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Mathematical Architects · Pre-Calculus · Foundations

Get Ready: Module 1 — Functions, Composition & Inverses

Everyone starts somewhere. Before we plug functions into each other and run them in reverse, we'll firm up the handful of skills Module 1 leans on — clearly, one step at a time, with nothing assumed. Go at your own pace. There's no rush and no wrong place to begin.

4 Foundation Skills Worked Examples Self-Paced

Module 1 takes the functions you already know and teaches them to combine (composition) and reverse (inverses). If \(f(x)\) notation, domain and range, or reflecting a point feel fuzzy right now, that's completely okay — we'll firm them up together below, then you'll be ready for the lab.

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Skills to build first

Four short skills. Read the idea, follow the worked example, then try one yourself — the answer is one click away whenever you want it.

1 Skill

Function notation \(f(x)\) & evaluation

What it is: \(f(x)\) names a function \(f\) and its input \(x\). To evaluate, you replace every \(x\) with the given input and simplify. Why you need it: composition is nothing but evaluation done twice — \((f\circ g)(x)\) means "evaluate \(g\) at \(x\), then evaluate \(f\) at that answer." If you can plug a number into a function, you can compose.

Worked example — if \(f(x) = 3x - 4\), find \(f(2)\) and \(f(a+1)\)
  1. Replace every \(x\) with \(2\): \(f(2) = 3(2) - 4 = 6 - 4 = 2\).
  2. For \(f(a+1)\), the whole input \(a+1\) replaces \(x\): \(f(a+1) = 3(a+1) - 4\).
  3. Distribute and simplify: \(3(a+1) - 4 = 3a + 3 - 4 = 3a - 1\).
  4. So \(f(a+1) = 3a - 1\). The input can be a number or a whole expression — exactly what composition needs.
Try it 1 — If \(g(x) = x^2 + 1\), find \(g(3)\) and \(g(-2)\).
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\(g(3) = 3^2 + 1 = 9 + 1 = 10\). \(g(-2) = (-2)^2 + 1 = 4 + 1 = 5\) (squaring kills the sign).

Try it 2 — If \(f(x) = 2x + 5\), find \(f(g)\) where the input is \(g\). (This is the composition idea!)
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Replace \(x\) with \(g\): \(f(g) = 2g + 5\). If later \(g = x - 1\), then \(f(g) = 2(x-1) + 5 = 2x + 3\).

Practice more (free) Khan Academy → IXL →
2 Skill

Domain & range

What it is: the domain is every input \(x\) a function is allowed to take; the range is every output \(y\) it can produce. Why you need it: when you build an inverse, the domain and range swap. And some functions (like \(y=x^2\)) only get an inverse after you restrict the domain — so you have to know what the domain is in the first place.

Worked example — domain & range of \(f(x) = \sqrt{x}\)
  1. Domain: you can't take the square root of a negative, so inputs must satisfy \(x \ge 0\). Domain \(= [0, \infty)\).
  2. Range: a square root is never negative, so outputs satisfy \(y \ge 0\). Range \(= [0, \infty)\).
  3. Bracket vs. parenthesis: \(0\) is reachable (\(\sqrt0 = 0\)) so it gets a bracket \([\,\); \(\infty\) is a direction, never a number, so it always gets a parenthesis \(\,)\).

Quick reminder: \(x \ge 0\) (inequality) \(=\) \([0,\infty)\) (interval) \(=\) \(\{x \mid x \ge 0\}\) (set-builder) — three ways to say the same set.

Try it 1 — What is the domain of \(f(x) = \tfrac{1}{x}\)?
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You can divide by anything except \(0\), so the domain is "all real numbers except \(0\)," i.e. \((-\infty, 0) \cup (0, \infty)\).

Try it 2 — What is the range of \(f(x) = x^2\)?
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Squaring never gives a negative, and \(0\) is reachable, so the range is \(y \ge 0\), i.e. \([0, \infty)\).

Practice more (free) Khan Academy → IXL →
3 Skill

Reflecting a point over the line \(y = x\)

What it is: reflecting the point \((a, b)\) over the diagonal line \(y = x\) simply swaps the coordinates to \((b, a)\). Why you need it: this single move is how an inverse graph is built. The graph of \(f^{-1}\) is the graph of \(f\) reflected over \(y=x\) — so if you can swap coordinates, you can draw any inverse.

Worked example — reflect \((3, 1)\) over \(y = x\)
  1. Swap the coordinates. \((3, 1)\) becomes \((1, 3)\). That's it — the \(x\) and \(y\) trade places.
  2. Why it works: the line \(y=x\) is the set of points where input equals output. Reflecting across it turns "input \(3\), output \(1\)" into "input \(1\), output \(3\)" — exactly what undoing a function does.
  3. Check a point on the line: \((2,2)\) reflects to \((2,2)\) — points already on \(y=x\) don't move.
Try it 1 — Reflect \((-4, 5)\) over \(y = x\).
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Swap the coordinates: \((-4, 5) \to (5, -4)\).

Try it 2 — The point \((0, 7)\) is on a graph. Where does it land on the inverse's graph?
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Reflect over \(y=x\) by swapping: \((0, 7) \to (7, 0)\). (Notice a \(y\)-intercept of \(f\) becomes an \(x\)-intercept of \(f^{-1}\).)

Practice more (free) Khan Academy → IXL →
4 Skill

The parent-function shapes

What it is: the basic "starter" graphs every family is built from — the line \(y=x\), the parabola \(y=x^2\), the cubic \(y=x^3\), the square root \(y=\sqrt{x}\), the reciprocal \(y=\tfrac1x\), and the V of \(y=|x|\). Why you need it: in the lab you'll pick these by name and watch them compose and reflect. Knowing each shape at a glance — and whether it's one-to-one — makes the whole module faster.

Worked example — which parents are one-to-one?
  1. \(y = x\) and \(y = x^3\): each output comes from exactly one input — they pass the horizontal line test, so they're one-to-one and invertible as-is.
  2. \(y = x^2\): both \(x=2\) and \(x=-2\) give \(4\), so it fails the horizontal line test. It's not one-to-one until you restrict to \(x \ge 0\).
  3. \(y = \sqrt{x}\): defined only for \(x \ge 0\); it's one-to-one, and it's exactly the inverse of \(y=x^2\) on \(x\ge 0\).
  4. Symmetry shortcut: \(y=x^2\) is even (mirror over the \(y\)-axis); \(y=x^3\) is odd (180° turn about the origin).
Try it 1 — Sketch (in your head) \(y = x^2\) and \(y = \sqrt{x}\). How are they related?
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For \(x \ge 0\) they are reflections of each other over \(y=x\) — \(\sqrt{x}\) is the inverse of \(x^2\) once the domain is restricted to \(x\ge 0\).

Try it 2 — Is \(y = |x|\) one-to-one? Why does it matter for inverses?
Show answer

No — \(|3| = |-3| = 3\), so it fails the horizontal line test. Without a domain restriction it has no inverse function (the reflection would fail the vertical line test).

Practice more (free) Khan Academy → IXL →
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Quick Readiness Check

Six short questions spanning all four skills. Try each one first, then reveal the answer to check yourself.

  1. If \(f(x) = 4x - 3\), find \(f(5)\).
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    \(f(5) = 4(5) - 3 = 20 - 3 = 17\).

  2. If \(g(x) = x^2 - 2\), find \(g(-3)\).
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    \(g(-3) = (-3)^2 - 2 = 9 - 2 = 7\).

  3. What is the domain of \(f(x) = \sqrt{x}\)?
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    \(x \ge 0\), i.e. \([0, \infty)\) — no negative inputs under a square root.

  4. Reflect the point \((6, 2)\) over the line \(y = x\).
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    Swap the coordinates: \((6, 2) \to (2, 6)\).

  5. Is \(y = x^3\) one-to-one (does it pass the horizontal line test)?
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    Yes — each output comes from exactly one input, so it has an inverse, \(\sqrt[3]{x}\), with no restriction needed.

  6. Is \(f(x) = x^2\) even, odd, or neither?
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    Even: \(f(-x) = (-x)^2 = x^2 = f(x)\), so it's symmetric over the \(y\)-axis.

If these feel comfortable, you're ready for the module. If one or two felt shaky, that's your cue — scroll back up to that skill, re-read the worked example, and try a couple from the free practice links. No score to hit, no clock running. Come back when it clicks.

Floor poured. When the skills above feel steady, step into the module — the Visual Lab lets you see composition and inverses move at once.

Module 1 Overview & Visual Lab Jump to the Visual Lab Student Support