This deck builds the single most important idea in College Algebra: a function is a machine with exactly one output for each input. It distinguishes relations from functions using mapping diagrams and the vertical line test, then covers function notation and evaluating at numbers and at expressions, the difference quotient, domain and range in interval notation, piecewise functions, increasing and decreasing intervals, relative extrema, average rate of change, and even against odd functions. It targets four classic errors: reading function notation as multiplication, substituting into only one copy of the variable, answering a domain question with a list of excluded values, and swapping domain with range.
Subject: College Algebra · 139 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
College Algebra - Deck 09
One input in, exactly one output out. The idea the whole rest of the course is built on.
Objectives
Everything after this deck - graphs, inverses, polynomials, logs - is a function wearing a costume. Get this idea clean and the rest gets easier.
Warm-up
Discussion prompt
Before we open Functions, Domain, and Function Notation: without looking back, what was the main idea of Complex Numbers, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck explains why the number system had to be extended, then introduces the imaginary unit and its one defining property, standard form, and the complex plane. It covers adding, subtracting, and multiplying, the four-step cycle of powers, the conjugate and division, complex solutions of quadratics with a negative discriminant, and the modulus. It targets the four errors that sink this unit: applying the radical product rule to two negative radicands, leaving the square of the imaginary unit unsimplified, slipping a sign when multiplying by a conjugate, and reporting the imaginary part with the unit still attached.
Section
Part 1
Concept
Before functions, something simpler. A relation is any rule or list that pairs inputs with outputs. That is the whole definition - no restrictions at all.
relation — Any set of ordered pairs. The first coordinate of each pair is an input; the second is an output.
Here is a relation written as a list of pairs. The first number in each pair is the input, the second is what it is paired with.
\[ \{(1,\,4),\ (2,\,7),\ (3,\,10)\} \]
Counterexample
Discussion prompt
Before functions, something simpler. A relation is any rule or list that pairs inputs with outputs. That is the whole definition - no restrictions at all.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Intuition
Picture a vending machine. You press B4 and a bag of pretzels drops out. You press B4 again tomorrow - pretzels again.
That is a function. Same button, same result, every single time. The machine is predictable.
Now picture a broken machine: press B4 and sometimes pretzels come out, sometimes gum. That machine is a relation, but it is not a function. You cannot answer the question 'what does B4 give you?'
Two different buttons giving the same snack is fine - that machine is still predictable. Only one button, two answers breaks it.
Analogy
Discussion prompt
Explain A function is a vending machine by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Picture a vending machine. You press B4 and a bag of pretzels drops out. You press B4 again tomorrow - pretzels again.
Concept
A function is a relation with one extra promise: each input gets exactly one output.
function — A relation in which every input is paired with exactly one output. No input is allowed to have two different partners.
Say it out loud in the form you will actually use: no input may repeat with different outputs. That single sentence answers every 'is it a function?' question in this deck.
Explain it
Discussion prompt
Explain The one rule that makes a relation a function to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
A function is a relation with one extra promise: each input gets exactly one output.
Picture it
Animation
Shows: What makes a relation a function — a rendered Manim animation.
Rendered with Manim.
Takeaway: Outputs may repeat; inputs may not.
Concept
Figure (svg): Two mapping diagrams. On the left, inputs 1, 2, 3 each send one arrow to outputs 4 and 5 - a function. On the right, input 1 sends two arrows to outputs 4 and 5 - not a function.
A mapping diagram puts the inputs in one bubble, the outputs in another, and draws an arrow for each pair.
The test becomes visual: look at the left bubble. If any input has two arrows leaving it, it is not a function.
On the left, inputs 2 and 3 both point at 5. That is fine - two buttons, same snack. On the right, the input 1 fires two arrows. That breaks the promise.
Ranking
Put in order
Put the moves of Are these sets of pairs functions? into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The inputs are the first coordinates.
Worked example
Decide whether each set of ordered pairs defines a function.
\[ A = \{(-2,\,5),\ (0,\,1),\ (3,\,5),\ (4,\,7)\} \]
List only the inputs of A
Why: The inputs are the first coordinates. Strip everything else away so nothing distracts you.
\[ -2,\ 0,\ 3,\ 4 \]
No input repeats, so A is a function
Why: Every one of the four inputs shows up exactly once, so each has exactly one partner. The repeated output 5 is irrelevant - the rule is about inputs.
\[ B = \{(1,\,2),\ (3,\,6),\ (1,\,8),\ (5,\,0)\} \]
List only the inputs of B
Why: Same first move every time. Here the inputs are 1, 3, 1, and 5.
\[ 1,\ 3,\ 1,\ 5 \]
The input 1 repeats with two different outputs, so B is not a function
Why: The pairs say the input 1 gives 2, and also that it gives 8. The machine has no single answer for the input 1.
Verify by asking the machine question of each input
Why: For A: what does -2 give? Only 5. What does 0 give? Only 1. What does 3 give? Only 5. What does 4 give? Only 7 - four clean answers, so A is a function. For B: what does 1 give? Both 2 and 8 - no single answer, so B is not.
\[ A \text{ is a function.}\qquad B \text{ is not.} \]
Concept
This is the asymmetry students trip over, so give it ten seconds of real attention.
| What repeats | Still a function? | Why |
|---|---|---|
| An output value | Yes | Two buttons can drop the same snack. |
| An input, same output both times | Yes | It is a duplicate entry, not a conflict. |
| An input, different outputs | No | The machine has two answers for one press. |
A real example of the middle row: the pairs listed as one and four, then one and four again. Same input, same output - harmless repetition.
Picture it
Animation
Shows: Range is what the outputs cover — a rendered Manim animation.
Rendered with Manim.
Takeaway: Read the domain across, and the range up.
Concept
A function can be handed to you in five costumes. Same object every time - only the packaging changes.
The 'is it a function?' test looks different in each costume, but it is always checking the same one thing: one output per input.
Step zero
Discussion prompt
Is this table a function? — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Identify which row holds the inputs
Answer:
Worked example
A table shows the number of practice problems a student finished and the quiz score that followed.
| problems done | 0 | 5 | 10 | 15 | 20 |
|---|---|---|---|---|---|
| quiz score | 62 | 71 | 79 | 84 | 84 |
Identify which row holds the inputs
Why: The problems done is what you choose; the score is what results. So the top row is the input row.
Scan the input row for a repeat
Why: The inputs 0, 5, 10, 15, 20 are all different, so no input can possibly have two outputs.
The repeated output 84 does not matter
Why: Two different amounts of practice landing on the same score is perfectly allowed - outputs are free to repeat.
Read a value off the table
Why: The column headed 10 pairs with 79, so the input 10 produces the output 79.
\[ S(10) = 79 \]
Verify against the definition
Why: Ask the machine question of every input: 0 gives only 62, 5 gives only 71, 10 gives only 79, 15 gives only 84, 20 gives only 84. Five inputs, five single answers - the table is a function.
Concept
Figure (svg): A sideways parabola opening to the right with a dashed vertical line crossing it at two points, marked with dots
When the function arrives as a graph, you do not have a list of pairs to scan. You have a picture. So the test becomes a picture too.
vertical line test — If any vertical line crosses the graph more than once, the graph is not a function. If every vertical line crosses at most once, it is.
The dashed line here hits the curve twice. That single vertical line is enough to disqualify the whole graph.
Picture it
Animation
Shows: The vertical line test — a rendered Manim animation.
Rendered with Manim.
Takeaway: Any vertical line meets this once, so it is a function.
Intuition
A vertical line is the set of all points sharing one input. Sliding along it, the input never changes - only the output does.
So asking 'how many times does this vertical line hit the graph?' is exactly asking 'how many outputs does this input have?'
Two hits means two outputs for one input. Same broken vending machine, drawn instead of listed. The vertical line test is not a new rule - it is the same rule in picture form.
Estimation
Predict first
Is this equation a function, with the horizontal axis as the input?
Commit before you compute: what does Testing a sideways parabola come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify both points really satisfy the equation
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Substituting the output 2: two squared is 4, matching the left side.
Worked example
Is this equation a function, with the horizontal axis as the input?
\[ x = y^{2} \]
Pick one input and solve for the outputs it produces
Why: The fastest way to break a function is to find a single input with two outputs. Try the input 4.
\[ 4 = y^{2} \]
Take the square root of both sides, keeping both signs
Why: Any positive number has two square roots. Dropping the negative one is exactly how students miss this.
\[ y = 2 \quad \text{or} \quad y = -2 \]
The input 4 produced two different outputs, so this is not a function
Why: The points shown as four and two, and four and negative two, both sit on the graph. A vertical line through the input 4 hits twice.
Verify both points really satisfy the equation
Why: Substituting the output 2: two squared is 4, matching the left side. Substituting the output -2: negative two squared is also 4, matching again. Both points are genuinely on the graph, so the double hit is real and the relation is not a function.
\[ (2)^{2} = 4 \quad \checkmark \qquad (-2)^{2} = 4 \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Testing a sideways parabola", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Substituting the output 2: two squared is 4, matching the left side. Substituting the output -2: negative two squared is also 4, matching again. Both points are genuinely on the graph, so the double hit is real and the relation is not a function.
Ranking
Put in order
These are the steps of Pattern: is it a function?, scrambled. Put them back in order before the next slide shows you.
Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.
Pattern
One question, four costumes. Find the inputs, then look for a conflict.
If you find no conflict anywhere, say yes, it is a function - and be ready to name the reason: every input has exactly one output.
Check
Scan the input row before you look at the choices.
| input | -2 | 0 | 1 | 0 |
|---|---|---|---|---|
| output | 5 | 3 | 7 | 8 |
Check your understanding
Does this table define a function?
Answer: B
Why: The input 0 appears twice, once paired with 3 and once paired with 8. One input with two different outputs breaks the definition, so this is a relation but not a function.
Section
Part 2
Concept
So far we have said 'the output'. Mathematicians got tired of that and invented a shorthand.
\[ f(x) = 3x + 1 \]
Read that as: the function named f, applied to the input x, gives three times the input plus one.
function notation — A naming scheme where the letter outside the parentheses is the function's name and the thing inside the parentheses is the input. The whole symbol stands for the output.
The huge win: the notation carries the input and the output in one symbol, so you can talk about a specific value without writing a sentence.
Picture it
Animation
Shows: f(x) is a name, not a multiplication — a rendered Manim animation.
Rendered with Manim.
Takeaway: Substitute the whole input, brackets included.
Intuition
Think of the parentheses as an empty slot on the front of the machine. Whatever you drop into the slot gets processed by the rule.
\[ f(\ \square\ ) = 3\,\square + 1 \]
Written that way, evaluating is mechanical: put the same thing in every box.
\[ f(2) = 3(2) + 1 = 7 \]
This is the mental picture worth keeping. Not 'f times something' - f of something.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The parentheses next to a letter look exactly like multiplication, so students distribute the name across a sum.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Treating f as a multiplier turns f of a sum into a sum of two pieces - the same move you would make with a number out front.
The letter is a name, not a factor. Simplify what is inside the slot first, then run the machine once.
Why: Treating f as a multiplier turns f of a sum into a sum of two pieces - the same move you would make with a number out front.
Trap
The parentheses next to a letter look exactly like multiplication, so students distribute the name across a sum.
\[ f(x) = 3x + 1 \]
Split the input as if f distributed
Why: Treating f as a multiplier turns f of a sum into a sum of two pieces - the same move you would make with a number out front.
\[ f(1+2) \;\overset{?}{=}\; f(1) + f(2) \]
Compute the two pieces and add
Why: The first gives 4 and the second gives 7, so this route lands on 11.
\[ 4 + 7 = 11 \]
The letter is a name, not a factor. Simplify what is inside the slot first, then run the machine once.
\[ f(x) = 3x + 1 \]
Simplify inside the parentheses first
Why: The parentheses hold the input. One plus two is three, so this is the function evaluated at the single input 3.
\[ f(1+2) = f(3) \]
Run the rule once, on 3
Why: Three times three plus one is ten. The true value is 10, not 11 - so the distributing move really does give a wrong number, every time.
\[ f(3) = 3(3) + 1 = 10 \]
Notation
Annotate
From Trap: reading the notation as multiplication — read this one piece at a time. What is each part doing?
On: \( f(1+2) \;\overset{?}{=}\; f(1) + f(2) \)
Concept
Evaluating is a search-and-replace. Find every appearance of the variable in the rule and swap in the new input.
\[ g(x) = x^{2} - 4x + 1 \]
There are two copies of the variable here. Miss either one and the answer is wrong.
A habit worth building: wrap the new input in parentheses as you substitute. It costs nothing and saves you from every sign error in this chapter.
\[ g(-3) = (-3)^{2} - 4(-3) + 1 \]
Fill the middle
Fill in the blanks
From Evaluating at a number — finish the line. Write what belongs on the right of the equals sign before you look.
f(x) = 2x^{2} - 5x + 3
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Parentheses keep the negative attached to the 3, which is exactly what the squaring and the multiplying need to see.
Worked example
Evaluate this function at negative three and at zero.
\[ f(x) = 2x^{2} - 5x + 3 \]
Substitute negative three into both copies, in parentheses
Why: Parentheses keep the negative attached to the 3, which is exactly what the squaring and the multiplying need to see.
\[ f(-3) = 2(-3)^{2} - 5(-3) + 3 \]
Square before multiplying
Why: Order of operations: exponents outrank multiplication, and negative three squared is positive nine.
\[ f(-3) = 2(9) - 5(-3) + 3 \]
Multiply, then add
Why: Two times nine is eighteen, and negative five times negative three is positive fifteen - a negative times a negative.
\[ f(-3) = 18 + 15 + 3 = 36 \]
Now evaluate at zero
Why: Both variable terms vanish, leaving only the constant. This is always the fastest input to test.
\[ f(0) = 2(0)^{2} - 5(0) + 3 = 3 \]
Verify by factoring the rule and re-evaluating
Why: The rule factors as the product of two binomials, and expanding gives back two x squared minus five x plus three. At the input negative three the factors are negative nine and negative four, whose product is 36 - matching the answer computed the long way.
\[ 2x^{2}-5x+3=(2x-3)(x-1)\ \Rightarrow\ (-9)(-4)=36 \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Evaluating at a number", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The rule factors as the product of two binomials, and expanding gives back two x squared minus five x plus three. At the input negative three the factors are negative nine and negative four, whose product is 36 - matching the answer computed the long way.
Missing information
Discussion prompt
Same machine, but now the thing you drop into the slot is itself an expression.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Two copies means two substitutions. The parentheses are not optional here - they hold the sum together.
Worked example
Same machine, but now the thing you drop into the slot is itself an expression.
\[ g(x) = x^{2} - 4x + 1, \qquad \text{find } g(a+2) \]
Replace every copy of the variable with the whole expression, in parentheses
Why: Two copies means two substitutions. The parentheses are not optional here - they hold the sum together.
\[ g(a+2) = (a+2)^{2} - 4(a+2) + 1 \]
Expand the square as a full binomial square
Why: Squaring a sum gives the first squared, plus twice the product, plus the last squared. The middle term is the one everybody drops.
\[ (a+2)^{2} = a^{2} + 4a + 4 \]
Distribute the negative four across both terms
Why: The four multiplies the whole parenthesis, so both the variable and the 2 get hit.
\[ -4(a+2) = -4a - 8 \]
Combine everything
Why: The four a and the negative four a cancel; four minus eight plus one is negative three.
\[ g(a+2) = a^{2} + 4a + 4 - 4a - 8 + 1 = a^{2} - 3 \]
Verify with a test number
Why: Let the input a be 1. The original route gives g of 3, which is nine minus twelve plus one, or negative two. The simplified formula gives one minus three, also negative two. The two agree, so the algebra is right.
\[ g(3) = 9 - 12 + 1 = -2 \qquad (1)^{2} - 3 = -2 \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Evaluating at an expression", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Let the input a be 1. The original route gives g of 3, which is nine minus twelve plus one, or negative two. The simplified formula gives one minus three, also negative two. The two agree, so the algebra is right.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The eye finds the first variable, substitutes carefully - and then gets lazy on the second one.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The second copy quietly becomes just a, dropping the plus two.
Count the copies before you substitute, then replace all of them.
Why: The second copy quietly becomes just a, dropping the plus two. Nothing looks broken on the page.
Trap
The eye finds the first variable, substitutes carefully - and then gets lazy on the second one.
\[ g(x) = x^{2} - 4x + 1, \qquad g(a+2) = ? \]
Substitute the full expression only in the squared term
Why: The second copy quietly becomes just a, dropping the plus two. Nothing looks broken on the page.
\[ (a+2)^{2} - 4a + 1 \]
Simplify to the wrong answer
Why: Expanding gives a squared plus four a plus four, minus four a, plus one. The a terms still cancel, so it even looks tidy - and it is wrong.
\[ a^{2} + 5 \qquad \text{(wrong)} \]
Count the copies before you substitute, then replace all of them.
\[ g(x) = x^{2} - 4x + 1 \qquad \text{(two copies of } x \text{)} \]
Substitute the full expression in both places
Why: Two copies, two parenthesized replacements. The rule does not care that the second copy is only linear.
\[ (a+2)^{2} - 4(a+2) + 1 \]
Simplify to the correct answer
Why: The extra negative eight from distributing is exactly what the wrong route lost. Testing the input a equal to 1 confirms it: the true value is negative two, and the wrong formula would have said six.
\[ a^{2} - 3 \qquad \checkmark \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
Two copies, two parenthesized replacements. The rule does not care that the second copy is only linear.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Answer:
The second copy quietly becomes just a, dropping the plus two. Nothing looks broken on the page.
Concept
A number, another variable, a whole expression, even the negative of the variable - the slot does not care. The rule runs the same way.
| You are asked for | What goes in every slot |
|---|---|
| The value at five | the number 5 |
| The value at t | the letter t |
| The value at twice t | the expression 2t |
| The value at the opposite of the input | the expression with a leading negative sign |
That last row is the one that unlocks even and odd functions at the end of this deck, so it is worth practicing now.
Comparison
Comparison matrix
From The slot accepts anything at all: refill the What goes in every slot column from what you know. The rest of the table is as it appeared.
| You are asked for | What goes in every slot |
|---|---|
| The value at five | the number 5 |
| The value at t | the letter t |
| The value at twice t | the expression 2t |
| The value at the opposite of the input | the expression with a leading negative sign |
Step zero
Discussion prompt
Evaluating at another expression — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Find the value at the opposite of the input
Answer:
Worked example
Two substitutions into the same machine.
\[ h(x) = x^{2} + 3x \]
Find the value at the opposite of the input
Why: Drop the negated variable into both slots, parentheses included, and let the exponent act on the whole thing.
\[ h(-x) = (-x)^{2} + 3(-x) \]
Square kills the sign, multiplication keeps it
Why: A negative squared is positive, so the first term is unchanged; the second term simply flips sign.
\[ h(-x) = x^{2} - 3x \]
Now find the value at twice the variable
Why: Same routine: substitute the expression into both slots and let the exponent act on the coefficient too.
\[ h(2t) = (2t)^{2} + 3(2t) \]
Apply the exponent to both factors inside
Why: Squaring a product squares each factor, so two t squared becomes four t squared - the 2 does not get to stay a 2.
\[ h(2t) = 4t^{2} + 6t \]
Verify both with test numbers
Why: For the first: at an input of 2, the original gives h of negative two, which is four minus six, or negative two; the formula gives four minus six, also negative two. For the second: at t equal to 1, the original gives h of 2, which is four plus six, or ten; the formula gives four plus six, also ten.
\[ h(-2) = -2 \quad \checkmark \qquad h(2) = 10 \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Evaluating at another expression", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: For the first: at an input of 2, the original gives h of negative two, which is four minus six, or negative two; the formula gives four minus six, also negative two. For the second: at t equal to 1, the original gives h of 2, which is four plus six, or ten; the formula gives four plus six, also ten.
Pattern
Step six takes fifteen seconds and catches nearly every algebra slip in this section. Do it every time.
Elimination
Eliminate the wrong options
Which expression equals the value of f at the input a minus one?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Substituting into both copies gives the quantity a minus one, squared, minus three times the quantity a minus one. Expanding: a squared minus two a plus one, minus three a plus three, which combines to a squared minus five a plus four. Test with a equal to 2: the original gives f of 1, which is one minus three, or negative two; the formula gives four minus ten plus four, also negative two.
Check
Work it on paper before you look. Count the copies first.
\[ f(x) = x^{2} - 3x, \qquad \text{find } f(a-1) \]
Check your understanding
Which expression equals the value of f at the input a minus one?
Answer: A
Why: Substituting into both copies gives the quantity a minus one, squared, minus three times the quantity a minus one. Expanding: a squared minus two a plus one, minus three a plus three, which combines to a squared minus five a plus four. Test with a equal to 2: the original gives f of 1, which is one minus three, or negative two; the formula gives four minus ten plus four, also negative two.
Concept
A graph is a picture of every input-output pair at once. Each point on the curve is one pair: the horizontal position is the input, the height is the output.
So evaluating from a graph is a two-move walk: go across to the input, then up or down to the curve, and read the height.
That is the same walk you make when you read a temperature off a weather chart. Nothing new - just named.
Estimation
Predict first
Use the graph to answer two different-looking questions.
Commit before you compute: what does Reading values off a graph, both directions come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify each answer by walking back the other way
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Start at the input 0 and go up: the height is three, matching.
Worked example
Figure (svg): A polygonal graph passing through the points negative two comma one, zero comma three, two comma negative one, and four comma three, with the endpoints marked by solid dots
Use the graph to answer two different-looking questions.
\[ \text{(a) } f(2) = ? \qquad \text{(b) solve } f(x) = 3 \]
For part (a), start on the horizontal axis at the input 2
Why: The input always lives on the horizontal axis, so an evaluate question starts there.
Move vertically to the curve and read the height
Why: The curve at that input sits one unit below the horizontal axis, so the output is negative one.
\[ f(2) = -1 \]
For part (b), start on the vertical axis at the output 3
Why: This question hands you the output and asks for the input, so the walk runs the other way - start on the vertical axis.
Move horizontally and mark every place the line meets the curve
Why: A horizontal line at a height of three touches the graph twice: once at the corner above zero and once at the right endpoint.
\[ x = 0 \quad \text{and} \quad x = 4 \]
Verify each answer by walking back the other way
Why: Start at the input 0 and go up: the height is three, matching. Start at the input 4 and go up: the height is three, matching. Start at the input 2 and go down: the height is negative one, matching part (a). All three round trips close.
Picture it
Animation
Shows: Each line of the worked example "Reading values off a graph, both directions", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Start at the input 0 and go up: the height is three, matching. Start at the input 4 and go up: the height is three, matching. Start at the input 2 and go down: the height is negative one, matching part (a). All three round trips close.
Concept
These two questions look almost identical on the page and are answered in opposite directions. Keeping them straight is worth real points.
| Question | You are given | You are looking for | Answer looks like |
|---|---|---|---|
| Find the value at 4 | an input | the output | a single number |
| Solve for the input giving 3 | an output | the input or inputs | possibly several numbers |
\[ f(4) = 3 \qquad \text{vs} \qquad f(x) = 3 \ \Rightarrow\ x = 0,\ 4 \]
Notice the second one had two answers. That is allowed - a function forbids one input from having two outputs, never the reverse.
Trade off
Comparison matrix
From Evaluate versus solve - two different jobs: every row here is a choice with a cost. Fill the Answer looks like column, then say which row you would actually pick and what you give up for it.
| Question | You are given | You are looking for | Answer looks like |
|---|---|---|---|
| Find the value at 4 | an input | the output | a single number |
| Solve for the input giving 3 | an output | the input or inputs | possibly several numbers |
Section
Part 3
Concept
This is the hardest-looking thing in the chapter and the most valuable. It is nothing but evaluate, subtract, divide - the skills you just built.
\[ \frac{f(x+h) - f(x)}{h}, \qquad h \ne 0 \]
difference quotient — The change in the output divided by the change in the input, between an input and an input a little farther along. The top is a difference of two outputs; the bottom is the gap between the two inputs.
Your first year of calculus opens by pushing the gap toward zero in this exact expression. Building it cleanly now is the whole point.
Intuition
Pick a point on the curve. Walk a little to the right - a gap of h - and land on a second point on the curve.
The rise between those two points is the difference of the two heights. The run is the gap you walked. Divide them and you get the slope of the straight line joining the two points.
\[ \text{slope} = \frac{\text{rise}}{\text{run}} = \frac{f(x+h) - f(x)}{(x+h) - x} = \frac{f(x+h) - f(x)}{h} \]
So the bottom is not a mystery - it is what is left when the two inputs are subtracted. And the answer must always simplify so the gap cancels out of the denominator. If it does not, you made an algebra mistake.
Hypothesis
Predict first
Difference quotient of a quadratic is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Evaluate at the shifted input first, on its own line
Why: Doing this piece separately keeps the substitution honest. Both copies of the variable get the whole shifted input.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Build and simplify the difference quotient for this function.
\[ f(x) = x^{2} - 4x \]
Evaluate at the shifted input first, on its own line
Why: Doing this piece separately keeps the substitution honest. Both copies of the variable get the whole shifted input.
\[ f(x+h) = (x+h)^{2} - 4(x+h) \]
Expand it completely
Why: The squared binomial gives three terms, and the negative four distributes over both pieces.
\[ f(x+h) = x^{2} + 2xh + h^{2} - 4x - 4h \]
Subtract the original, keeping it in parentheses
Why: The whole original expression is being subtracted, not just its first term. The parentheses are what protect the sign of every term inside.
\[ \bigl(x^{2} + 2xh + h^{2} - 4x - 4h\bigr) - \bigl(x^{2} - 4x\bigr) \]
Distribute the subtraction and cancel
Why: The x squared terms cancel, and the negative four x cancels against the positive four x. Everything left contains the gap h - that is the sign you did it right.
\[ 2xh + h^{2} - 4h \]
Factor the gap out of the top, then divide
Why: Every surviving term has a factor of h, so it comes out cleanly and cancels the h in the denominator. Dividing is legal because the gap is never zero.
\[ \frac{h(2x + h - 4)}{h} = 2x + h - 4 \]
Verify with real numbers
Why: Take the input 3 and a gap of 1. The original outputs are f of 4, which is sixteen minus sixteen, or zero, and f of 3, which is nine minus twelve, or negative three. The quotient is zero minus negative three, over one, which is 3. The simplified formula gives six plus one minus four, also 3.
\[ \frac{f(4)-f(3)}{1} = \frac{0-(-3)}{1} = 3 \qquad 2(3)+1-4 = 3 \quad \checkmark \]
Picture it
Animation
Shows: The difference quotient, simplified — a rendered Manim animation.
Rendered with Manim.
Takeaway: The h cancels, which is what makes the calculus limit possible.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The original function is being subtracted as a whole. Writing it without parentheses quietly subtracts only its first term.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The minus sign gets applied to the x squared and then the second term is copied down unchanged - with its original sign.
Wrap the whole original function in parentheses before you subtract, every single time.
Why: The minus sign gets applied to the x squared and then the second term is copied down unchanged - with its original sign.
Trap
The original function is being subtracted as a whole. Writing it without parentheses quietly subtracts only its first term.
\[ f(x) = x^{2} - 4x \]
Write the numerator without parentheses
Why: The minus sign gets applied to the x squared and then the second term is copied down unchanged - with its original sign.
\[ x^{2} + 2xh + h^{2} - 4x - 4h - x^{2} - 4x \]
Combine and divide
Why: The two negative four x terms add instead of cancelling, leaving a term with no factor of h. Dividing leaves a fraction that does not simplify away.
\[ \frac{2xh + h^{2} - 8x - 4h}{h} = 2x + h - \frac{8x}{h} - 4 \]
Notice the alarm bell
Why: A gap left in the denominator means the answer blows up as the gap shrinks. A correct difference quotient never does that - the h always cancels.
Wrap the whole original function in parentheses before you subtract, every single time.
\[ f(x) = x^{2} - 4x \]
Write the numerator with parentheses
Why: Now the minus sign is attached to a package, and it will flip the sign of everything in that package when you distribute.
\[ \bigl(x^{2} + 2xh + h^{2} - 4x - 4h\bigr) - \bigl(x^{2} - 4x\bigr) \]
Distribute the minus and divide
Why: The negative four x becomes a positive four x and cancels its twin. Every surviving term carries an h, so the denominator cancels completely.
\[ \frac{2xh + h^{2} - 4h}{h} = 2x + h - 4 \]
Use the cancellation as your check
Why: If the gap does not cancel out of the bottom, stop and hunt for a dropped sign. That single habit catches this error before it reaches the answer line.
Translation
\( f(x) = x^{2} - 4x \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Ranking
Put in order
Put the moves of Difference quotient of a reciprocal into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The whole shifted input drops into the single slot in the denominator.
Worked example
Same three moves, but the subtraction now needs a common denominator.
\[ f(x) = \frac{1}{x} \]
Evaluate at the shifted input
Why: The whole shifted input drops into the single slot in the denominator.
\[ f(x+h) = \frac{1}{x+h} \]
Write the numerator of the quotient as one subtraction
Why: These are two unlike fractions, so nothing can be simplified until they share a denominator.
\[ \frac{1}{x+h} - \frac{1}{x} \]
Build both fractions up to the common denominator
Why: The least common denominator is the product of the two denominators, since neither one is a factor of the other.
\[ \frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} \]
Subtract the numerators, keeping the second one in parentheses
Why: The minus applies to the whole second numerator. The x terms cancel and the negative gap survives.
\[ \frac{x - (x+h)}{x(x+h)} = \frac{-h}{x(x+h)} \]
Divide by the gap by multiplying by its reciprocal
Why: Dividing by h is multiplying by one over h, and the h on top cancels it - exactly the cancellation you expect.
\[ \frac{-h}{x(x+h)} \cdot \frac{1}{h} = \frac{-1}{x(x+h)} \]
Verify with real numbers
Why: Take the input 2 and a gap of 1. The outputs are one third and one half, and one third minus one half is negative one sixth; divided by 1 that is negative one sixth. The formula gives negative one over two times three, which is also negative one sixth.
\[ \frac{\frac{1}{3}-\frac{1}{2}}{1} = -\frac{1}{6} \qquad \frac{-1}{2(3)} = -\frac{1}{6} \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Difference quotient of a reciprocal", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Take the input 2 and a gap of 1. The outputs are one third and one half, and one third minus one half is negative one sixth; divided by 1 that is negative one sixth. The formula gives negative one over two times three, which is also negative one sixth.
Constraint
Discussion prompt
Run Pattern: the four-step difference quotient with this step confiscated:
Check that every surviving term carries the gap. If one does not, a sign was dropped in step two.
Is it still possible? If it is, say what takes its place and what it costs you. If it is not, say exactly what that step was providing that nothing else does.
Hint: A step you can drop for free was never load-bearing. If you cannot drop it, name the thing that goes wrong the moment it is gone.
Answer:
Pattern
Then finish the way you finish everything in this deck: pick a number for the input and a number for the gap, and confirm the original and the simplified form agree.
\[ \frac{f(x+h)-f(x)}{h} \ \longrightarrow \ \text{no } h \text{ left in a denominator} \]
Edge cases
Discussion prompt
Pattern: the four-step difference quotient works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".
Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.
Answer:
Then finish the way you finish everything in this deck: pick a number for the input and a number for the gap, and confirm the original and the simplified form agree.
Prediction
Predict first
Simplify the difference quotient for this function.
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: two x plus h plus five
Why: The shifted evaluation is x squared plus two x h plus h squared plus five x plus five h. Subtracting the original leaves two x h plus h squared plus five h, and factoring the gap out gives h times the quantity two x plus h plus five, so the gap cancels. Test with the input 1 and a gap of 2: f of 3 is 24 and f of 1 is 6, so the quotient is 18 divided by 2, which is 9 - and the formula gives 2 plus 2 plus 5, also 9.
Check
Expand first, subtract in parentheses, then cancel.
\[ f(x) = x^{2} + 5x \]
Check your understanding
Simplify the difference quotient for this function.
Answer: A
Why: The shifted evaluation is x squared plus two x h plus h squared plus five x plus five h. Subtracting the original leaves two x h plus h squared plus five h, and factoring the gap out gives h times the quantity two x plus h plus five, so the gap cancels. Test with the input 1 and a gap of 2: f of 3 is 24 and f of 1 is 6, so the quotient is 18 divided by 2, which is 9 - and the formula gives 2 plus 2 plus 5, also 9.
Section
Part 4
Concept
Every function comes with two sets attached to it: what it will accept, and what it can produce.
domain — The set of all legal inputs - every value you are allowed to feed the function.
range — The set of all outputs the function actually produces as the input runs over the whole domain.
Two words worth burning in: domain is input, range is output. Alphabetical order even helps - d comes before r, and input comes before output.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of relation, function, vertical line test, function notation, domain as Functions, Domain, and Function Notation uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Intuition
Back to the vending machine. The domain is the list of buttons that actually exist - press anything else and nothing happens.
The range is the collection of snacks that can land in the tray. If the machine has never been stocked with gum, gum is not in the range no matter how hard you press.
So a domain question is 'what can I put in?' and a range question is 'what can come out?'. Ask yourself which one the problem is asking before you write anything down.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The squaring function is the classic place this goes wrong, because one of the two answers really is restricted - just not the one students pick.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The restriction is real, but it lives on the output side.
Ask the two questions in order: what may I put in, then what comes out.
Why: The restriction is real, but it lives on the output side. Pinning it to the inputs is the swap.
Trap
The squaring function is the classic place this goes wrong, because one of the two answers really is restricted - just not the one students pick.
\[ f(x) = x^{2} \]
Notice that squaring never produces a negative, then attach that fact to the domain
Why: The restriction is real, but it lives on the output side. Pinning it to the inputs is the swap.
\[ \text{domain} = [0, \infty) \qquad \text{(wrong)} \]
Report every real number as the range
Why: The two sets have been traded. This answer even claims the function outputs negative numbers, which it never does.
\[ \text{range} = (-\infty, \infty) \qquad \text{(wrong)} \]
Ask the two questions in order: what may I put in, then what comes out.
\[ f(x) = x^{2} \]
Ask what inputs are legal
Why: You can square anything at all - negatives, zero, fractions, huge numbers. Nothing is forbidden, so the domain is all real numbers.
\[ \text{domain} = (-\infty, \infty) \]
Ask what outputs actually appear
Why: A square is never negative, and every non-negative number is the square of its own square root, so zero and up is exactly what shows up in the tray.
\[ \text{range} = [0, \infty) \]
Sanity-check against the picture
Why: The parabola stretches left and right forever - that is the domain being all reals - but never dips below the horizontal axis, which is the range starting at zero.
Notation
Annotate
From Trap: swapping domain and range — read this one piece at a time. What is each part doing?
On: \( \text{range} = [0, \infty) \)
Concept
When the function is a finite list, both sets are finite too. Collect the first coordinates for one and the second coordinates for the other.
\[ \{(-1,\,6),\ (2,\,0),\ (5,\,6),\ (7,\,3)\} \]
\[ \text{domain} = \{-1,\, 2,\, 5,\, 7\} \qquad \text{range} = \{0,\, 3,\, 6\} \]
Two details: the range is written without repeating the 6 even though it appeared twice, and both sets are conventionally listed in increasing order.
Finite lists get set braces, not intervals. Intervals are for unbroken stretches of the number line, which a list of four points is not.
Concept
For a graph, imagine the curve casting a shadow. Shine a light from directly above and below: the shadow on the horizontal axis is the domain.
Now shine the light from the left and right: the shadow on the vertical axis is the range.
| To find | Squash the graph onto | Read |
|---|---|---|
| Domain | the horizontal axis | leftmost input to rightmost input |
| Range | the vertical axis | lowest output to highest output |
Endpoints matter: a solid dot includes that value and gets a square bracket; an open circle excludes it and gets a parenthesis; an arrow means the graph runs forever and gets an infinity symbol.
Comparison
Comparison matrix
From Domain and range from a graph: refill the Read column from what you know. The rest of the table is as it appeared.
| To find | Squash the graph onto | Read |
|---|---|---|
| Domain | the horizontal axis | leftmost input to rightmost input |
| Range | the vertical axis | lowest output to highest output |
Fill the middle
Fill in the blanks
From Domain and range from a graph — finish the line. Write what belongs on the right of the equals sign before you look.
\text[-3,\, 4] = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The shadow starts under the left endpoint and ends under the right endpoint, with no gaps in between - the graph is one connected piece.
Worked example
Figure (svg): A V-shaped graph running from a solid dot at negative three comma four down to a corner at one comma negative two and back up to a solid dot at four comma three
State the domain and range of the graph shown. Both endpoints are solid dots, so the graph stops there.
Squash the graph down onto the horizontal axis
Why: The shadow starts under the left endpoint and ends under the right endpoint, with no gaps in between - the graph is one connected piece.
\[ \text{leftmost input } -3, \quad \text{rightmost input } 4 \]
Write the domain with brackets on both ends
Why: Both endpoints are solid dots, so both values are included and both get square brackets.
\[ \text{domain} = [-3,\, 4] \]
Now squash the graph sideways onto the vertical axis
Why: The lowest point of the graph is the corner and the highest is the left endpoint - not the right endpoint, which only climbs back to three.
\[ \text{lowest output } -2, \quad \text{highest output } 4 \]
Write the range with brackets on both ends
Why: Both extremes are actually reached by points on the graph, so both are included.
\[ \text{range} = [-2,\, 4] \]
Verify by testing a value in and a value out of each set
Why: The input 0 sits inside the horizontal shadow and the graph does have a point there, so 0 belongs to the domain; the input 6 is past the right endpoint and the graph has nothing there, so it does not. On the output side, the height 3 is hit twice by the graph, while the height 5 is above every point on it - so 3 is in the range and 5 is not.
Picture it
Animation
Shows: Each line of the worked example "Domain and range from a graph", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The input 0 sits inside the horizontal shadow and the graph does have a point there, so 0 belongs to the domain; the input 6 is past the right endpoint and the graph has nothing there, so it does not. On the output side, the height 3 is hit twice by the graph, while the height 5 is above every point on it - so 3 is in the range and 5 is not.
Concept
Interval notation is how every domain answer in this course is written. It has exactly three moving parts.
| Symbol | Means | Use it when |
|---|---|---|
| square bracket | the endpoint is included | the inequality allows equality, or the dot is solid |
| parenthesis | the endpoint is excluded | the inequality is strict, or the dot is open |
| infinity, always with a parenthesis | no endpoint on that side | the set runs forever; infinity is a direction, not a number you can include |
\[ x \ge 2 \ \Rightarrow\ [2, \infty) \qquad x < 5 \ \Rightarrow\ (-\infty, 5) \]
When a set comes in disconnected pieces, glue them with a union symbol - one interval per unbroken stretch.
\[ (-\infty,\, 1) \cup (1,\, \infty) \]
Comparison
Comparison matrix
From Interval notation, in one slide: refill the Means column from what you know. The rest of the table is as it appeared.
| Symbol | Means | Use it when |
|---|---|---|
| square bracket | the endpoint is included | the inequality allows equality, or the dot is solid |
| parenthesis | the endpoint is excluded | the inequality is strict, or the dot is open |
| infinity, always with a parenthesis | no endpoint on that side | the set runs forever; infinity is a direction, not a number you can include |
Concept
When a function arrives as a formula with no picture, the domain is 'everything, except what would break the arithmetic'. Only three things break it.
If a formula has none of those three features - just adding, subtracting, multiplying, odd roots, or whole-number powers - the domain is all real numbers.
\[ (-\infty,\, \infty) \]
Matching
Match the pairs
From The three domain rules — match each one to what it actually does. The descriptions have been shuffled.
Why: No dividing by zero, No even root of a negative, No log of zero or a negative are easy to tell apart while they are sitting next to their descriptions and much harder afterwards, which is what this checks.
Step zero
Discussion prompt
Domain of a rational function — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Look only at the denominator
Answer:
Worked example
Find the domain, in interval notation.
\[ f(x) = \frac{x+3}{x^{2} - x - 6} \]
Look only at the denominator
Why: The numerator can be anything at all, including zero. Only the bottom of a fraction can break the arithmetic.
Factor the denominator
Why: Two numbers multiplying to negative six and adding to negative one are negative three and positive two.
\[ x^{2} - x - 6 = (x-3)(x+2) \]
Set each factor equal to zero and solve
Why: A product is zero exactly when one of its factors is zero, so these two inputs are the only ones that kill the denominator.
\[ x = 3 \quad \text{or} \quad x = -2 \]
Remove those two points from the number line and write the pieces left over
Why: Punching two holes in the line leaves three unbroken stretches, and each one becomes its own interval joined by unions.
\[ (-\infty,\, -2) \cup (-2,\, 3) \cup (3,\, \infty) \]
Verify the excluded values really break it and a nearby value does not
Why: At the input negative two the denominator is four plus two minus six, which is zero - undefined, correctly excluded. At the input 3 it is nine minus three minus six, also zero - correctly excluded. At the input 0 it is negative six, so the function is fine there, and 0 does sit inside the middle interval.
\[ f(0) = \frac{3}{-6} = -\frac{1}{2} \quad \checkmark \]
Picture it
Animation
Shows: Where a rational function is undefined — a rendered Manim animation.
Rendered with Manim.
Takeaway: The denominator's zeros are exactly the excluded inputs.
Fill the middle
Fill in the blanks
From Domain of a radical function — finish the line. Write what belongs on the right of the equals sign before you look.
g(x) = \sqrt{7 - 2x}
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. A square root of a negative is not a real number, but a square root of zero is perfectly fine - so the inequality is non-strict.
Worked example
Find the domain, in interval notation.
\[ g(x) = \sqrt{7 - 2x} \]
Set the expression under the square root greater than or equal to zero
Why: A square root of a negative is not a real number, but a square root of zero is perfectly fine - so the inequality is non-strict.
\[ 7 - 2x \ge 0 \]
Subtract seven from both sides
Why: Standard inequality move; adding or subtracting never changes the direction of the symbol.
\[ -2x \ge -7 \]
Divide both sides by negative two and flip the symbol
Why: Dividing an inequality by a negative number reverses the direction. This is the step that is skipped most often on this problem.
\[ x \le \frac{7}{2} \]
Translate to interval notation
Why: Everything from negative infinity up to and including seven halves. Infinity always takes a parenthesis; the seven halves takes a bracket because equality was allowed.
\[ \left(-\infty,\ \tfrac{7}{2}\right] \]
Verify with one input inside, one at the edge, and one outside
Why: At the input 0 the radicand is 7, giving a real output. At seven halves the radicand is exactly zero, giving the output zero, so the endpoint truly belongs. At the input 4 the radicand is seven minus eight, or negative one, which has no real square root - correctly excluded.
\[ g(0)=\sqrt{7},\quad g\!\left(\tfrac{7}{2}\right)=0,\quad g(4)=\sqrt{-1}\ \text{undefined} \quad \checkmark \]
Picture it
Animation
Shows: A square root needs a non-negative inside — a rendered Manim animation.
Rendered with Manim.
Takeaway: The graph simply does not exist to the left of the restriction.
Picture it
Figure (svg): A number line with open circles at negative two and three, splitting it into three shaded stretches
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
The question said 'in interval notation'. The work was right and the answer still loses points.
Trap
The question said 'in interval notation'. The work was right and the answer still loses points.
\[ f(x) = \frac{x+3}{(x-3)(x+2)} \]
Stop at the list of forbidden inputs
Why: This describes the domain correctly in words, but it is not interval notation - it is the answer to a different question.
\[ x \ne 3, \quad x \ne -2 \qquad \text{(not interval notation)} \]
Or convert carelessly and lose the middle piece
Why: Two holes in the line leave three stretches, not two. Writing only the outer two silently throws away every input between negative two and three.
\[ (-\infty, -2) \cup (3, \infty) \qquad \text{(wrong)} \]
Draw the number line, mark the holes, and count the pieces you have left.
\[ f(x) = \frac{x+3}{(x-3)(x+2)} \]
Mark the two excluded inputs on a number line
Why: Two marks cut the line into three unbroken stretches. Counting the pieces before writing anything prevents the missing-middle error.
Figure (svg): A number line with open circles at negative two and three, splitting it into three shaded stretches
Write one interval per piece, joined by unions
Why: Three pieces means three intervals. Every endpoint is a parenthesis because both marked values are excluded.
\[ (-\infty, -2) \cup (-2, 3) \cup (3, \infty) \]
Spot-check the middle interval
Why: The input 0 lies between the two holes, and the function is perfectly defined there - proof that the middle piece belongs in the answer.
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Ranking
Put in order
These are the steps of Pattern: domain straight from a formula, scrambled. Put them back in order before the next slide shows you.
Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.
Pattern
Then verify: test one input from each surviving piece and one from each hole. The good ones should evaluate; the bad ones should break.
Picture it
Animation
Shows: Two things restrict a domain — a rendered Manim animation.
Rendered with Manim.
Takeaway: Scan the formula for those two, and you have the domain.
Prediction
Predict first
What is the domain of this function in interval notation?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: the interval from negative four included to one excluded, union one excluded to infinity
Why: The radical needs x plus four greater than or equal to zero, so the input must be at least negative four, and the endpoint is included because the square root of zero is defined. The denominator is zero at the input 1, so that single point is punched out. Together that gives negative four included up to 1 excluded, union 1 excluded to infinity.
Check
Two troublemakers in one formula. Handle each, then combine.
\[ f(x) = \frac{\sqrt{x+4}}{x-1} \]
Check your understanding
What is the domain of this function in interval notation?
Answer: A
Why: The radical needs x plus four greater than or equal to zero, so the input must be at least negative four, and the endpoint is included because the square root of zero is defined. The denominator is zero at the input 1, so that single point is punched out. Together that gives negative four included up to 1 excluded, union 1 excluded to infinity.
Section
Part 5
Concept
Some machines behave differently depending on what you feed them. A phone plan charges one rate up to a data cap and a different rate after it - one function, two rules.
piecewise-defined function — A single function given by different formulas on different parts of its domain. Each formula comes with a condition saying which inputs it applies to.
\[ f(x) = \begin{cases} 2x + 1, & x < -1 \\ x^{2}, & -1 \le x < 3 \\ 5, & x \ge 3 \end{cases} \]
The conditions never overlap and together they cover everything. That is not decoration - it is exactly what keeps this a function: every input matches one rule, so it gets one output.
Explain it
Discussion prompt
Explain One function, several rules to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Some machines behave differently depending on what you feed them. A phone plan charges one rate up to a data cap and a different rate after it - one function, two rules.
Missing information
Discussion prompt
Find four values of this function. The whole job is choosing the right rule first.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Negative four is less than negative one, so the first rule applies and the others do not.
Worked example
Find four values of this function. The whole job is choosing the right rule first.
\[ f(x) = \begin{cases} 2x + 1, & x < -1 \\ x^{2}, & -1 \le x < 3 \\ 5, & x \ge 3 \end{cases} \]
For the input negative four, check the conditions top to bottom
Why: Negative four is less than negative one, so the first rule applies and the others do not.
\[ f(-4) = 2(-4) + 1 = -7 \]
For the input negative one, read the inequality symbols carefully
Why: The first condition is strict, so negative one is not less than negative one. The second condition allows equality, so negative one belongs to the squaring rule.
\[ f(-1) = (-1)^{2} = 1 \]
For the input 2, use the middle rule
Why: Two sits between negative one and three, so the squaring rule applies.
\[ f(2) = 2^{2} = 4 \]
For the input 3, notice the middle condition is strict on the right
Why: Three is not less than three, so the middle rule does not reach it. The third condition allows equality, and that rule ignores the input entirely.
\[ f(3) = 5 \]
Verify that each input matched exactly one condition
Why: Negative four satisfies only the first line. Negative one fails the strict first line and satisfies the second. Two satisfies only the second. Three fails the second and satisfies the third. Four inputs, four single matches - the definition holds and every value above is the only possible answer.
\[ f(-4)=-7,\quad f(-1)=1,\quad f(2)=4,\quad f(3)=5 \]
Worked example
Figure (svg): A rising line segment ending in an open circle at zero comma three, and a horizontal ray at height negative one starting from a solid dot at zero comma negative one
Graph this function. Two rules means two separate pieces drawn on the same axes.
\[ g(x) = \begin{cases} x + 3, & x < 0 \\ -1, & x \ge 0 \end{cases} \]
Graph the first rule, but only to the left of zero
Why: The line has slope one and a height of three at the boundary, but the condition stops it there - so draw the segment coming in from the left and cut it off at the boundary.
Put an open circle at the boundary point of the first piece
Why: The condition is strict, so the input zero does not belong to this rule. The open circle says 'the curve heads here but this point is not part of the graph'.
\[ \text{open circle at } (0,\, 3) \]
Graph the second rule to the right of zero
Why: The rule ignores the input and always returns negative one, so it draws a horizontal ray one unit below the axis.
Put a solid dot at the boundary point of the second piece
Why: This condition allows equality, so the input zero really does belong to this rule and its point is genuinely on the graph.
\[ \text{solid dot at } (0,\, -1) \]
Verify with the vertical line test at the boundary
Why: The vertical line at the input zero passes through the open circle, which is not a point of the graph, and through the solid dot, which is. That is exactly one point, so the graph is a function and the value is negative one. Checking a nearby input confirms the left piece: at negative two the first rule gives 1, and the segment does pass through the point negative two comma one.
\[ g(0) = -1, \qquad g(-2) = -2 + 3 = 1 \quad \checkmark \]
Picture it
Animation
Shows: A piecewise rule picks by input — a rendered Manim animation.
Rendered with Manim.
Takeaway: Choosing the wrong branch is the usual error.
Elimination
Eliminate the wrong options
What is the value of this function at the input 2?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The first condition allows equality, so the input 2 belongs to the first rule. Three minus two is 1. The second rule never gets used here because its condition is strict and 2 is not greater than 2.
Check
Read the inequality symbols before you compute anything.
\[ g(x) = \begin{cases} 3 - x, & x \le 2 \\ x^{2} - 4, & x > 2 \end{cases} \]
Check your understanding
What is the value of this function at the input 2?
Answer: A
Why: The first condition allows equality, so the input 2 belongs to the first rule. Three minus two is 1. The second rule never gets used here because its condition is strict and 2 is not greater than 2.
Section
Part 6
Concept
Read a graph the way you read a sentence: left to right. Ask what the height is doing as you walk that direction.
| Walking left to right, the graph... | The function is |
|---|---|
| climbs | increasing |
| falls | decreasing |
| stays level | constant |
The answer is always given as intervals of inputs, never of outputs - you are naming the stretch of the horizontal axis where the behavior happens.
By convention these intervals are written with parentheses, because at the exact turning point the graph is neither climbing nor falling.
Pattern
Step through it
Step through Increasing, decreasing, constant one row at a time. What is driving the change, and what would the row after the last one be?
Picture it
Animation
Shows: Reading increase and decrease — a rendered Manim animation.
Rendered with Manim.
Takeaway: Read left to right: rising, then falling, then rising again.
Concept
A turning point is where a climb becomes a fall, or a fall becomes a climb. Those points get names.
relative maximum — A point that is higher than everything immediately around it - the top of a local hill. It need not be the highest point on the whole graph.
relative minimum — A point that is lower than everything immediately around it - the bottom of a local valley.
Watch the wording, because tests punish it: the maximum value is the height, an output. The place where it occurs is the input. Two different numbers, and the question always wants one specific one.
Analogy
Discussion prompt
Explain Relative maxima and minima by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
A turning point is where a climb becomes a fall, or a fall becomes a climb. Those points get names.
Estimation
Predict first
The graph is shown for inputs from negative three to four. Name the increasing and decreasing intervals and both relative extrema.
Commit before you compute: what does Reading behavior and extrema off a graph come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify each interval by comparing two heights inside it
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. On the first interval, the height at negative two is below the height at negative one point five, confirming a climb.
Worked example
Figure (svg): A curve rising to a peak at negative one comma four, falling to a valley at two comma negative three, then rising again
The graph is shown for inputs from negative three to four. Name the increasing and decreasing intervals and both relative extrema.
Walk from the left edge to the peak
Why: Over this stretch the height keeps climbing, from about negative two up to four, so the function is increasing there.
\[ \text{increasing on } (-3,\, -1) \]
Keep walking from the peak down to the valley
Why: The height falls the whole way, from four down to negative three, so the function is decreasing on that stretch.
\[ \text{decreasing on } (-1,\, 2) \]
Finish from the valley to the right edge
Why: The height climbs again all the way to the end of the picture.
\[ \text{increasing on } (2,\, 4) \]
Name the relative maximum, value first
Why: The peak is higher than everything nearby. Its height is the maximum value; the input underneath it is where that value occurs.
\[ \text{relative maximum value } 4 \text{ at } x = -1 \]
Name the relative minimum the same way
Why: The valley is lower than everything nearby, so the minimum value is its height and the input underneath it is where it happens.
\[ \text{relative minimum value } -3 \text{ at } x = 2 \]
Verify each interval by comparing two heights inside it
Why: On the first interval, the height at negative two is below the height at negative one point five, confirming a climb. On the middle interval, the height at zero is above the height at one, confirming a fall. On the last, the height at three is above the height at two point five, confirming a climb. And the peak sits above its neighbors on both sides while the valley sits below its neighbors on both sides, which is exactly the definition of the two extrema.
Concept
Increasing and decreasing say whether the output moves. Average rate of change says how fast, on average, over a stretch.
\[ \frac{f(b) - f(a)}{b - a} \]
That is the same rise-over-run you built in the difference quotient, with the two endpoints named instead of a starting point and a gap. It is the slope of the straight line joining the two points on the curve.
In context it always carries units: output units per input unit. Dollars per shirt, centimetres per week, miles per hour. Say the units out loud and the answer usually interprets itself.
Picture it
Animation
Shows: Average rate of change is a slope — a rendered Manim animation.
Rendered with Manim.
Takeaway: The idea calculus later takes to a limit.
Step zero
Discussion prompt
Average rate of change in context — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Evaluate at the earlier endpoint
Answer:
Worked example
A seedling's height in centimetres after a number of weeks is modelled by this function. Find the average rate of change from week one to week four and say what it means.
\[ H(t) = t^{2} - 2t + 10 \]
Evaluate at the earlier endpoint
Why: One squared is one, minus two, plus ten. The plant is nine centimetres tall at week one.
\[ H(1) = 1 - 2 + 10 = 9 \]
Evaluate at the later endpoint
Why: Four squared is sixteen, minus eight, plus ten. The plant is eighteen centimetres tall at week four.
\[ H(4) = 16 - 8 + 10 = 18 \]
Divide the change in height by the change in time
Why: Nine centimetres of growth spread over three weeks. Put the outputs on top and the inputs on the bottom - never the other way around.
\[ \frac{18 - 9}{4 - 1} = \frac{9}{3} = 3 \]
State the answer with units and a sentence
Why: The number alone is not the answer to a modelling question. Between week one and week four the seedling grew at an average of three centimetres per week.
\[ 3 \ \text{cm per week} \]
Verify against the difference quotient
Why: For this rule the difference quotient simplifies to two t plus the gap minus two. Using a starting input of one and a gap of three gives two plus three minus two, which is 3 - matching the answer exactly, since average rate of change is just the difference quotient with the gap written as an endpoint.
\[ 2(1) + 3 - 2 = 3 \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Average rate of change in context", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: For this rule the difference quotient simplifies to two t plus the gap minus two. Using a starting input of one and a gap of three gives two plus three minus two, which is 3 - matching the answer exactly, since average rate of change is just the difference quotient with the gap written as an endpoint.
Concept
One last classification, and it is the payoff for all that practice evaluating at the opposite of the input.
even function — A function whose value at the opposite of an input equals its value at the input - feeding in the negative changes nothing.
odd function — A function whose value at the opposite of an input is the negative of its value at the input - feeding in the negative flips the sign of the output.
\[ \text{even: } f(-x) = f(x) \qquad \text{odd: } f(-x) = -f(x) \]
Most functions are neither. That is a legitimate answer and often the right one - do not force a function into a box.
Counterexample
Discussion prompt
One last classification, and it is the payoff for all that practice evaluating at the opposite of the input.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
Most functions are neither. That is a legitimate answer and often the right one - do not force a function into a box.
Intuition
Figure (svg): Left panel: a parabola symmetric about the vertical axis, labelled even. Right panel: an S-shaped curve symmetric about the origin, labelled odd.
Even means the vertical axis is a mirror. Fold the paper along it and the two halves land on each other exactly.
Odd means the origin is a pivot. Spin the whole picture half a turn about the origin and it lands back on itself.
That is why the names feel arbitrary until you see where they come from: a power of the input with an even exponent gives the mirror, and a power with an odd exponent gives the spin.
Ranking
Put in order
Put the moves of Even, odd, or neither into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Both exponents are even, so both negatives are destroyed by the squaring and the fourth power.
Worked example
Classify each of these three functions. The move is always the same: evaluate at the opposite of the input and compare.
\[ f(x) = 3x^{4} - x^{2} + 2 \]
Substitute the negated variable into every copy
Why: Both exponents are even, so both negatives are destroyed by the squaring and the fourth power.
\[ f(-x) = 3(-x)^{4} - (-x)^{2} + 2 = 3x^{4} - x^{2} + 2 \]
Compare with the original: identical, so it is even
Why: Nothing changed at all, which is precisely the even condition.
\[ f(-x) = f(x) \ \Rightarrow\ \text{even} \]
\[ g(x) = x^{3} - 4x \]
Substitute and simplify
Why: An odd power keeps the negative, and the linear term flips too, so every term changed sign.
\[ g(-x) = (-x)^{3} - 4(-x) = -x^{3} + 4x \]
Factor a negative one out and compare
Why: Pulling out the negative reveals the original expression inside, which is exactly the odd condition.
\[ -x^{3} + 4x = -\bigl(x^{3} - 4x\bigr) = -g(x) \ \Rightarrow\ \text{odd} \]
\[ k(x) = x^{2} + x \]
Substitute and compare with both conditions
Why: One term kept its sign and the other flipped, so the result matches neither the original nor its negative.
\[ k(-x) = x^{2} - x \ \ne \ k(x), \qquad k(-x) \ne -k(x) \]
Verify all three with a test number
Why: At the input 1 and its opposite: the first function gives 4 both times, matching even. The second gives negative three and positive three, opposites, matching odd. The third gives 2 and 0, which are neither equal nor opposite - so neither is the correct classification.
\[ f(1)=f(-1)=4 \quad\checkmark \qquad g(1)=-3,\ g(-1)=3 \quad\checkmark \qquad k(1)=2,\ k(-1)=0 \quad\checkmark \]
Picture it
Animation
Shows: An even function is mirror-symmetric — a rendered Manim animation.
Rendered with Manim.
Takeaway: f(-x) = f(x) means the y-axis is a mirror.
Pattern
Given any function in any costume, these are the questions this chapter can ask - and the one move that answers each.
| Question | The move |
|---|---|
| Is it a function? | Look for one input with two outputs; on a graph, sweep a vertical line. |
| Find the value at something | Wrap it in parentheses and replace every copy of the variable. |
| Solve for the input giving a value | Set the rule equal to that value and solve, or read across from the vertical axis. |
| Build the difference quotient | Expand the shifted evaluation, subtract the original in parentheses, cancel the gap. |
| Question | The move |
|---|---|
| Find the domain | Exclude denominator zeros and negative radicands; write intervals joined by unions. |
| Find the range | Squash the graph onto the vertical axis, or reason about what the rule can output. |
| Increasing or decreasing where? | Walk left to right and name intervals of inputs. |
| Even, odd, or neither? | Evaluate at the negated variable and compare with the original and its negative. |
And the habit underneath all of them: verify with a test number. Pick an input, run it through the original and through your answer, and make them agree.
Real world
Discussion prompt
Outside this lesson: where does Functions, Domain, and Function Notation actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Pattern: the full function checklist is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
The single most important idea in College Algebra: a function as a machine with exactly one output per input. Relations versus functions, mapping diagrams and the vertical line test, function notation and evaluating at numbers and expressions, the difference quotient, domain and range in interval notation, piecewise functions, increasing and decreasing intervals, relative extrema, average rate of change, and even versus odd.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Relations and Functions · Function Notation · The Difference Quotient · Domain and Range · Piecewise Functions · Reading Behavior off a Graph. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
A function is a machine with one promise: one input, exactly one output. Every rule in this deck is that promise wearing a different outfit.
| The four traps you now know | The fix |
|---|---|
| Reading the notation as multiplication | Simplify inside the slot, then run the rule once. |
| Substituting into only one copy | Count the copies before you substitute. |
| Answering a domain with excluded values | Mark the holes on a number line and write one interval per surviving piece. |
| Swapping domain and range | Domain is what goes in; range is what comes out. |
Next up: transformations, where you take these same functions and slide, stretch, and flip their graphs on purpose.
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