Learning Outcomes
By the end of this lesson, you will be able to:
- State the formal definition of a function and explain why it requires exactly one output per input
- Distinguish a function from a general relation, using both ordered pairs and equations
- Find the domain of a function algebraically, accounting for division, even roots, and logarithms — including combinations of these restrictions
- Find the range of a function using algebraic and graphical reasoning
- Classify a function as even, odd, or neither using the symmetry test, and recognize useful shortcuts
- Find all $x$-intercepts and the $y$-intercept of a function, and explain why a function can have many $x$-intercepts but never more than one $y$-intercept
This lesson introduces the formal definition of a function and four core properties used to describe one: its domain and range, its symmetry, and its intercepts. These ideas are not just definitions to memorize — they are the language used throughout the rest of the course. When we later study limits, we will constantly ask “is this defined here?” (a domain question). When we study derivatives of trigonometric functions, knowing that $\sin(x)$ is odd and $\cos(x)$ is even will simplify many calculations. So this lesson is worth understanding deeply, not just memorizing.
1. Function Definition
1.1 What is a Relation?
Before defining a function, it helps to understand the broader idea of a relation, since a function is really just a relation with one extra restriction.
A relation is simply any set of ordered pairs $(x, y)$ that connects elements from one set (inputs) to another set (outputs). A relation can be given in several different forms:
- As a list of ordered pairs, e.g. ${(1, 2), (2, 4), (3, 6)}$
- As a table of values
- As a graph on the coordinate plane
- As an equation, e.g. $y = 2x$ or $x^2 + y^2 = 9$
- As a verbal description, e.g. “the age of each student in a class”
A relation places no restriction on how many outputs one input can have. The input $1$ could be connected to five different outputs, and that would still be a perfectly valid relation.
Examples of relations:
- ${(1, 2), (1, 5), (3, 4)}$ — here the input $1$ is paired with two different outputs, $2$ and $5$. This is allowed for a relation.
- ${(2, 7), (4, 3), (6, 9)}$ — here each input happens to have exactly one output.
- A circle drawn on a graph, such as $x^2 + y^2 = 25$ — for a given $x$ value (say $x=3$), there are generally two $y$-values ($y = 4$ and $y = -4$).
1.2 What is a Function?
A function is a special type of relation that follows one strict rule, and this rule is what makes functions so useful in mathematics: they behave predictably.
Key Definition: Every input must map to exactly one output. A function $f$ from set $A$ to set $B$ is a relation in which every element in $A$ (the domain) is paired with exactly one element in $B$ (the range).
$$f : A \to B \qquad \text{or} \qquad y = f(x)$$
Think of a function as a machine: you feed in a number (the input $x$), the machine processes it according to a fixed rule, and you get exactly one result back (the output $y$). Press the same button twice, and you always get the same result — a function never behaves randomly or ambiguously.
Why does this restriction matter? In real life and in mathematics, we often want a quantity that depends predictably on another. For example, the area of a circle depends on its radius in one specific way: $A(r) = \pi r^2$. For any given radius, there is only ever one correct area — it would not make sense for the same circle to simultaneously have two different areas. This is exactly the behavior a function is designed to capture.
The key rule to remember: one input → exactly one output.
Two important clarifications:
- It is perfectly fine for different inputs to share the same output. For example, $f(x) = x^2$ sends both $x = 2$ and $x = -2$ to the same output, $4$. This does not break the function rule, because the rule is about each input having one output — not about each output coming from only one input.
- It is never allowed for one input to produce two different outputs at the same time. That is the one and only thing that disqualifies a relation from being a function.
1.3 Identifying Functions from Ordered Pairs
To check whether a set of ordered pairs represents a function, look at all the $x$-values (the first number in each pair). The check is purely about whether any $x$-value repeats with a different $y$-value:
- Scan through the list and note every $x$-value.
- If any $x$-value appears more than once with different $y$-values attached, it is not a function.
- If an $x$-value happens to repeat with the same $y$-value (e.g. $(2,5)$ appearing twice in a list), that is just a repeated point — it does not break anything.
- If every distinct $x$-value is paired with only one $y$-value, the relation is a function.
Example 1.1.1 — Is it a function?
Relation A: ${(1, 2), (2, 4), (3, 6)}$ Each $x$-value (1, 2, 3) appears only once, each mapping to exactly one $y$. → This is a function. ✓
Relation B: ${(1, 2), (1, 5), (2, 3)}$ The input $x = 1$ maps to both $y = 2$ and $y = 5$. As soon as we see one input with two different outputs, we can stop checking — the relation already fails the test. → This is not a function. ✗
Relation C: ${(1, 3), (2, 3), (5, 3)}$ Different inputs ($1, 2, 5$) all give the same output $y = 3$. Re-check the rule: it says nothing about outputs needing to be unique, only that each input needs exactly one output. Here, each of $1$, $2$, and $5$ does have exactly one output (which happens to be $3$ for all of them). That is completely valid. → This is a function. ✓
Relation D: ${(0, 1), (1, 0), (-1, 0), (2, 3)}$ Check the $x$-values: $0, 1, -1, 2$ — all different from each other. Even though the $y$-value $0$ appears twice (for $x=1$ and $x=-1$), that is allowed, since the repetition is in the outputs, not the inputs. → This is a function. ✓
[ IMAGE PLACEHOLDER — Mapping Diagrams. Suggested visual: arrow diagrams (domain → range) for Relations A, B, C, and D, with Relation B showing one input pointing to two outputs, and clear “Function ✓ / Not a Function ✗” labels under each. ]
1.4 Identifying Functions from Equations
When a relation is given as an equation rather than a list of pairs, the method changes slightly: solve the equation for $y$, and check whether there is only one possible $y$-value for each $x$ you might plug in.
This works because solving “for $y$” tells you exactly how many outputs a given input produces. If the algebra naturally produces a $\pm$ sign (such as from taking a square root), that is a strong signal that the relation is not a function, since it means two different $y$-values come from the same $x$.
Example 1.1.2 — Functions from Equations
a) $y = 3x + 1$ For any value of $x$ you substitute, there is exactly one resulting value of $y$ — multiplying by $3$ and adding $1$ never produces two different answers. → Function ✓
b) $y = x^2$ For any $x$, squaring it gives exactly one number. Even though two different inputs (like $2$ and $-2$) can give the same output, each individual input still only gives one output. → Function ✓
c) $y^2 = x$ Solve for $y$: taking the square root of both sides gives $y = \pm\sqrt{x}$. The $\pm$ here is the key warning sign. For $x = 4$: $y = 2$ or $y = -2$ — two different outputs for the same input. → Not a function ✗
d) $x^2 + y^2 = 25$ (a circle of radius 5) Solve for $y$: $y^2 = 25 – x^2 \Rightarrow y = \pm\sqrt{25 – x^2}$. Again a $\pm$ appears. For $x = 0$: $y = 5$ or $y = -5$. → Not a function ✗
e) $y = \sqrt{x}$ (note: this is different from part c!) Here the square root symbol itself, by convention, refers only to the positive root. So for any $x \geq 0$, there is exactly one value of $y$ (the non-negative one). → Function ✓
This last example is worth sitting with: $y^2 = x$ and $y = \sqrt{x}$ look similar but behave completely differently. The first allows two $y$-values per $x$; the second is restricted by definition to only the positive root, making it a function.
Common Mistake: $f(x)$ is the name of the output, not multiplication. $f(2)$ means “evaluate the function at $x = 2$” — it never means “$f$ multiplied by $2$.” If $f(x) = 3x + 1$, then $f(2) = 7$, not “$2f$.”
2. Domain and Range
2.1 Definitions
| Term | Meaning |
|---|---|
| Domain | The set of all valid input values ($x$-values) the function can accept |
| Range | The set of all output values ($y$-values) the function actually produces |
| Co-domain | The larger set the function maps into — the range is always a subset of it |
It’s worth being precise about the difference between range and co-domain, since students often confuse them. The co-domain is the set you declare the outputs will live in (often “all real numbers,” written $\mathbb{R}$), while the range is the set of outputs the function actually achieves. For example, $f(x) = x^2$ could be said to map into the co-domain $\mathbb{R}$, but its actual range is only $[0, +\infty)$ — it never produces a negative output, even though negative numbers are part of the co-domain it maps into.
[ IMAGE PLACEHOLDER — Domain / Co-domain / Range Diagram. Suggested visual: two ovals labeled “Domain” and “Co-domain,” with a smaller highlighted region inside the co-domain labeled “Range,” and arrows from domain elements landing only inside the range region. ]
2.2 Finding the Domain
Unless the domain is explicitly stated, we assume it is all real numbers, except for values that cause one of these three restrictions. It helps to understand why each restriction exists:
- Division by zero — the denominator cannot equal zero, because division by zero is mathematically undefined; there is no number that correctly answers “how many times does zero fit into this value?”
- Even root of a negative number — the expression inside an even root (square root, fourth root, etc.) must be $\geq 0$, because no real number squared (or raised to any even power) gives a negative result. (Odd roots, like cube roots, do not have this restriction — you can take the cube root of a negative number just fine.)
- Logarithm of zero or a negative number — the argument of a logarithm must be strictly $> 0$, because a logarithm answers “what power do I raise the base to, to get this number?” and no real power of a positive base ever produces zero or a negative result.
If a function has more than one restriction at once, all conditions must hold simultaneously — you find each restriction separately, then combine them with “and.”
Example 1.1.3 — Domain from a Single Restriction
a) $f(x) = \dfrac{5}{x – 3}$ The denominator is $x – 3$. Set it not equal to zero: $x – 3 \neq 0 \Rightarrow x \neq 3$. Domain: $(-\infty, 3) \cup (3, +\infty)$
b) $f(x) = \sqrt{x – 5}$ The expression under the root is $x – 5$. Set it greater than or equal to zero: $x – 5 \geq 0 \Rightarrow x \geq 5$. Domain: $[5, +\infty)$
c) $f(x) = \ln(x – 2)$ The argument of the log is $x – 2$. Set it strictly greater than zero: $x – 2 > 0 \Rightarrow x > 2$. Domain: $(2, +\infty)$
d) $f(x) = \sqrt[3]{x – 1}$ (a cube root — an odd root) Since this is an odd root, there is no restriction at all — cube roots are defined for every real number, positive, negative, or zero. Domain: $(-\infty, +\infty)$, i.e., all real numbers
That last example is included deliberately: it’s a common error to apply the “expression $\geq 0$” rule to every root. That rule only applies to even roots (square root, 4th root, 6th root, …). Odd roots (cube root, 5th root, …) have no domain restriction from the root itself.
[ IMAGE PLACEHOLDER — Domain Number Lines. Suggested visual: number lines for examples (a), (b), and (c) above, using an open circle at x=3 for the excluded value, a filled circle at x=5 for the included boundary, and an open circle at x=2 for the excluded boundary of the log domain. ]
Example 1.1.4 — Domain with a Combined Restriction
$f(x) = \dfrac{\sqrt{x + 2}}{x – 1}$
This function has two separate restrictions because it has both a square root and a denominator. Handle them one at a time, then combine.
Condition 1 (from the square root): $x + 2 \geq 0 \Rightarrow x \geq -2$
Condition 2 (from the denominator): $x – 1 \neq 0 \Rightarrow x \neq 1$
Both conditions must hold together: $x \geq -2$ and $x \neq 1$. In other words, start with everything from $-2$ onward, then remove the single point $x=1$ from that set.
Domain: $[-2, 1) \cup (1, +\infty)$
Example 1.1.5 — A Trickier Combined Domain
$g(x) = \dfrac{\sqrt{x^2 – 4}}{x}$
Condition 1 (root): $x^2 – 4 \geq 0$. Factor: $(x-2)(x+2) \geq 0$. This is true when $x \leq -2$ or $x \geq 2$ (testing values shows the expression is positive outside the roots of the factored expression and negative between them).
Condition 2 (denominator): $x \neq 0$. Note that $x = 0$ is already excluded by Condition 1, since $0$ falls strictly between $-2$ and $2$. So this condition doesn’t remove anything additional here.
Domain: $(-\infty, -2] \cup [2, +\infty)$
This example shows that sometimes one condition automatically takes care of another — always check both, but don’t assume they will always change the final answer independently.
2.3 Finding the Range
Finding the range is generally harder than finding the domain, because there isn’t one single rule to apply — it requires thinking about the function’s behavior. Common strategies:
- Algebraic method: Let $y = f(x)$, then rearrange the equation to solve for $x$ in terms of $y$. Whatever values of $y$ keep $x$ real and valid form the range.
- Graphical / behavioral method: Think about the shape of the graph. Does it have a minimum or maximum value? Does it grow without bound? Square roots, for instance, never produce negative outputs; squares never produce negative outputs either.
- Domain-based reasoning: If you already know the domain, sometimes you can directly reason about what outputs result from the extreme and typical values in that domain.
Example 1.1.6 — Finding the Range
a) $f(x) = x^2$ Since squaring any real number always gives a result $\geq 0$, and every non-negative number can be achieved (e.g., to get output $9$, just use input $3$), the range is $[0, +\infty)$.
b) $g(x) = \sqrt{9 – x^2}$ First, recall the domain is $[-3, 3]$ (from $9 – x^2 \geq 0$). Within this domain, the smallest output happens at the domain’s edges: at $x = \pm 3$, $g(x) = \sqrt{0} = 0$. The largest output happens in the middle, at $x = 0$: $g(0) = \sqrt{9} = 3$. Every value between $0$ and $3$ is achieved as $x$ moves from the edges to the center. Range: $[0, 3]$
c) $h(x) = \dfrac{1}{x – 2}$ Use the algebraic method. Let $y = \frac{1}{x-2}$. Solve for $x$: multiply both sides by $(x-2)$ to get $y(x-2) = 1$, so $x – 2 = \frac{1}{y}$, so $x = 2 + \frac{1}{y}$. This expression is defined for any $y$ except $y = 0$ (since dividing by $y$ would be undefined). So every real number except $0$ is achievable. Range: $(-\infty, 0) \cup (0, +\infty)$
d) $p(x) = x^2 + 3$ This is the basic parabola $x^2$ shifted upward by 3 units. Since $x^2 \geq 0$ always, adding 3 means $p(x) \geq 3$ always, and every value of $3$ or greater is achievable. Range: $[3, +\infty)$
[ IMAGE PLACEHOLDER — Domain and Range Visual for g(x) = √(9 − x²). Suggested visual: the graph of g(x) (an upper semicircle), with the domain [-3, 3] shaded along the x-axis, and the range [0, 3] marked with a bracket along the y-axis. ]
3. Symmetry
Functions can be classified by symmetry, a property that is always tested algebraically by comparing $f(-x)$ to $f(x)$ — never by simply “eyeballing” a graph, since graphs can be misleading or hard to read precisely.
| Type | Algebraic Condition | Geometric Symmetry |
|---|---|---|
| Even | $f(-x) = f(x)$ | Symmetric about the $y$-axis |
| Odd | $f(-x) = -f(x)$ | Symmetric about the origin |
| Neither | Neither condition holds | No symmetry |
Why does this work geometrically? If a function is even, then the point $(x, f(x))$ and the point $(-x, f(-x)) = (-x, f(x))$ both lie on the graph, and these two points are mirror images of each other across the $y$-axis. That’s exactly what “symmetric about the $y$-axis” means. For an odd function, the point $(-x, f(-x)) = (-x, -f(x))$ is the 180° rotation of $(x, f(x))$ about the origin — that’s “symmetric about the origin.”
3.1 How to Test, Step by Step
- Write down $f(x)$ exactly as given.
- Replace every occurrence of $x$ with $(-x)$ — be careful with exponents and signs — to get $f(-x)$.
- Simplify $f(-x)$ completely. Do not stop halfway; an unsimplified expression can hide whether it matches $f(x)$ or $-f(x)$.
- Compare the simplified $f(-x)$ to the original $f(x)$:
- If $f(-x) = f(x)$ exactly → Even
- If $f(-x) = -f(x)$ exactly (i.e., every term flipped sign) → Odd
- If it matches neither → Neither
Example 1.1.7 — Classifying Functions
a) $f(x) = x^4 – 2x^2$ $f(-x) = (-x)^4 – 2(-x)^2$. Now simplify carefully: $(-x)^4 = x^4$ (an even power of a negative is positive), and $(-x)^2 = x^2$. So $f(-x) = x^4 – 2x^2 = f(x)$ → Even ✓
b) $g(x) = x^3 + x$ $g(-x) = (-x)^3 + (-x)$. Simplify: $(-x)^3 = -x^3$ (an odd power of a negative is negative), and $(-x) = -x$. So $g(-x) = -x^3 – x = -(x^3 + x) = -g(x)$ → Odd ✓
c) $h(x) = x^2 + x$ $h(-x) = (-x)^2 + (-x) = x^2 – x$. Compare: is $x^2 – x$ equal to $h(x) = x^2 + x$? No (unless $x=0$, but it must hold for all $x$). Is it equal to $-h(x) = -x^2 – x$? Also no. Since it matches neither pattern → Neither ✓
d) $k(x) = 5$ (a constant function) $k(-x) = 5 = k(x)$ → Even ✓ (Every constant function, other than $f(x)=0$ itself only being odd too in the trivial case, is even — its graph is a horizontal line, which is symmetric about the $y$-axis.)
e) $m(x) = x^3 – 2x^2$ $m(-x) = (-x)^3 – 2(-x)^2 = -x^3 – 2x^2$. Compare to $m(x) = x^3 – 2x^2$: not equal. Compare to $-m(x) = -x^3 + 2x^2$: also not equal (the signs on the $2x^2$ term don’t match). → Neither ✓
[ IMAGE PLACEHOLDER — Even / Odd / Neither Graphs. Suggested visual: three side-by-side graphs of f(x) = x⁴ – 2x², g(x) = x³ + x, and h(x) = x² + x, clearly showing y-axis mirror symmetry, 180° origin symmetry, and no symmetry respectively. ]
Useful Shortcut: If $f(0)$ is defined and $f(0) \neq 0$, the function cannot be odd. Here’s why: the odd condition $f(-x) = -f(x)$ must hold for every $x$, including $x=0$. Plugging in $x=0$ gives $f(0) = -f(0)$, which rearranges to $2f(0) = 0$, so $f(0)$ must equal $0$. If you already know $f(0) \neq 0$, you can immediately rule out “odd” without doing any more algebra — saving time on an exam.
3.2 A Special Case
The zero function $f(x) = 0$ (the function that outputs $0$ for every input) satisfies both conditions at once:
- $f(-x) = 0 = f(x)$ → satisfies the even condition
- $f(-x) = 0 = -0 = -f(x)$ → also satisfies the odd condition
It is the only function that is both even and odd simultaneously. Every other function is either even, odd, or neither — never two of these at once.
3.3 A Note on Sums and Products
Although not always tested directly, it’s useful to know:
- Even $\times$ Even = Even, and Odd $\times$ Odd = Even
- Even $\times$ Odd = Odd
- The sum of two even functions is even; the sum of two odd functions is odd
- The sum of an even and an odd function (neither being the zero function) is generally neither
These rules can save time when working with combinations of known functions like $\sin(x)$ (odd) and $\cos(x)$ (even), which you will use frequently in later chapters.
4. Intercepts
Intercepts are the points where a graph crosses (or touches) the coordinate axes. They are some of the easiest and most useful points to find on a graph, since they require simple substitution rather than deep analysis, and they are very often the first points plotted when sketching a curve by hand.
- $x$-intercept(s): the point(s) where the graph crosses the $x$-axis. At any such point, the height (the $y$-value) is zero. So: set $y = 0$ in the function’s equation and solve for $x$. A function can have zero, one, several, or even infinitely many $x$-intercepts (think of $\sin(x) = 0$, which happens at infinitely many points).
- $y$-intercept: the point where the graph crosses the $y$-axis. At this point, the input is zero. So: set $x = 0$ and compute $f(0)$. A function has at most one $y$-intercept. This follows directly from the definition of a function: since $x=0$ is a single input, it can only produce one output, so there can only be one point where the graph meets the $y$-axis.
Example 1.1.8 — Finding Intercepts
For $f(x) = x^2 – 4$:
$y$-intercept: substitute $x=0$: $f(0) = 0^2 – 4 = -4$ → point $(0, -4)$
$x$-intercepts: set $f(x) = 0$: $x^2 – 4 = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$ → points $(2, 0)$ and $(-2, 0)$
[ IMAGE PLACEHOLDER — Graph of f(x) = x² − 4 with Intercepts Marked. Suggested visual: the parabola with the y-intercept (0, −4) and both x-intercepts (±2, 0) clearly marked with dots and labels. ]
Example 1.1.9 — A Cubic with Three x-intercepts
For $g(x) = x^3 – 4x$:
$y$-intercept: $g(0) = 0^3 – 4(0) = 0$ → point $(0, 0)$ (note: here the $x$-intercept and $y$-intercept happen to be the same point)
$x$-intercepts: set $g(x)=0$: $x^3 – 4x = 0$. Factor out the common factor of $x$ first: $x(x^2 – 4) = 0$. Then factor the difference of squares: $x(x-2)(x+2) = 0$.
By the zero-product property, this product is zero exactly when one of the factors is zero: $x = 0$, or $x – 2 = 0 \Rightarrow x = 2$, or $x + 2 = 0 \Rightarrow x = -2$
So the $x$-intercepts are $(0,0)$, $(2, 0)$, and $(-2, 0)$.
Example 1.1.10 — A Function with No x-intercepts
For $h(x) = x^2 + 1$:
$y$-intercept: $h(0) = 0 + 1 = 1$ → point $(0, 1)$
$x$-intercepts: set $h(x) = 0$: $x^2 + 1 = 0 \Rightarrow x^2 = -1$. There is no real number whose square is negative, so there is no real solution.
This function has no $x$-intercepts at all — its graph (an upward parabola shifted up by 1) never touches the $x$-axis. This is a useful reminder that not every function needs to have an $x$-intercept; always check whether the resulting equation actually has real solutions before assuming intercepts exist.
Lesson Summary
| Concept | Key Idea |
|---|---|
| Relation | Any set of input-output pairs; no restriction on repeated inputs |
| Function | Each input maps to exactly one output |
| Domain | Valid $x$-values — exclude division by zero, negative even roots, log of non-positive numbers |
| Range | All $y$-values the function actually produces — found algebraically or graphically |
| Even | $f(-x) = f(x)$ — symmetric about the $y$-axis |
| Odd | $f(-x) = -f(x)$ — symmetric about the origin; requires $f(0) = 0$ |
| $x$-intercept | Set $y = 0$, solve for $x$ — can be zero, one, or many |
| $y$-intercept | Set $x = 0$, solve for $y$ — at most one, and it may not exist if $x=0$ isn’t in the domain |
Common Mistakes to Avoid:
- $f(x)$ is not multiplication — always read it as “the output of $f$ at $x$”
- Apply every domain restriction simultaneously, not separately, when a function has more than one
- The “$\geq 0$” rule for roots only applies to even roots — odd roots (cube root, etc.) have no such restriction
- Always test symmetry algebraically — never just by glancing at a graph
- A function may have several $x$-intercepts but never more than one $y$-intercept
- Not every function has $x$-intercepts — always check whether the equation $f(x)=0$ actually has real solutions
Practice Problems
Level 1 — Direct Application
- Determine which of the following relations is a function. Justify your answer. a) ${(2,5), (3,7), (4,9)}$ b) ${(1,4), (2,4), (1,6)}$ c) ${(0,1),(1,0),(-1,0),(2,3)}$
- Find the domain of $f(x) = \dfrac{1}{x^2 – 9}$. Write your answer in interval notation.
- Find the domain of $g(x) = \sqrt{2x – 6}$.
- Find the domain of $k(x) = \sqrt[3]{x+5}$ (note the type of root carefully).
- Classify $f(x) = x^6 – 4x^2$ as even, odd, or neither. Show full algebraic work.
- Find all intercepts of $f(x) = x^2 – 5x + 6$.
Level 2 — Intermediate
- Find the domain of $h(x) = \dfrac{\sqrt{x+4}}{x-3}$.
- Find the domain and range of $p(x) = \sqrt{16 – x^2}$.
- Classify $g(x) = x^5 – 3x$ as even, odd, or neither.
- Find all $x$-intercepts and the $y$-intercept of $g(x) = x^3 – 4x$.
- Find the range of $h(x) = \dfrac{1}{x+3}$ using the algebraic method shown in this lesson.
- Determine whether $r(x) = x^2 – 6$ has any $x$-intercepts. If so, find them; if not, explain why.
Level 3 — Challenge
- A function $f$ satisfies $f(-x) = f(x)$ for all $x$, and $f(3) = 7$. Find $f(-3)$.
- A function $f$ satisfies $f(x) + f(-x) = 10$ for every $x$ in its domain. Is $f$ even, odd, or neither — or is this not enough information? Justify your reasoning, and give one example of such a function.
- For $f(x) = \dfrac{x^2 – 4}{x – 2}$ (this simplifies to $x + 2$, but has a hole at $x = 2$): state the domain, and find all intercepts.
- Find the domain of $w(x) = \dfrac{\sqrt{x^2 – 9}}{x – 5}$, being careful to combine both restrictions correctly.
AB Courses · Engineering Foundation Course · Calculus · Lesson 1.1