for a delivery of $x = 12$ km.
c) Invert the demand equation $p = 200 - 4x$ to express quantity demanded $x$ as a function of price $p$ .

Cheat-sheet & the 4-step method

· Cheat-sheet: algebraic restrictions, difference quotient, and piecewise functions

Function type / Question	Condition or Formula	When to use it	How to write the exam answer
Even root $\sqrt{g(x)}$	$g(x) \ge 0$	Finding domain with radical in numerator	State inequality, solve for $x$ , write in interval notation e.g. $[a, \infty)$
Denominator $\dfrac{1}{g(x)}$	$g(x) \ne 0$	Finding domain with fraction	Set denominator $\ne 0$ , exclude values: $(-\infty, c) \cup (c, \infty)$
Root in denominator $\dfrac{1}{\sqrt{g(x)}}$	$g(x) \gt 0$	Finding domain with radical in denominator	Strictly greater than 0 (no equality, since $0$ in denominator is invalid)
Economic domain	$x \ge 0$ (or $x \gt 0$ ), $x \le \text{capacity}$	Applied business / cost / demand problems	State physical restriction: $(0, \text{capacity}]$
Difference quotient	$\dfrac{f(x+h) - f(x)}{h}$	Rate of change, prep for derivative	Expand $f(x+h)$ , cancel non- $h$ terms, factor out $h$ , simplify
Piecewise function	$f(x) = f_i(x)$ on interval $I_i$	Multi-tier pricing, taxes, piecewise formulas	Check which interval $x$ falls in, evaluate only that branch
Inverse function	Solve $y = f(x)$ for $x$	Inverting demand $p = D(x)$ or cost functions	Rearrange terms to get $x = f^{-1}(y)$

Key idea · The 4-step method for any exam question on functions

Step 1: Identify the function family and constraints.
- Is there a denominator? Set $D(x) \ne 0$ .
- Is there an even root? Set inside $\ge 0$ (or $\gt 0$ if in denominator).
- Is it an economic quantity? Impose $x \gt 0$ and $x \le \text{capacity}$ .
Step 2: Set up the algebraic equation or substitution.
- For difference quotients: write $f(x+h)$ with parentheses around $(x+h)$ .
- For piecewise functions: test which interval $x$ satisfies.
- For inverse functions: set $y = f(x)$ and isolate $x$ .
Step 3: Carry out arithmetic one step at a time.
- Expand powers and distribute signs carefully (watch $-3(x+h) = -3x - 3h$ ).
- Cancel common terms in numerators.
Step 4: State the final answer clearly with units or interval brackets.
- Use $[ \dots ]$ for included endpoints, $( \dots )$ for excluded endpoints.
- Run a quick boundary check to ensure no zero denominator was admitted.

What a function is

A function assigns to each input exactly one output. Total cost depends on output: for each quantity $x$ there is one cost $C(x)$ , not two.

Function notationFormula sheet →

y = f(x)

$x$: input (independent variable)
$y$: f(x) = output (dependent variable)

Use it when you need to name a relationship, or evaluate it at a specific input.

Domain and range

The domain is the set of allowed inputs, the range the set of outputs actually produced.

Inside the formula	Rule	Example
Denominator	must not be 0	$\frac{1}{x - 3}$ : $x \ne 3$
Square root $\sqrt{u}$	$u \ge 0$	$\sqrt{x - 4}$ : $x \ge 4$
Root in a denominator $\frac{1}{\sqrt{u}}$	$u \gt 0$	$\frac{1}{\sqrt{5 - x}}$ : $x \lt 5$
Economic quantity	output $\ge 0$ , capacity	$0 \le x \le 500$

Evaluating and the difference quotient

Barnett asks for $f(a + h)$ and the difference quotient because it becomes the derivative later. Brackets are everything.

Difference quotientKey formulaFormula sheet →

\frac{f(x + h) - f(x)}{h}, \quad h \ne 0

$f(x+h)$: f with x replaced by (x + h)
$h$: a small change in x

Use it when a question asks you to simplify the difference quotient, or the average rate of change from x to x + h.

The basic functions

Six graphs you must draw from memory (plus $1/x$ ). For each: three points, the shape, the domain and range.

Function	Key points	Domain	Range
$x$	$(0,0)$ , $(1,1)$	all reals	all reals
$x^2$	$(0,0)$ , $(\pm1, 1)$ , $(\pm2, 4)$	all reals	$[0, \infty)$
$x^3$	$(0,0)$ , $(\pm1, \pm1)$	all reals	all reals
$\sqrt{x}$	$(0,0)$ , $(1,1)$ , $(4,2)$	$[0, \infty)$	$[0, \infty)$
$\sqrt[3]{x}$	$(0,0)$ , $(\pm1, \pm1)$	all reals	all reals
$\lvert x \rvert$	$(0,0)$ , $(\pm2, 2)$	all reals	$[0, \infty)$

Transformations

TransformationsRulesFormula sheet →

y = A\,f(x - h) + k

$h$: horizontal shift (right if h > 0)
$k$: vertical shift (up if k > 0)
$A$: vertical stretch, flip across the x-axis if A < 0

Use it when a graph is a basic function moved, flipped or stretched, such as y = -(x - 2)^2 + 3 or y = \sqrt{x + 1} - 2.

Cost, revenue and profit functions

Cost, revenue and profitFormula sheet →

C(x) = \text{fixed cost} + (\text{cost per unit})\,x \qquad R(x) = p\,x \qquad P(x) = R(x) - C(x)

$C(x)$: total cost
$R(x)$: total revenue
$P(x)$: total profit
$p$: price per unit

Use it when you are given a fixed cost, a per-unit cost and a price, and need any of the three functions.

Piecewise-defined functions

Tiered prices, tax brackets and bulk discounts use a different formula on different parts of the domain.

Graphing a piecewise function is the classic exam question. Watch the dots:

Composition of functions

$(f \circ g)(x) = f(g(x))$ : apply $g$ first, then feed the result into $f$ .

CompositionFormula sheet →

(f \circ g)(x) = f\big(g(x)\big)

$g$: the inner function, applied first
$f$: the outer function, applied to g's output

Use it when one quantity depends on a second, which depends on a third, such as cost in TL from cost in dollars from output.

Inverse functions

If $y = f(x)$ , the inverse gives $x = f^{-1}(y)$ . Solving a demand function for $x$ gives quantity as a function of price.

Inverse functionFormula sheet →

f^{-1}\big(f(x)\big) = x

$f^{-1}$: the inverse of f, which exists only where f is one-to-one

Use it when a demand or supply function is solved for one variable and you need it solved for the other.

Exam practice

Summary and review

A function gives one output per input; vertical line test on a graph.
Domain: no zero denominators, no negative numbers under even roots; add economic limits.
Difference quotient: expand $f(x+h)$ , cancel, factor $h$ , divide.
Six basic graphs plus $1/x$ ; $y = A f(x - h) + k$ moves them (inside: opposite sign, outside: same sign).
Piecewise: choose the piece by the inequality; open dot for $\lt$ , $\gt$ , closed for $\le$ , $\ge$ .
$P(x) = R(x) - C(x)$ ; $(f \circ g)(x) = f(g(x))$ ; invert demand by solving for $x$ .