Blok uses only the cookies needed to run the site and remember your preferences. See our Privacy Policy.
Set notation shows up in every domain, every range, and every inequality answer this block. Learn five symbols here and the rest of the course stops looking like code.
A set is a collection of objects. The objects are its elements. That's the whole concept.
The difficulty is never the idea — it's the notation. So this topic is really about reading five symbols fluently.
Once you can, {x ∈ ℝ | x > 3} reads as a sentence: “the real numbers x, such that x is bigger than 3”.
Four conventions. Match them exactly — the exam marks notation.
A ⊆ B when every element of A is also an element of B. A set is always a subset of itself: A ⊆ A.
Mixing up ∈ and ⊆ is the most common notation error in Week 1.
∈ relates an ELEMENT to a SET. ⊆ relates a SET to a SET.
Take A = {1, 2, 3}: • 1 ∈ A — true • 1 ⊆ A — meaningless, 1 isn't a set • {1} ⊆ A — true • {1} ∈ A — false, the element 1 is in A, not the set {1}
∈ is “is in”. ⊆ is “is inside”. Curly braces on the left? You want ⊆.
Three operations build new sets from old ones. Each is one English word:
The three formulas below are just those words in symbols.
Union. In A, or in B, or in both. The “or” is inclusive — never “one but not the other”.
Intersection. In both at once. If nothing qualifies, A ∩ B = ∅ and the sets are disjoint.
Difference. A with anything that's also in B thrown away. Not symmetric — A \ B and B \ A usually differ.
Swapping ∪ and ∩ under time pressure loses more marks than misunderstanding them. Use the shape.
∪ is a cup — open at the top, holds everything you pour in. UNION. ∩ is a cap — closed at the top, only the overlap gets through. INTERSECTION.
A = {1, 2, 3, 4} and B = {3, 4, 5}. Find A ∪ B, A ∩ B, A \ B and B \ A.
A ∪ B — in A or in B. Take all of A, then add anything new from B: 5. Sets never repeat elements, so 3 and 4 appear once. → {1, 2, 3, 4, 5}
A ∩ B — in both. Walk through A: 1 no, 2 no, 3 yes, 4 yes. → {3, 4}
A \ B — in A, not in B. Start with A, delete 3 and 4. → {1, 2}
B \ A — in B, not in A. Start with B, delete 3 and 4. → {5}
A ∪ B = {1, 2, 3, 4, 5} · A ∩ B = {3, 4} · A \ B = {1, 2} · B \ A = {5}
The last two differ. That's the point — difference is not symmetric.
For any sets A, B and C:
Don't memorise all seven.
The commutative ones are obvious: “x in A or x in B” and “x in B or x in A” are the same sentence.
The last two — the difference laws — are the ones that get examined.
Notice the operation FLIPS coming out of the bracket. ∩ inside becomes ∪ outside. ∪ inside becomes ∩ outside.
Remember “the symbol flips” and you can rebuild both.
Calculus 1 lives inside ℝ, the real numbers. Domains and ranges are stated in these terms, so know the family tree.
Each sits inside the next: ℕ ⊆ ℤ ⊆ ℚ ⊆ ℝ
No. ℚ′ is defined as ℝ \ ℚ — the reals with the rationals removed.
So ℚ ∩ ℚ′ = ∅. They're disjoint by construction.
This isn't deep. It's the definition restated, which is why it's a good test of whether you can read the notation.
Interval notation is the compressed way to write a run of consecutive reals. Most answers this block are expected in this form.
Round bracket EXCLUDES the endpoint. Square bracket INCLUDES it.
Infinity never takes a square bracket. [3, ∞] is always wrong and always costs a mark.
Not a convention to memorise: ∞ isn't a real number, so it can't be an element of a set of reals, so nothing can “include” it.
Write (3, ∞) or [3, ∞) depending on whether 3 is in. The infinity side stays round.
A = [−2, 5) and B = (1, 8]. Express A ∩ B and A \ B in interval notation.
Intervals are sets, so ∩ and \ mean exactly what they meant above. Nothing new is being asked.
Draw the number line. Open circle for a round bracket, filled for a square one. Reading the answer off a sketch beats juggling four endpoints in your head.
The only real trap is the endpoints. Decide each one deliberately.
: −2 included, up to 5 excluded. : 1 excluded, up to 8 included.
— the overlap. Starts at the later left endpoint, 1; ends at the earlier right endpoint, 5. Is 1 in? It'd have to be in , and excludes it. No → round. Is 5 in? It'd have to be in , and excludes it. No → round.
— take , delete anything in . Everything in above 1 is also in , so that goes. What survives runs from −2 to 1. Is −2 in? In , not in . Stays → square. Is 1 in? In , and excludes 1 — so it was never deleted. Stays → square.
Endpoints score as much as ranges — or is partial credit at best.
Commonest error: writing , dropping 1 because it feels like 's. is open at 1, so 1 survives the deletion.
Every question this block starts by asking what a function is allowed to eat and what it can produce. Domain and range are those two answers, and they're written in the set notation you just learned.
A function from set to set assigns to each element exactly one element .
The load-bearing words are “exactly one”.
One input can't give two outputs. That single restriction is what separates a function from any old rule.
The vocabulary:
The codomain Y is where outputs are allowed to land. The range is where they actually land.
So ⊆ cdom(f) — always a subset, not always equal.
Range and codomain get treated as the same thing. They aren't.
Take f(x) = x² with codomain ℝ. The codomain says outputs may be any real. The range is only [0, ∞) — no square is ever negative.
Codomain is declared. Range is earned. Only the range needs working out.
In practice, finding a domain means asking what breaks.
Only three things break in this course:
The domain is ℝ with whatever breaks removed.
Find the domain of f(x) = √(x − 3) ⁄ (x − 5).
Two things can break here. Handle them separately.
The even root needs a non-negative inside:
The denominator can't be zero:
Domain questions are subtraction, not construction.
Start with all of ℝ. Take away what breaks. Whatever's left is the domain.
Four function families carry the rest of Week 1:
No. At x = 0 it gives both y = 1 and y = −1.
That breaks “exactly one output per input”, so it's a relation, not a function.
This is what the vertical line test checks: a vertical line crossing the graph twice means two outputs for one input.
Polynomials are the friendliest functions in the course — domain ℝ, no breaking. The work is finding their roots, and there are two tools for that: the integral root theorem and long division.
A function with
where and all coefficients .
That last pair is worth pausing on.
Odd degree guarantees a root because the graph runs from −∞ to +∞ and must cross the axis somewhere.
Even degree guarantees nothing — x² + 1 never touches it.
If is a polynomial of degree and is a root — that is, for some — then there exists a polynomial of degree with for all .
If a polynomial with integer coefficients has an integer root , then divides — the constant term.
These two theorems are a procedure, not trivia.
The integral root theorem gives you a short list of candidates to test. Once one works, the factorising theorem says you can divide it out and drop the degree by one.
Repeat until you're at a quadratic, then use the formula.
Facing a cubic with no obvious factorisation?
List the divisors of the constant term. Test them. One will be a root. Then long-divide.
Factorise x³ − x² − 7x − 5 using long division.
A cubic with no common factor and no obvious grouping. That's the signal for the integral root theorem.
The constant term is −5, so any integer root divides −5. The candidates are the divisors of −5: {−5, −1, 1, 5}. Four numbers to test, and testing is just substitution.
Start with the small ones — they're quicker to evaluate and more often the answer.
Test :
The discriminant D = b² − 4ac decides the nature of the roots:
The sign of a decides which way it opens: a > 0 opens upward, a < 0 opens downward.
Combine the two and you can sketch any quadratic from its coefficients without plotting a single point.
Complete the square on x² + 4x + 13 and on x² − x + 1.
Take half the -coefficient, square it, add and subtract it.
Half of 4 is 2, squared is 4:
At most 4 — degree n gives at most n roots.
At least 0. The “at least one root” guarantee only applies to odd degree, and x⁴ + 1 has none.
One polynomial over another. The denominator is where everything interesting happens — it removes points from the domain and creates the vertical lines the graph runs alongside but never touches.
where and are polynomials of degree and respectively.
Everything you need follows from p and q:
Read that list as one question asked five ways: what does the denominator do here?
Zero denominator, non-zero numerator → vertical asymptote. Zero numerator, non-zero denominator → root. Both zero → neither. The factor cancels and you get a hole.
Not every zero of the denominator is a vertical asymptote.
If the same factor is in the numerator it cancels, and you get a hole in the graph — a single missing point — not an asymptote.
The point still leaves the domain either way. That part doesn't change.
Always factorise both parts and cancel BEFORE reading off asymptotes.
The horizontal asymptote rules are about which part wins as x grows.
Work down the list rather than staring at the finished curve. Each box adds one step’s worth of information, and by the time all of them are ticked the shape is forced — there is only one curve that can have those intercepts, those asymptotes, and those two sign patterns.
. Tick the asymptotes box.
The denominator vanishes at while the numerator does not, so the curve runs away vertically there — and it runs UP on one side and DOWN on the other, which is what a vertical asymptote looks like and why the graph is in two pieces.
Find the domain, roots, and vertical and horizontal asymptotes of f(x) = (x³ − x) ⁄ (x³ − x² − 7x − 5).
Nothing can be read off this as written. Both parts have to be factorised first, because a shared factor would change every answer.
The numerator is easy — take out x. The denominator is the cubic factorised in the previous topic, so it's (x + 1)(x² − 2x − 5).
Once both are factorised, cancel, then answer the four questions in order. Degrees are equal (3 and 3), so the horizontal asymptote is a ratio of leading coefficients, not zero.
Factorise both parts.
Cancelling changes the graph but not the domain.
A cancelled factor leaves a hole, not an asymptote — but the point is still gone from the domain.
No. The numerator has degree 2 and the denominator degree 1, so n > m.
Horizontal asymptotes only exist when n < m (giving y = 0) or n = m (giving y = aₙ⁄bₘ). With n > m the function grows without bound.
Power functions are one formula, ax^n, whose domain swings wildly with n. Absolute value is the last family of Week 1, and the properties here are the ones the inequalities topic runs on.
A function is a power function if , where .
One definition, four very different behaviours. What changes is n — and what it changes most is the domain.
n ∈ ℕ — a polynomial of degree n. e.g. f(x) = −(5⁄2)x³, dom(f) = (−∞, ∞)
n = −1 — the reciprocal function. e.g. f(x) = x⁻¹ = 1⁄x, dom(f) = (−∞, 0) ∪ (0, ∞)
n = 1⁄m with m ≥ 2 even — an even root function. e.g. f(x) = = 3√x, dom(f) = [0, ∞)
n = 1⁄m with m ≥ 2 odd — an odd root function. e.g. f(x) = − = −∛x, dom(f) = (−∞, ∞)
The even/odd root split is the only part of this worth memorising, and there's a reason behind it.
Even roots need a non-negative input — no real number squares to −4. Odd roots take anything — ∛(−8) = −2 is perfectly real.
Two readings of the same thing:
Work down the list rather than staring at the finished curve. Each box adds one step’s worth of information, and by the time all of them are ticked the shape is forced — there is only one curve that can have those intercepts, those asymptotes, and those two sign patterns.
. Tick the intercepts box.
The V has its corner where the inside of the bars changes sign, at — that is always where an absolute value bends, and it is the first thing to find.
The two arms are the two cases of the definition: to the right of 2 the graph is , to the left it is . Reflecting the negative part upward is what “ if ” looks like, and why that minus sign does not produce a negative number.
“|x| = −x if x < 0” looks like it produces a negative. It doesn't.
If x is negative then −x is positive. Try x = −3: |−3| = −(−3) = 3.
The minus sign flips a negative into a positive — that's the whole job.
√(x²) = |x|, not x. Dropping the bars is wrong whenever x could be negative.
Properties, for all x, y ∈ ℝ:
Two of those earn their keep.
|a − b| = |b − a| means distance doesn't care about direction — which is why |x − 3| < 2 reads as “x is within 2 of 3”.
The triangle inequality is ≤, not =. Equality only holds when x and y point the same way.
Show that |x + y| ≤ |x| + |y| fails to be an equality when x = 3 and y = −5.
Left side: |3 + (−5)| = |−2| = 2
Right side: |3| + |−5| = 3 + 5 = 8
So 2 ≤ 8. True, but a long way from equal.
2 ≤ 8 — the inequality holds strictly.
Opposite signs cause cancellation on the left but never on the right, because the bars strip the signs first. Equality needs x and y to share a sign.
is a fourth root — m = 4 is even — so dom(f) = [0, ∞).
is a fifth root — m = 5 is odd — so dom(f) = (−∞, ∞).
Two rules do all the work here, and the only thing to get right is which one gives “and” and which gives “or”. The quadratic case has a trap the lecture flags explicitly.
Every absolute value inequality becomes an ordinary one the moment you strip the bars. There are exactly two ways to strip them.
Less than — an AND. x must be under a and above −a at once. One interval.
Greater than — an OR. Two separate pieces, joined with a union.
Which one is “and”, which is “or”? Think about what the inequality means as distance.
|x| ≤ a is “close to zero” — one region, so AND, an intersection. |x| ≥ a is “far from zero” — you can be far in either direction, so OR, a union.
Solve |x − 5| < 2.
Less than, so it's the AND rule — one chain.
Add 5 to all three parts:
Solve |3x + 2| ≥ 4.
Greater than, so it's the OR rule — two branches, kept separate.
Quadratic absolute value inequalities use the same two rules. The temptation is to factorise inside the bars first — and that's where the lecture stops to warn you.
The lecture presents this solution to |x² − 2x| ≥ 1 and asks you to spot the error:
|x² − 2x| ≥ 1 |x(x − 2)| ≥ 1 |x|·|x − 2| ≥ 1 |x| ≥ 1 or |x − 2| ≥ 1 ← INCORRECT
That last step gives x ∈ (−∞, ∞), which is wrong.
Apply the ≥ rule to the WHOLE expression first. Don't split a product inside the bars.
Why that step fails is worth understanding, because it looks reasonable.
The implication only runs one way. If |x|·|x − 2| ≥ 1 then yes, at least one factor must be ≥ 1 — two numbers below 1 can't multiply to 1 or more.
But the reverse is false. Take x = 1.5: |x| = 1.5 ≥ 1, yet the product is 1.5 × 0.5 = 0.75, which is less than 1.
So the “or” statement is implied by the original, not equivalent to it. Treating ⟹ as ⟺ lets in solutions that don't belong, which is why the bad answer is the whole real line.
Solve |x² − 2x| ≥ 1.
Do not factorise inside the bars. Apply the ≥ rule to x² − 2x as a single object, which splits into two ordinary quadratic inequalities.
Both branches get solved by completing the square — it's more reliable here than the formula, because the second branch turns into a perfect square and that's the whole point of it.
Expect the second branch to behave strangely. A squared quantity that must be ≤ 0 can only be exactly 0, so it contributes a single point rather than an interval.
One. It's a ≤, so it's the AND rule, and that always produces a single interval.
An answer with a union in it means the wrong rule was used.
Two kinds of sequence, told apart by one question: do you add the same thing each step, or multiply by it? Each has a formula for the nth term and a formula for the sum, and that's the whole topic.
a, a + d, a + 2d, …, a + (n − 1)d, … — a sequence where the difference between any term and the one before it is a fixed constant d. For example: 5, 7, 9, 11, 13, 15, …
Recursive form, then the closed form. Use the closed form unless asked otherwise.
That sum formula isn't arbitrary. Pair the first term with the last, the second with the second-last, and so on — every pair adds to the same total.
There are n⁄2 such pairs, each worth (a₁ + aₙ). That's the formula.
a, ar, ar², …, arⁿ, … — a sequence where the quotient of any term with the one before it is a fixed ratio r. For example: 2, 4, 8, 16, 32, 64, …
The two most expensive errors here are both about counting, not algebra.
Off-by-one on n: a sum from k = 5 to k = 24 has 24 − 5 + 1 = 20 terms, not 19.
Wrong a: in the sum formula a means the first term of the sum you're computing. If the sum starts at k = 5, then a is the k = 5 term.
Write down n and a explicitly before touching either formula. Both errors survive perfect algebra.
Calculate Σ from k = 5 to 24 of (2k − 1).
The terms go up by 2 each step, so it's arithmetic with .
Count the terms:
First and last:
Calculate Σ from k = −1 to 9 of (5k + 3).
Arithmetic again, . The negative starting index is the only awkward part, and it only affects the count.
Calculate Σ from k = 0 to 8 of 3·2ᵏ.
Geometric with .
The sum runs , so terms. The first term is at : , so .
Calculate the finite sum 2 − 1 + 1⁄2 − 1⁄4 + … − 1⁄16.
Alternating signs with halving magnitudes means geometric with a negative ratio. Divide any term by the one before to confirm: −1 ÷ 2 = −1⁄2, so r = −1⁄2.
Nothing is given as a summation, so the term count has to be worked out by listing. That's the step to be careful with — the final term is stated as −1⁄16, so keep listing until it appears.
The negative r makes the denominator 1 − (−1⁄2) = 3⁄2, not 1⁄2. Getting that sign wrong is the main risk.
Identify and :
Geometric, with r = 2.
Subtract consecutive terms: 3, 6, 12 — not constant, so not arithmetic. Divide consecutive terms: 2, 2, 2 — constant, so geometric.
One subtraction and one division settles it every time.
Every input the function is allowed to take.
Every output the function actually produces.
Now intersect the two conditions — both must hold at once. From we get ; remove 5:
Note 3 is included — is fine. Only the denominator forces an exclusion.
So −1 is a root, and is a factor.
Now long-divide by :
Remainder 0, which confirms −1 really was a root:
Show the substitution that finds the root — the marks are for justifying which factor you divided by, not just for the division.
A zero remainder is the check. A non-zero remainder means you picked a number that isn't a root; go back to the candidate list rather than pressing on.
The quadratic formula, for f(x) = ax² + bx + c with a ≠ 0.
Half of −1 is , squared is :
Both have a positive constant left over, so both sit entirely above the axis — no real roots. Completing the square shows that without computing a discriminant.
The degrees match top and bottom, so far out in either direction the ratio settles at the ratio of leading coefficients: . Tick the intercepts box for the root at , where the numerator vanishes and the denominator does not.
Long division, with deg(r) < deg(q). Use it when the numerator's degree is the larger.
The cancels:
Solve by completing the square:
Now read off:
(1) Domain — the denominator must not vanish, and was removed before cancelling:
(2) Roots — numerator zero, denominator not:
(3) Vertical asymptotes:
(4) Horizontal asymptote — degrees equal, leading coefficients both 1:
excludes 1 ± √6 and −1 · roots x = 0, 1 · vertical asymptotes x = 1 ± √6 · horizontal asymptote y = 1.
The mark most often dropped is x ≠ −1. It cancels, so it produces a hole rather than an asymptote — but it still leaves the domain, and the domain is asked for separately.
Claiming x = −1 as a vertical asymptote is the other half of the same error.
An even root of a negative isn't real; an odd root is.
Reads as “ is within 2 of 5”, which is exactly what says. The answer is centred on 5 and 2 wide each way.
Branch one:
Branch two:
Complete the square on both:
Branch one: . Factorise as a difference of squares:
A product of two factors is outside the roots:
Branch two: . A square is never negative, so the only possibility is
Union the branches.
The isolated point is the mark most often lost. Branch two doesn't vanish — it contributes exactly one value, and dropping it loses a solution.
Splitting the product inside the bars scores nothing even if the arithmetic afterwards is clean, because it gives .
The finite sum: number of terms, halved, times the sum of the first and last.
Same pair of forms. Note the exponent is n − 1, not n — the first term has r⁰.
The finite geometric sum. a is the FIRST term of the sum, not necessarily the sequence's a₁.
Apply the sum formula:
The count is where this one is won or lost: , not 10. Subtracting a negative index adds.
List the terms to find :
That's 6 terms, so .
Apply the formula:
Numerator — an even power kills the sign, so :
Denominator:
So:
Three places marks go: writing instead of ; counting 5 terms instead of 6; and mishandling as rather than .
The even power making positive is worth stating explicitly — it shows the sign was handled deliberately rather than dropped.