Finds the domain of a rule by hunting for what it forbids — a zero denominator and a negative even radicand cover nearly every case — then writes the survivors in interval and set-builder notation. Reads both domain and range off a graph as its two shadows, and handles piecewise definitions, whose conditions must partition the domain without overlapping or leaving a gap.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 1 — Functions
§1.2 Domain and Range, pp. 41-66
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §1.2 Domain and Range §1.2, pp. 41-66 — the pages these objectives are drawn from
Warm-up
The domain question is not 'which numbers work'. It is 'which numbers break the rule', and then everything else.
Discussion prompt
Give a number you may not put into the reciprocal rule, one you may not put into the square root rule, and one you may not put into the cube root rule.
Hint: For the third one, try hard, and then consider that you might not be able to.
Answer:
The reciprocal rule forbids 0, because dividing by zero has no answer. Every other real number is fine.
The square root rule forbids every negative number, because no real number squares to a negative. So it forbids infinitely many inputs, all in one block.
The cube root rule forbids nothing. Negative numbers have perfectly good cube roots: the cube root of negative 8 is negative 2. This is the difference between even and odd roots, and it is why the domain rules single out even ones.
Concept
The domain of a function given by a formula is every real number, minus the inputs that would make the formula do something undefined. So the work is not finding what is allowed; it is finding the small set of things that are not.
domain — The set of inputs a function accepts. When a function is given by a formula with no context attached, its domain is taken to be every real number for which the formula produces a real output — sometimes called the implied domain.
\[ \text{domain} = \{x \in \mathbb{R} : f(x) \text{ is defined}\} \]
At this level exactly two things are forbidden: a denominator equal to zero, and an even root of a negative number. Logarithms will add a third in Chapter 4. Everything else — adding, multiplying, powers, odd roots — accepts every real number without complaint.
Figure (svg): A card listing the two operations that restrict a domain at this level: division, which forbids a zero denominator, and even roots, which forbid a negative radicand, each with the inequality it produces
OpenStax, Precalculus, §1.2 Domain and Range §1.2, pp. 41-44
Section
Section 1
Concept
Scan the formula for a denominator and for an even root. Each one contributes a condition, and the domain is what survives all of them at once.
The word 'simultaneously' in the third bullet is where marks are lost. A rule with a square root in its denominator must have a radicand that is both nonnegative and nonzero, which together means strictly positive — a stricter condition than either produces alone.
Figure (svg): A card listing the two operations that restrict a domain at this level: division, which forbids a zero denominator, and even roots, which forbid a negative radicand, each with the inequality it produces
OpenStax, Precalculus, §1.2 Domain and Range §1.2, pp. 42-46
Picture it
The two forbidden operations pull in opposite directions, which is why running them on autopilot goes wrong.
Figure (svg): A card listing the two operations that restrict a domain at this level: division, which forbids a zero denominator, and even roots, which forbid a negative radicand, each with the inequality it produces
Notice the asymmetry: from a denominator you solve an equation and throw the answers away; from a root you solve an inequality and keep the answers. Confusing the two produces a domain that is exactly the complement of the right one.
Worked example
One denominator, so one condition.
\[ \text{Find the domain of } f(x) = \frac{x+3}{x^2 - 9}. \]
Locate the denominator
Why: It is the only thing that can be forbidden here.
\[ \text{denominator } x ^{2} - 9 \]
Set it equal to zero
Why: These are the inputs to remove.
\[ x ^{2} - 9 = 0 \]
Solve
Why: Factor as a difference of squares.
\[ x = 3\text{ and } x = -3 \]
Remove them from the reals
Why: Everything else is legal.
\[ \text{all reals except } 3\text{ and } -3 \]
Figure (svg): The solution to Worked example a rational rule shown as a ladder of expressions, one row per legal move
\[ (-\infty, -3) \cup (-3, 3) \cup (3, \infty) \]
Verify: resist the cancellation
Why: The numerator also vanishes at negative 3, so the fraction simplifies to one over x minus 3. But simplifying happens after the domain is fixed: the original rule was never defined at negative 3, so negative 3 stays excluded. This becomes a hole in the graph rather than a vertical asymptote, a distinction §3.7 develops.
OpenStax, Precalculus, §1.2 Domain and Range §1.2, pp. 44-45
Sorting
Before solving anything, classify what kind of restriction each formula imposes.
Sort into buckets
Sort each rule by the restriction it carries.
Worked example
Two forbidden operations in one rule, so two conditions at once.
\[ \text{Find the domain of } g(x) = \frac{1}{\sqrt{x-4}}. \]
The even root needs a nonnegative radicand
Why: Otherwise the root is not real.
\[ x - 4 \ge 0 \]
The denominator may not be zero
Why: The root itself is the denominator.
\[ \sqrt{x - 4} \ne 0 \]
Combine them
Why: Nonnegative and nonzero together means strictly positive.
\[ x - 4 > 0 \]
Solve
Why: Add four to both sides.
\[ x > 4 \]
Figure (svg): The solution to Worked example a root inside a denominator shown as a ladder of expressions, one row per legal move
\[ (4, \infty) \]
Verify: test the boundary
Why: At x equal to 4 the radicand is 0, the root is 0, and the rule asks for one divided by zero, which is undefined — so 4 is correctly excluded. At x equal to 5 the rule gives one over one, which is fine. The endpoint is the whole subtlety here, and it is why the bracket is round rather than square.
OpenStax, Precalculus, §1.2 Domain and Range §1.2, pp. 46-47
Trap
\[ f(x) = \frac{x^2-1}{x-1} = x+1 \;\Longrightarrow\; \text{domain is all reals} \]
Simplify the fraction first
Why: The numerator factors and the common factor cancels.
The simplified rule has no denominator, so the conclusion drawn is that every real number is allowed.
The domain belongs to the rule you were given, not to the one you rewrote. The original has a denominator that vanishes at 1, so 1 is not in the domain, and no amount of algebra afterwards puts it back.
The two expressions agree at every input except 1, where one is defined and the other is not. They are therefore different functions, and the graph of the original is the line with a hole punched in it at that point.
Find the domain from the original expression, before simplifying. Then simplify if it helps. Doing it the other way round silently enlarges the domain, and this is the single most common domain error in the chapter.
Prediction
A rule is a single fraction whose denominator is a quadratic with two distinct real roots.
Predict first
What will its domain look like as a union of intervals?
Correct: Three intervals, split at the two roots.
Why: Removing two isolated points from the real line leaves three pieces: everything below the smaller root, everything between the two, and everything above the larger. This is why the previous worked example produced a union of three intervals. The numerator is irrelevant to the domain, since a numerator can never be undefined — it only affects whether an excluded point shows as a hole or an asymptote.
Faded example
Find the domain of the square root of the quantity seven minus two x.
Fill in the blanks
7 - 2x \ge 0 \;\Longrightarrow\; -2x \ge -7 \;\Longrightarrow\; x \le 7/2
Why: Subtracting 7 from both sides gives negative 2x at or above negative 7. Dividing by negative 2 reverses the inequality, giving x at or below seven halves. The reversal is the whole difficulty: dividing an inequality by a negative number flips it, and forgetting to flip produces exactly the wrong half of the line.
Step zero
You are asked for the domain of a rule that is a fraction with a square root in the numerator and a polynomial in the denominator.
Discussion prompt
Before doing any algebra, what are the two conditions you will need, and how will you combine them?
Hint: How many forbidden operations are present, and does an input have to satisfy one condition or both?
Answer:
Two conditions. From the square root in the numerator: its radicand must be at or above zero. From the denominator: it must not be zero.
They are combined by intersection, not union. An input is legal only if it satisfies every condition at once, because breaking any single one makes the whole expression undefined.
In practice: solve the inequality to get an interval, then punch out of it any point where the denominator vanishes. Points where the denominator vanishes outside that interval were already excluded and need no mention.
Section
Section 2
Concept
Interval notation names a set by its endpoints and says with a bracket whether each endpoint is included. Set-builder notation names it by the condition its members satisfy.
A useful reflex when reading interval notation aloud: say 'included' or 'excluded' at each end rather than 'bracket'. It converts a typographic detail into the mathematical claim it stands for, and makes a wrong bracket sound wrong.
Figure (svg): Four number lines showing the four bracket combinations: a closed interval with filled endpoints, an open interval with hollow endpoints, a half-open interval, and a ray to infinity always with a parenthesis
OpenStax, Precalculus, §1.2 Domain and Range §1.2, pp. 47-52
Picture it
A filled dot on the number line and a square bracket in the notation say the same thing.
Figure (svg): Four number lines showing the four bracket combinations: a closed interval with filled endpoints, an open interval with hollow endpoints, a half-open interval, and a ray to infinity always with a parenthesis
The bottom row is the one to memorise: infinity is a direction, not a destination, so it never gets a square bracket no matter what the inequality looked like.
Worked example
One set, written three ways.
\[ \text{Write } -3 < x \le 5 \text{ in interval and set-builder notation.} \]
Read the left end
Why: Strictly greater than negative 3, so negative 3 is out.
\[ \text{round bracket at } -3 \]
Read the right end
Why: At or below 5, so 5 is in.
\[ \text{square bracket at } 5 \]
Assemble the interval
Why: Left endpoint first.
\[ (-3, 5) \]
Write the set-builder form
Why: The condition is the inequality itself.
\[ {x: - 3 < x \le 5} \]
Figure (svg): The solution to Worked example inequality to interval to set-builder shown as a ladder of expressions, one row per legal move
\[ (-3, 5] \qquad \text{or} \qquad \{x \mid -3 < x \le 5\} \]
Verify: test both endpoints
Why: Is negative 3 in the set? The inequality says strictly greater, so no — and the round bracket agrees. Is 5 in the set? The inequality allows equality, so yes — and the square bracket agrees. Checking the two endpoints catches every bracket error there is.
OpenStax, Precalculus, §1.2 Domain and Range §1.2, pp. 48-49
Translation
The same set, said in two languages.
Match the pairs
Why: The first two differ only in which endpoint is included, and swapping them is the most common slip. The third shows infinity taking a round bracket while the finite endpoint takes a square one. The fourth is how a single removed point is written, and it is worth recognising on sight because every rational function's domain looks like it.
Worked example
Excluding a single interior point splits one interval into two.
\[ \text{Write the domain } x \ge 2, \; x \ne 5 \text{ in interval notation.} \]
Start from the unbroken interval
Why: Everything from 2 upward, with 2 included.
\[ [2, \infty] \]
Cut it at the removed point
Why: The point 5 splits it into a lower and an upper piece.
\[ \text{up to } 5,\text{ then past } 5 \]
Give the cut round brackets on both sides
Why: The point 5 itself belongs to neither piece.
\[ [2, 5]\text{ and } (5, \infty) \]
Join with a union
Why: The domain is both pieces together.
\[ [2, 5] U(5, \infty) \]
Figure (svg): The solution to Worked example a domain with a point removed shown as a ladder of expressions, one row per legal move
\[ [2, 5) \cup (5, \infty) \]
Verify: check the three interesting inputs
Why: The input 2 is included, so its bracket is square. The input 5 is excluded, and it carries a round bracket on both sides of the union — this is the detail most often got wrong, since only one of the two usually gets fixed. The input 6 sits in the second piece, as it should.
OpenStax, Precalculus, §1.2 Domain and Range §1.2, pp. 50-51
Error analysis
A student writes the domain of the square root rule.
Annotate
On: \( \text{domain} = [0, \infty] \)
This is a small error with a clear rule and no edge cases: infinity and negative infinity always take round brackets. It is worth fixing as a reflex, because it appears on nearly every domain answer in the course.
Prediction
A domain is found by solving the inequality that a radicand be at or above zero, giving x at or above 3. The root then sits in a denominator.
Predict first
What happens to the bracket at 3?
Correct: It becomes round, because 3 is now excluded.
Why: Putting the root in a denominator adds the condition that it not be zero, which excludes exactly the input where the radicand is zero. So the combined condition is strictly greater than 3 rather than at or above it, and the bracket opens. This is the second worked example of the previous section, seen from the notation side.
Fill the middle
The domain is every real number except 1 and 4.
Fill in the blanks
(-\infty, 1) \cup (1, 4) \cup (4, \infty)
Why: Both 1 and 4 are excluded, so the middle interval opens at both ends. Every excluded point appears twice in a union like this, once as the right endpoint of one piece and once as the left endpoint of the next, and both occurrences must be round. Fixing only one of the two is the usual half-correction.
Socratic
Interval notation is shorter, so it is fair to ask what set-builder adds.
Discussion prompt
Describe a set that interval notation handles badly and set-builder handles easily.
Hint: What if the members are not a run of consecutive numbers?
Answer:
Any set that is not made of intervals. The whole numbers, for instance, or the numbers whose square is less than 2 and which are also rational. Interval notation can only describe unions of runs, so it has nothing to say about these.
Set-builder handles them because it names the condition rather than the shape. Anything you can state as a condition, you can write.
The trade-off is directness: for a run of consecutive numbers, interval notation shows the endpoints at a glance and set-builder makes you read an inequality. Most domains in this course are unions of runs, which is why interval notation dominates here — but Chapter 9 and Chapter 11 both need the more general tool.
Section
Section 3
Concept
The domain is what the graph covers along the horizontal axis; the range is what it covers along the vertical axis. Both are read by flattening the curve onto an axis and seeing what is hit.
The commonest mistake here is reading the range off the visible portion of a plot rather than off the function. If the curve carries arrows, it keeps going, and the range extends past the edge of the picture even though the ink does not.
Figure (svg): A graph of a function with its domain marked as a shaded band along the horizontal axis and its range marked as a shaded band along the vertical axis, showing that the domain is the shadow cast downward and the range the shadow cast sideways
OpenStax, Precalculus, §1.2 Domain and Range §1.2, pp. 52-58
Picture it
The two shaded bands are the domain and the range, and each is a shadow of the same curve.
Figure (svg): A graph of a function with its domain marked as a shaded band along the horizontal axis and its range marked as a shaded band along the vertical axis, showing that the domain is the shadow cast downward and the range the shadow cast sideways
The curve starts at a solid point and runs up and to the right forever. So the domain has a square bracket at its left end and infinity at its right, and the range does the same at its own start.
Worked example
The curve shown on the visual slide starts at a solid point and rises forever.
\[ \text{From the graph of } f(x) = \sqrt{x+2} - 1, \text{ state the domain and range.} \]
Find the leftmost point of the curve
Why: It begins at a solid dot and nothing lies to its left.
\[ \text{starts at } x = -2 \]
Check whether it ends on the right
Why: The curve carries on rising with no endpoint.
Find the lowest height reached
Why: The starting point is also the lowest.
\[ \text{lowest } y = -1 \]
Check whether the height is bounded above
Why: It keeps climbing, so no.
Figure (svg): The solution to Worked example read both from a graph shown as a ladder of expressions, one row per legal move
\[ \text{domain } [-2, \infty), \qquad \text{range } [-1, \infty) \]
Verify: confirm from the formula
Why: The radicand x plus 2 must be at or above zero, giving x at or above negative 2, which matches. A square root outputs values at or above zero, so subtracting 1 gives outputs at or above negative 1, which also matches. Graph and algebra agree, and each is a check on the other.
OpenStax, Precalculus, §1.2 Domain and Range §1.2, pp. 54-55
Matching
Every visual feature corresponds to one piece of notation.
Match the pairs
Why: These four cover essentially every graph-reading question in the section. The first two are the same distinction as strict versus non-strict inequality, drawn. The third is why a straight line has all the reals as its range. The fourth is how a removed point shows itself in a picture rather than in algebra.
Worked example
Sometimes the graph is not given and the range has to be reasoned out.
\[ \text{Find the range of } f(x) = x^2 - 4x + 1. \]
Recognise the shape
Why: A quadratic with a positive leading coefficient opens upward.
Find the turning point by completing the square
Why: Half of negative 4 is negative 2; its square is 4.
\[ (x - 2) ^{2} - 3 \]
Read the minimum output
Why: A square is at least zero, so the whole thing is at least negative 3.
\[ \text{minimum is } -3 \]
State the range
Why: Every height from the minimum upward is attained.
\[ y \ge - 3 \]
Figure (svg): The solution to Worked example a range that needs the rule shown as a ladder of expressions, one row per legal move
\[ \text{range } = [-3, \infty) \]
Verify: check the minimum is attained
Why: Setting x equal to 2 gives 4 minus 8 plus 1, which is negative 3, so the value is genuinely reached and the bracket is square rather than round. An upward parabola has no largest output, so the other end runs to infinity, as stated.
OpenStax, Precalculus, §1.2 Domain and Range §1.2, pp. 56-57
Error analysis
A student reads the range of a rising line from a plot drawn on the window from negative 5 to 5.
Annotate
On: \( f(x) = 2x + 1 \;\Longrightarrow\; \text{range} = [-9, 11] \)
Ask what the function does, not what the plot shows. A graphing window is a viewport, and its edges are almost never features of the function.
Prediction
A downward parabola has its highest point at a height of 7.
Predict first
What is its range?
Correct: All heights at or below 7.
Why: A downward parabola climbs to its vertex and then falls away forever on both sides, so 7 is the largest output and every height below it is attained on the way down. The bracket at 7 is square because the vertex is a genuine point on the curve. Note the contrast with the domain, which for any parabola is every real number — the two questions have quite different answers here.
Sorting
Each observation limits one of the two, not both.
Sort into buckets
Sort each fact.
Explain it to yourself
Finding a domain from a formula is usually routine; finding a range from a formula is often hard.
Discussion prompt
Explain why the range is the harder of the two to compute directly from a rule.
Hint: What do you have to know about the rule's behaviour, not just its ingredients?
Answer:
The domain is decided by local, syntactic facts: is there a denominator, is there an even root. You can find it by inspecting the formula's parts without understanding what the function does.
The range depends on the function's global behaviour — where it turns, whether it is bounded, what it approaches far out. Those are not visible in the ingredients, which is why the quadratic above needed completing the square before its range could be stated.
This is why range questions are so often answered from a graph rather than from a formula at this stage, and it is one of the things calculus is for: Chapter 12 begins the machinery that finds maxima and minima without having to recognise a special form first.
Section
Section 4
Concept
A small set of basic functions recurs throughout the course, as building blocks and as the objects Section 1.5 will transform. Their domains and ranges are worth knowing without derivation.
The pattern behind the table is worth noticing: odd behaviour — cubing, the identity — keeps the whole line in both columns, while even behaviour — squaring, absolute value — folds the negatives onto the positives and halves the range. That single observation predicts most of the table.
| rule | domain | range |
|---|---|---|
| the identity, y = x | all reals | all reals |
| the square, y = x squared | all reals | at or above 0 |
| the cube, y = x cubed | all reals | all reals |
| the square root | at or above 0 | at or above 0 |
| the reciprocal, one over x | all reals except 0 | all reals except 0 |
| the absolute value | all reals | at or above 0 |
Figure (svg): Six small graphs of the toolkit functions with their domains and ranges written under each: the identity, the square, the cube, the square root, the reciprocal, and the absolute value
OpenStax, Precalculus, §1.2 Domain and Range §1.2, pp. 58-61
Picture it
Every function in the next four chapters is one of these, transformed.
Figure (svg): Six small graphs of the toolkit functions with their domains and ranges written under each: the identity, the square, the cube, the square root, the reciprocal, and the absolute value
Only two of the six restrict their domain, and they are the two with a forbidden operation in them. The other four accept every real number, which is exactly what the big idea predicted.
Worked example
Recognising the parent rule makes the domain and range immediate.
\[ \text{State the domain and range of } f(x) = \frac{1}{x-3}. \]
Recognise the parent
Why: It is the reciprocal rule with the input shifted.
Apply the parent's domain restriction to the new input
Why: The denominator, not x, must avoid zero.
\[ x - 3 \ne 0 \]
Solve
Why: The excluded input has moved with the shift.
\[ x \ne 3 \]
Read the range from the parent
Why: A reciprocal never outputs zero, and shifting the input does not change that.
\[ y \ne 0 \]
Figure (svg): The solution to Worked example identify the toolkit function shown as a ladder of expressions, one row per legal move
\[ \text{domain } (-\infty,3)\cup(3,\infty), \qquad \text{range } (-\infty,0)\cup(0,\infty) \]
Verify: ask whether the output zero is attainable
Why: Setting the rule equal to zero gives the equation one equals zero after multiplying up, which has no solution. So no input produces the output 0 and the range genuinely excludes it. Note that the excluded input moved from 0 to 3 while the excluded output stayed at 0 — shifting the input does not shift the outputs.
OpenStax, Precalculus, §1.2 Domain and Range §1.2, pp. 59-60
Sorting
Only a forbidden operation can restrict a domain.
Sort into buckets
Sort the six toolkit rules.
Worked example
Both the domain and the range move, and they move for different reasons.
\[ \text{State the domain and range of } g(x) = \sqrt{x} + 4. \]
Find the domain from the radicand
Why: Nothing was done to the input.
\[ x \ge 0 \]
Find the parent's range
Why: The principal square root returns nonnegative values.
\[ \sqrt{x} \ge 0 \]
Apply the outside operation to those outputs
Why: Adding 4 lifts every output by 4.
\[ \sqrt{x} + 4 \ge 4 \]
State both
Why: The domain is untouched; the range has moved up.
\[ D: x \ge 0, R: y \ge 4 \]
Figure (svg): The solution to Worked example a shifted square root shown as a ladder of expressions, one row per legal move
\[ \text{domain } [0,\infty), \qquad \text{range } [4,\infty) \]
Verify: check the corner point
Why: At the input 0 the rule gives the square root of 0 plus 4, which is 4 — the lowest output, and it is attained, so its bracket is square. Compare with the previous example, where the change was inside the rule and moved the domain instead. Inside changes the domain, outside changes the range, which is the whole content of Section 1.5.
OpenStax, Precalculus, §1.2 Domain and Range §1.2, pp. 60-61
Trap
\[ f(x) = \frac{1}{x-3} \;\Longrightarrow\; \text{range} = \{y : y \ne 3\} \]
Note that the domain excludes 3
Why: The denominator vanishes there, so 3 is not a legal input.
The conclusion drawn is that the range must therefore exclude 3 as well, by symmetry.
The excluded input and the excluded output are unrelated numbers. The domain excludes 3 because that is where the denominator vanishes. The range excludes 0 because a fraction with numerator 1 can never equal zero.
The output 3 is perfectly attainable: solving the rule equal to 3 gives x equal to ten thirds, a legal input. So 3 is in the range, and the claim is simply false.
Find the range by asking which outputs are attainable, never by copying the domain. The two sets answer different questions and coincide only by accident.
Matching
The ranges differ more than the domains do.
Match the pairs
Why: Cubing reaches every height because it is increasing and unbounded in both directions. Squaring folds the negatives up, so it reaches nothing below zero. The reciprocal misses only zero, which no fraction with numerator 1 can equal. The square root matches the squaring rule's range but from only half the inputs, which is exactly the restriction §1.7 will need to invert it.
Prediction
The reciprocal rule has its input replaced by x minus 5, and then 2 is added to the whole thing.
Predict first
Which of the domain and the range moves?
Correct: The domain moves to exclude 5, and the range to exclude 2.
Why: The change inside the rule shifts which input is illegal, from 0 to 5, because the denominator now vanishes there. The change outside lifts every output by 2, so the output that was unattainable moves from 0 to 2. Both move, and each is moved by the change on its own side — which is the rule of thumb the next section makes precise.
Analogy
The toolkit splits into pairs that behave alike.
Match the pairs
Why: The first pair are the odd ones: they preserve sign, so they climb through every height exactly once, which is what makes them invertible without restriction. The second pair are the even ones: each sends a number and its negative to the same output, which is exactly the failure of the horizontal line test from Section 1.1. The third pair are the only two with a forbidden operation, one an even root and one a denominator.
Section
Section 5
Concept
A piecewise function applies one formula on part of its domain and another elsewhere. It is still a single function, because each input is matched by exactly one condition.
The overlap rule is Section 1.1's definition doing its work again. If two conditions both claimed the input 1 with different formulas, that input would have two outputs and the thing would not be a function at all. So the conditions are not a stylistic choice; they are what keeps the object legal.
Figure (svg): A piecewise definition written with a brace beside the graph it produces, showing the two rules meeting at the changeover point with a solid dot on the piece that includes it
OpenStax, Precalculus, §1.2 Domain and Range §1.2, pp. 61-65
Picture it
The brace on the left produces the two-part curve on the right, and the dots record which piece owns the changeover.
Figure (svg): A piecewise definition written with a brace beside the graph it produces, showing the two rules meeting at the changeover point with a solid dot on the piece that includes it
Exactly one dot at the changeover is solid. If both were solid the input would have two outputs; if neither were, it would have none.
Worked example
Choose the piece first, substitute second. Doing it the other way round is the whole difficulty.
\[ \text{For } f(x) = \begin{cases} 2x+1 & x < 1 \\ 4-x & x \ge 1 \end{cases} \text{ find } f(0), f(1), f(3). \]
For the input 0, test the conditions
Why: Zero is below 1, so the first condition holds.
\[ \text{use } 2 x + 1 \]
Substitute into that piece only
Why: Twice zero plus one.
\[ f(0) = 1 \]
For the input 1, test again carefully
Why: The second condition allows equality, so it claims 1.
\[ \text{use } 4 - x \]
Substitute for both remaining inputs
Why: Four minus one, and four minus three.
\[ f(1) = 3, f(3) = 1 \]
Figure (svg): The solution to Worked example evaluate a piecewise rule shown as a ladder of expressions, one row per legal move
\[ f(0) = 1, \quad f(1) = 3, \quad f(3) = 1 \]
Verify: check the changeover deliberately
Why: The input 1 is the one that matters. The first condition is strict, so it does not claim 1; the second allows equality, so it does. Using the wrong piece there would have given 3 rather than 3 — here the two pieces happen to agree at 1, which is why the graph joins up. Had they disagreed, the graph would have a visible jump.
OpenStax, Precalculus, §1.2 Domain and Range §1.2, pp. 62-63
Discrimination
The conditions must not overlap. Gaps are allowed; collisions are not.
Sort into buckets
Sort each pair of conditions.
Worked example
The domain is the union of the conditions, restricted further by each piece's own formula.
\[ \text{Find the domain of } g(x) = \begin{cases} \frac{1}{x} & x < 0 \\ \sqrt{x-1} & x \ge 1 \end{cases} \]
Take the first condition and check its formula there
Why: The reciprocal is fine for every negative input.
\[ \text{all } x < 0\text{ legal} \]
Take the second and check its formula there
Why: The radicand x minus 1 is nonnegative exactly when x is at or above 1.
\[ \text{all } x \ge 1\text{ legal} \]
Notice what neither condition claims
Why: Nothing covers the inputs from 0 up to but not including 1.
\[ \text{gap on } [0, 1] \]
Union what remains
Why: The domain is the two claimed stretches.
\[ (-\infty, 0) U [1, \infty] \]
Figure (svg): The solution to Worked example find the domain of a piecewise rule shown as a ladder of expressions, one row per legal move
\[ (-\infty, 0) \cup [1, \infty) \]
Verify: test one input from the gap
Why: The input 0.5 satisfies neither condition: it is not below 0, and it is not at or above 1. So the rule says nothing about it and it is genuinely outside the domain. A piecewise definition is allowed to leave gaps; it is only forbidden to overlap.
OpenStax, Precalculus, §1.2 Domain and Range §1.2, pp. 64-65
Error analysis
A student writes a piecewise rule with conditions that both claim the same input.
Annotate
On: \( h(x) = \begin{cases} x^2 & x \le 2 \\ 3x & x \ge 2 \end{cases} \)
Check the changeover points of every piecewise definition you write. Exactly one condition must claim each of them — no more, and, if that input is meant to be in the domain, no fewer.
Faded example
Using the same rule as the worked example, evaluate at negative 2.
Fill in the blanks
-2 < 1 \;\Longrightarrow\; \text-2 2x+1 \;\Longrightarrow\; f(-2) = 2(-3)+1 = ___
Why: Negative 2 is below 1, so the first piece applies. Substituting gives negative 4 plus 1, which is negative 3. Testing the condition before substituting is the habit worth building: substituting into whichever formula is written first is the standard way this goes wrong, and it produces a plausible-looking wrong answer.
Prediction
A piecewise rule has pieces that give 5 and 8 at the changeover input, and the second piece's condition is the one that includes it.
Predict first
What does the graph look like there?
Correct: A jump, with a hollow dot at 5 and a solid dot at 8.
Why: The piece that owns the input gets the solid dot, and that is the second piece, giving 8. The first piece approaches the height 5 but never reaches it there, so its end is drawn hollow. The result is a visible jump, and the function is perfectly well defined at that input — its value is 8. §12.3 will call this a jump discontinuity and treat it as a central object rather than an oddity.
Real world
Piecewise definitions are not artificial; most pricing is one.
Discussion prompt
A parking garage charges 3 pounds for up to 2 hours, then 2 pounds per additional hour. Write this as a piecewise rule and say what its changeover condition must be.
Hint: Which piece should claim exactly 2 hours, and why does it matter to a customer?
Answer:
For a stay of at most 2 hours the charge is the flat 3 pounds. Beyond that it is 3 pounds plus 2 pounds for each hour past the second.
The changeover must be claimed by exactly one piece, and the customer-friendly reading gives it to the first: a stay of exactly 2 hours costs 3 pounds. So the first condition is 'at most 2' and the second is 'strictly more than 2'.
If both conditions included 2, a two-hour stay would have two prices — which is not a mathematical nicety but a genuine ambiguity a customer would notice. The definition's ban on overlap is doing real work here.
Comparison
Fill the blanks from memory. These two are asked about together so often that their differences are easy to blur.
Comparison matrix
| domain | range | |
|---|---|---|
| what it collects | the legal inputs | the attainable outputs |
| on a graph | the shadow cast onto the x axis | the shadow cast onto the y axis |
| from a formula | inspect for denominators and even roots | reason about the rule's behaviour |
| difficulty | usually routine and syntactic | often needs the shape of the graph |
| changed by | changes inside the rule | changes outside the rule |
The last row is the one that pays off in Section 1.5. Everything done to the input moves the domain; everything done to the output moves the range.
Pattern
The same four steps handle every formula you will meet before Chapter 4 adds logarithms to the list.
Step 1 is worth taking literally. A great many domain questions are answered correctly in five seconds by noticing there is nothing to restrict, and answered wrongly in five minutes by looking for a restriction that is not there.
OpenStax Algebra and Trigonometry 2e, §3.2 Domain and Range §3.2
Check
Find what the formula forbids.
Check your understanding
What is the domain of the rule that takes the square root of the quantity x minus 6?
Answer: A
Why: An even root needs a radicand at or above zero, so x minus 6 must be at or above 0, giving x at or above 6. The bracket at 6 is square because the radicand may be exactly zero — the square root of zero is zero, a perfectly good output.
Check
Two conditions at once.
Check your understanding
What is the domain of the rule that divides 1 by the quantity x squared minus 4?
Answer: A
Why: The denominator vanishes when x squared equals 4, which happens at both 2 and negative 2. Both are excluded, leaving three intervals joined by unions. The negative root is the one most often forgotten.
Check
Read the definition carefully at the changeover.
Check your understanding
For the piecewise rule giving x squared when x is below 3 and giving 2x when x is at or above 3, what is the value at the input 3?
Answer: A
Why: The second condition allows equality, so it is the piece that claims the input 3. Substituting into that piece gives twice 3, which is 6. The first piece is strict and does not apply at 3 at all.
Real world
Every input field in a piece of software is a domain, whether or not its author thought of it that way.
Discussion prompt
A web form asks for a date of birth and computes the person's age. Describe its domain, and say what the software must do about inputs outside it.
Hint: Which dates are legal, and what happens to a rule fed an illegal input?
Answer:
The domain is roughly dates in the past, and not absurdly far in the past — the rule has no sensible output for a date next year, or for the year 1200.
Software that fails to check its domain does exactly what a formula does at an illegal input: it produces nonsense, or it crashes. Validation is domain checking, and a negative age displayed on a screen is the same error as taking the square root of a negative number.
The piecewise idea appears here too. Age-based pricing, tax brackets and shipping bands are all piecewise rules, and the bugs in them cluster at exactly the place this lesson warns about: the changeover, where an off-by-one condition either claims a value twice or claims it not at all.
Commit first
State your confidence along with your answer.
Predict first
What is the range of the rule that adds 2 to the square root of x?
Correct: At or above 2.
Why: The square root returns values at or above 0, and adding 2 lifts every one of them by 2, so the outputs start at 2. The value 2 is attained, at the input 0, so the bracket is square rather than round. The second option is the parent's range with the shift forgotten, which is the most common wrong answer here.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why finding a domain is mostly mechanical while finding a range often is not, and give one example of each.
Hint: What can you see by looking at the parts of a formula, and what can you not?
Answer:
A good explanation names the two forbidden operations and points out that you can spot them by looking. A denominator and an even root are visible in the formula's shape, so the domain falls out of inspection.
The range, by contrast, asks which outputs actually occur, and that depends on how the function behaves overall — where it turns, whether it is bounded. None of that is visible in the ingredients list.
Good examples: the domain of a fraction with a quadratic denominator takes one factorisation, while the range of the same rule may take a graph. And a quadratic's domain is every real number at a glance, but its range needs the vertex found first.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The second is the one that costs the most marks for the least conceptual difficulty, and is worth ten minutes of deliberate practice. The fourth is the one that returns in Chapter 12, where the changeover point becomes the whole subject rather than a detail.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Draw a flowchart that starts at 'a formula' and ends at 'a domain in interval notation'. Put a decision diamond for 'is there a denominator?' and another for 'is there an even root?', and write on each branch what you do with the condition it produces. Then, on the same page, sketch the six toolkit functions with their domains and ranges written beneath.
If your flowchart has the denominator branch throwing inputs away and the root branch keeping them, you have the part that most often goes wrong.
Recap
Five things, and the fifth one returns as a central object in Chapter 12.
| if you remember one thing | it should be this |
|---|---|
| about domains | start from everything, then remove what the rule forbids |
| about the two forbidden things | a denominator excludes points; an even root keeps a ray |
| about brackets | square means included, and infinity is never included |
| about piecewise rules | exactly one condition claims each input |
Section 1.3 stops asking where a function is defined and starts asking how fast it changes there — the first step towards the derivative.
OpenStax, Precalculus, §1.2 Domain and Range §1.2, pp. 41-66 — everything on these slides traces back here
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