A vending machine takes a button code and gives you a snack. Press B4 and you get crisps. Press B4 again tomorrow and you get crisps again. If B4 sometimes gave crisps and sometimes a chocolate bar, you would call the machine broken.

That reliability (same input, same output, every time) is the whole idea of a function. It is such a simple idea that it is easy to underrate. But once you have it, a huge amount of mathematics becomes one story: a line is a function, a parabola is a function, the area of a circle as its radius changes is a function, a sorted list is a function from positions to values, and a database index is a function from keys to rows.

The questions this section answers are: what exactly counts as a function, and how do we write one down?

There are two pictures worth keeping in your head.

The machine picture. A function is a box. You feed in an input; it returns exactly one output. You do not need to see inside the box; you only need to know that it is deterministic.

The mapping picture. Draw the set of allowed inputs on the left, the set of possible outputs on the right, and an arrow from each input to its output. The rule is: every input has exactly one arrow leaving it. An output may receive several arrows, or none at all. That asymmetry is the heart of the definition.

Notice what the right-hand picture says. "Give me a number whose square is 9" has two honest answers, +3 and −3. A rule with two answers is a perfectly good relation, but it is not a function. This is exactly why mathematicians define \(\sqrt{9}\) to mean the non-negative root only, 3: they wanted the square root to be a function.

So "\(f : \mathbb{R} \to \mathbb{R},\ x \mapsto x^2\)" and "\(f(x) = x^2\) for real \(x\)" say the same thing; the first is the style of university texts, the second of school texts.

Let \(f(x) = 2x^2 - 3x + 1\). Find \(f(3)\), \(f(-1)\) and \(f(a+1)\).

The rule is "whatever is in the parentheses, put it everywhere you see \(x\)". Use brackets when substituting; they prevent sign errors.

Check with \(a = 2\): \(f(3)\) should equal \(2(2)^2 + 2 = 10\). It does.

The same function can be shown in four ways, and fluency means moving between them freely:

A formula is compact and exact; a table is what a computer actually stores; a graph shows behaviour at a glance; words carry the meaning. None is "the" function. The function is the input-to-output assignment that all four describe.

Your code has a function sqrt(x). What happens if you call sqrt(-4)? In Python, math.sqrt(-4) raises ValueError: math domain error. The error message uses exactly the right mathematical word. Every function has a set of inputs it accepts, and asking for anything outside that set is meaningless (or, in the case of computers, an exception).

The codomain versus range distinction is exactly the distinction between a function’s declared return type and the values it really returns. A Python function annotated → int that only ever returns 0 or 1 has codomain int and range \(\{0, 1\}\).

When a function is given by a formula with no domain stated, the convention is to use the natural domain: every real number for which the formula makes sense. In real-number algebra, only a few things can go wrong:

A domain may also be restricted by the situation: a length cannot be negative, a number of people must be a whole number, a time might run only from launch to landing. These are called contextual or practical domains.

Example 1. \(g(x) = \dfrac{1}{x^2 - 4}\).

Denominator zero: \(x^2 - 4 = 0 \iff (x-2)(x+2) = 0 \iff x = \pm 2\). No roots or logs. Domain: \((-\infty, -2) \cup (-2, 2) \cup (2, \infty)\).

Example 2. \(f(x) = \dfrac{\sqrt{x - 3}}{x - 5}\).

Spot check: \(f(3) = 0/(-2) = 0\) is fine; \(f(5)\) would divide by zero; \(f(2)\) would need \(\sqrt{-1}\).

Example 3 (domain and range). \(h(x) = \sqrt{4 - x^2}\).

Domain: \(4 - x^2 \ge 0 \iff x^2 \le 4 \iff -2 \le x \le 2\), so \(\lbrack -2, 2]\).

Range: as \(x\) runs over \(\lbrack -2, 2]\), \(x^2\) runs over \(\lbrack 0, 4]\), so \(4 - x^2\) runs over \(\lbrack 0, 4]\) and its square root runs over \(\lbrack 0, 2]\). The range is \(\lbrack 0, 2]\). (The graph is the upper half of the circle of radius 2, which you can check against Chapter 10.)

Ranges are harder than domains; there is no universal recipe. Three methods cover most school cases:

Example of the third method: the range of \(f(x) = \dfrac{x - 1}{x - 2}\).

Every \(y \ne 1\) gives a valid \(x\) (and that \(x\) is never 2, because \(x = 2\) would need \(2y - 2 = 2y - 1\), impossible). So the range is \(\mathbb{R} \setminus \{1\}\). Hold on to this calculation: it is also how you find an inverse function, in Section 11.7.

A table of values shows a few points. A formula hides behaviour inside algebra. The graph shows everything at once: where the function is positive, where it rises, where it turns, where it is undefined. How do you know whether a curve someone draws is the graph of a function at all?

Since each input \(x\) has exactly one output, the graph contains exactly one point above (or below) each \(x\) in the domain, and none above \(x\) values outside it. A vertical line is the set of all points with a fixed \(x\), so:

The circle is still a perfectly good curve, described by an equation in \(x\) and \(y\). It just is not the graph of one function of \(x\). Solving for \(y\) gives \(y = \pm\sqrt{4 - x^2}\): two functions, the upper semicircle \(y = \sqrt{4 - x^2}\) (Example 3 above) and the lower semicircle \(y = -\sqrt{4 - x^2}\). Splitting a curve into function-pieces like this is a routine move in calculus.

When you meet the graph of a new function, ask in this order:

There are infinitely many functions, but a small number of families do most of the work in science and computing. If you know the shape of each family’s basic member (its parent function), you can sketch almost anything by shifting and stretching (the next section). This section is a field guide.

\(f(x) = mx + b\). Graph: a straight line with slope \(m\) and \(y\)-intercept \(b\) (all developed in Chapter 5). The defining property: equal steps in \(x\) give equal steps in \(y\). Increasing \(x\) by 1 always adds \(m\). Linear functions model anything with a constant rate: a taxi’s per-kilometre charge, a constant-speed journey, a linear cost.

\(f(x) = ax^2 + bx + c\) with \(a \ne 0\). Graph: a parabola, opening up if \(a > 0\) and down if \(a < 0\), with a vertex at \(x = -b/(2a)\) (see Chapter 8). Here equal steps in \(x\) give linearly growing steps in \(y\): for \(x^2\) the outputs 0, 1, 4, 9, 16 have differences 1, 3, 5, 7. Projectiles, areas, and the running time of a double loop are quadratic.

\(f(x) = x^n\) (or \(kx^n\)). The shape depends on \(n\):

Power laws appear everywhere: \(A = \pi r^2\), \(V = \tfrac{4}{3}\pi r^3\), gravity proportional to \(r^{-2}\), algorithm costs \(n^2\), \(n^3\).

The definition above is our first piecewise formula, and the "\(-x\)" branch is the one that surprises people: if \(x = -3\) then \(-x = 3\). The minus sign makes a negative number positive.

The example above jumps at \(x = 1\): approaching from the left, the values approach 2; at \(x = 1\) itself, the value is \(1 - 3 = -2\). A jump like this is a discontinuity, a concept made precise in Chapter 16.

Piecewise functions are just if/elif/else written in mathematics. Every function in your code with a branch is piecewise.

When you have data or a description and need a model, the family is chosen by how the output changes when the input changes:

You know the graph of \(y = x^2\). What does \(y = -2(x - 3)^2 + 4\) look like? Or \(y = \lvert 2x + 6 \rvert - 1\)? You could plot a table of points each time, but there is a far better way: every one of these is the parent graph moved and stretched, and the formula tells you exactly how.

Start with a function \(f\) whose graph you know. There are four basic changes, two to the output and two to the input:

The output changes do what you expect. The input changes feel backwards: \(x - 3\) moves the graph right, and \(2x\) makes it narrower. Why?

Let \(g(x) = f(x - 3)\). Ask: where does \(g\) do what \(f\) does at 0? That is, when is the input to \(f\) equal to 0? When \(x - 3 = 0\), so at \(x = 3\). Everything \(f\) did at 0, \(g\) does at 3; everything \(f\) did at 1, \(g\) does at 4. The whole graph has moved 3 to the right. The subtraction is the delay: \(g\) has to wait until \(x\) reaches 3 before it gets the input 0.

Similarly, let \(g(x) = f(2x)\). When is the input to \(f\) equal to 6? When \(2x = 6\), so at \(x = 3\). \(g\) gets through \(f\)'s inputs twice as fast, so its graph is squashed horizontally by a factor of 2.

The rule of thumb: changes outside \(f\) act on \(y\) directly; changes inside \(f\) act on \(x\) inversely. To find where a point goes, solve "inside = old input" for \(x\).

All four moves combine into one formula:

If \((x_0, y_0)\) is on the graph of \(f\), then the corresponding point on the graph of \(g\) is where the input to \(f\) is again \(x_0\):

On the \(x\) side: scale by \(1/b\), then shift by \(h\). On the \(y\) side: scale by \(a\), then shift by \(k\). In each direction, scale first, then shift.

Sketch \(g(x) = -2(x - 3)^2 + 4\). Try it: Graph g(x) = −2(x − 3)² + 4.

Parent: \(f(x) = x^2\). Here \(a = -2\), \(b = 1\), \(h = 3\), \(k = 4\). Key points on the parent: \((-1, 1), (0, 0), (1, 1), (2, 4)\).

So: a parabola opening down (since \(a < 0\)), twice as steep as \(x^2\), with vertex \((3, 4)\). Check directly: \(g(5) = -2(2)^2 + 4 = -8 + 4 = -4\). Correct. This is exactly the vertex form of a quadratic from Chapter 8; now you can see that vertex form is the transformation rule.

Sketch \(g(x) = \lvert 2x + 6 \rvert - 1\). Try it: Graph g(x) = |2x + 6| − 1.

Step 1, factor the inside: \(2x + 6 = 2(x + 3) = 2(x - (-3))\). So \(a = 1\), \(b = 2\), \(h = -3\), \(k = -1\). Parent key points of \(\lvert x \rvert\): \((-2, 2), (0, 0), (2, 2)\).

A V with its corner at \((-3, -1)\), with sides of slope \(\pm 2\). Check: \(g(-2) = \lvert 2 \rvert - 1 = 1\). Correct.

The four moves have everyday meanings once the input is time:

In science, recognising a transformed parent saves work: once you know the shape of \(y = x^2\), you know the shape of every projectile path, and once you know \(1/x\) you know Boyle’s law (\(P = k/V\)) for every gas sample.

A shop converts a price in dollars to euros, then adds 20% tax. Converting is one function; adding tax is another. The whole process (dollars in, final euro price out) is a third function made by running the first, then feeding its output into the second. That is composition, and it is how complicated functions are built from simple ones.

Let \(f(x) = x^2\) and \(g(x) = x + 3\).

At \(x = 2\): \((f \circ g)(2) = 25\) but \((g \circ f)(2) = 7\). "Add 3 then square" is not "square then add 3". Composition is not commutative. (It is, however, associative: \(f \circ (g \circ h) = (f \circ g) \circ h\), both meaning "do \(h\), then \(g\), then \(f\)".)

Let \(f(x) = \sqrt{x}\) and \(g(x) = x - 4\).

A subtler case: with \(p(x) = x^2\) and \(q(x) = \sqrt{x}\), \((p \circ q)(x) = (\sqrt{x})^2 = x\). The simplified formula \(x\) looks defined everywhere, but \(p \circ q\) is defined only for \(x \ge 0\), because \(q\) must be able to act first. As with holes, find the domain before simplifying.

Often the useful move is the reverse: seeing a complicated function as a composition of simple ones.

\(h(x) = (3x - 1)^5\) is \(f(g(x))\) with inner function \(g(x) = 3x - 1\) and outer function \(f(u) = u^5\).

\(k(x) = \sqrt{x^2 + 1}\) is \(f(g(x))\) with \(g(x) = x^2 + 1\) and \(f(u) = \sqrt{u}\).

The question to ask is "what would I compute last on a calculator?" That is the outer function. This skill is exactly what the chain rule needs in Chapter 17, where the derivative of a composition is built from the derivatives of the pieces.

The transformation rule of the last section is itself a composition: \(a f(b(x - h)) + k\) is "subtract \(h\), multiply by \(b\), apply \(f\), multiply by \(a\), add \(k\)", a five-stage pipeline.

Fahrenheit from Celsius is \(F = \tfrac{9}{5}C + 32\). A visitor wants the other direction: given \(F\), what is \(C\)? We want a function that undoes the first one. When does such an undo function exist, and how do you find it?

In the mapping picture, inverting a function means reversing every arrow. For the result to be a function again, two things must hold:

On a graph, "two inputs with the same output" means two points at the same height. So:

Lines with nonzero slope pass. \(x^3\) passes. \(x^2\) fails (the line \(y = 4\) meets it at \(x = \pm 2\)), as does \(\lvert x \rvert\). Any function that is strictly increasing or strictly decreasing everywhere on its domain passes automatically (see Section 11.8).

If \((a, b)\) is on the graph of \(f\), then \(f(a) = b\), so \(f^{-1}(b) = a\), so \((b, a)\) is on the graph of \(f^{-1}\). Swapping the coordinates of every point is the same as reflecting the plane in the line \(y = x\). The graph of \(f^{-1}\) is the mirror image of the graph of \(f\) in the line \(y = x\).

\(x^2\) on all of \(\mathbb{R}\) has no inverse. But on the restricted domain \(\lbrack 0, \infty)\) it is one-to-one, and its inverse is \(\sqrt{x}\). On \((-\infty, 0]\) it is also one-to-one, with inverse \(-\sqrt{x}\). The square root function exists because we chose the right half. The same choice gives the inverse trigonometric functions in Chapter 13, and the logarithm is the inverse of the exponential (which needs no restriction) in Chapter 12.

Some books swap \(x\) and \(y\) first and then solve; the two orders are equivalent. Solving first keeps the meaning clearer ("given the output, recover the input"), which matters in applications where \(x\) and \(y\) stand for different quantities like Celsius and Fahrenheit.

Find the inverse of \(f(x) = \dfrac{2x + 3}{5}\).

Check: \(f(6) = (12 + 3)/5 = 3\), and \(f^{-1}(3) = (15 - 3)/2 = 6\). Correct. Notice the inverse undoes the steps in reverse order: \(f\) multiplies by 2, adds 3, divides by 5; \(f^{-1}\) multiplies by 5, subtracts 3, divides by 2. Like taking off socks and shoes: the last thing on is the first thing off. In symbols, \((f \circ g)^{-1} = g^{-1} \circ f^{-1}\).

For Celsius and Fahrenheit, the same steps turn \(F = \tfrac{9}{5}C + 32\) into \(C = \tfrac{5}{9}(F - 32)\).

Find the inverse of \(f(x) = \dfrac{3x + 1}{x - 2}\), \(x \ne 2\).

The range of \(f\) is everything except 3, which becomes the domain of \(f^{-1}\). Check: \(f(3) = 10/1 = 10\); \(f^{-1}(10) = 21/7 = 3\). Correct.

Some graphs have a symmetry you can exploit: compute half, get the other half free. Others always go up, which guarantees an inverse exists and allows binary search. These properties can be tested algebraically, without drawing.

Worked example. Classify \(f(x) = x^4 - 3x^2\), \(g(x) = x^3 + x\), \(h(x) = x^2 + x\).

For \(h\), the result \(x^2 - x\) is neither \(h(x) = x^2 + x\) nor \(-h(x) = -x^2 - x\) (try \(x = 1\): \(h(-1) = 0\) while \(h(1) = 2\)). So \(h\) is neither even nor odd. One counterexample is enough to rule a property out; to establish it, you need the algebra for all \(x\).

Example: \(f(x) = x^2\) is decreasing on \((-\infty, 0]\) and increasing on \(\lbrack 0, \infty)\), with a minimum at 0. The parabola \(-2(x - 3)^2 + 4\) is increasing on \((-\infty, 3]\), decreasing on \(\lbrack 3, \infty)\), with a maximum of 4 at \(x = 3\).

Two important facts:

The average cost per item of a print run, with a fixed setup cost of $500 and $2 per copy, is \(C(n) = \dfrac{500 + 2n}{n}\). For small runs the cost is enormous; for huge runs it settles towards $2 but never gets there. Graphs like this, which blow up near some inputs and level off far away, belong to rational functions.

The parent is \(y = 1/x\). Near \(x = 0\), dividing 1 by a tiny number gives a huge number: \(1/0.001 = 1000\), \(1/(-0.001) = -1000\). So the graph shoots up on the right of 0 and down on the left: a vertical asymptote at \(x = 0\). Far out, dividing 1 by a huge number gives almost 0: a horizontal asymptote at \(y = 0\).

Now use transformations. \(y = \dfrac{1}{x - 2} + 1\) is \(1/x\) shifted right 2 and up 1, so its asymptotes are \(x = 2\) and \(y = 1\). And that is the function from the range example, since

This explains, geometrically, why its range was everything except 1.

Why the degree rule works. Far from the origin, the highest power dominates each polynomial. For \(R(x) = \dfrac{2x^2 + 1}{x^2 - 9}\), divide top and bottom by \(x^2\):

Analyse \(R(x) = \dfrac{2x^2 + 1}{x^2 - 9}\).

Also: \(R(0) = 1/(-9) = -1/9\) (the \(y\)-intercept), and since the numerator is never zero there are no \(x\)-intercepts. \(R\) is even, since only even powers appear.

A hole. \(S(x) = \dfrac{x^2 - 1}{x - 1} = \dfrac{(x - 1)(x + 1)}{x - 1}\). The factor \(x - 1\) cancels, so there is a hole at \(x = 1\), at height \(1 + 1 = 2\), and no vertical asymptote. The graph is the line \(y = x + 1\) with the point \((1, 2)\) removed.

For the print-run example, \(C(n) = (500 + 2n)/n\) has equal degrees, leading coefficients 2 and 1, so the horizontal asymptote is \(y = 2\): the cost per copy approaches $2. Writing it as \(C(n) = 2 + 500/n\) makes that obvious: the setup cost is spread ever more thinly.

A function is also a claim about the world: "tax depends on income like this", "height depends on time like that". Building such a model means choosing variables, choosing a family, fitting parameters, and (the step most often skipped) stating the domain where the model is valid.

A (simplified, fictional) tax system: 10% on income up to $10,000, and 20% on every dollar above $10,000. Write tax \(T\) as a function of income \(x\) and compute \(T(25{,}000)\).

On the first $10,000 the tax is \(0.1x\). Above that, the first $10,000 contributes \(0.1 \times 10{,}000 = 1{,}000\), and the excess \(x - 10{,}000\) is taxed at 20%:

Check continuity at the boundary: the first branch gives \(T(10{,}000) = 1{,}000\), and the second branch, as \(x\) approaches 10,000 from above, approaches \(1{,}000 + 0 = 1{,}000\). No jump. That is the meaning of marginal rates: moving into a higher bracket only raises the tax on the extra dollars, so earning more never leaves you with less. (A system with a jump in tax at the boundary would create exactly that perverse effect, which is why real systems are designed as continuous piecewise-linear functions.)

A courier charges $3.00 for the first kilogram (or part of one) and $1.50 for each additional kilogram or part. Model the cost and price a 2.3 kg parcel.

"Or part" means round up. A parcel of weight \(w\) counts as \(\lceil w \rceil\) kilograms, of which the first is charged at $3.00:

Sanity checks: \(C(0.4) = 3 + 1.5(1 - 1) = 3.00\); \(C(1) = 3.00\); \(C(1.01) = 3 + 1.5 = 4.50\).

A ball is thrown upwards from 1.5 m with speed 20 m/s. Ignoring air resistance, its height after \(t\) seconds is \(h(t) = -4.9t^2 + 20t + 1.5\) metres. What is the practical domain, the maximum height, and the range?

The model only makes sense from the throw (\(t = 0\)) until the ball lands (\(h(t) = 0\), \(t > 0\)). Using the quadratic formula (Chapter 8) on \(4.9t^2 - 20t - 1.5 = 0\):

(The other root is negative, before the throw, so we discard it.) The domain is \(\lbrack 0, 4.155]\) approximately.

The vertex is at \(t = -b/(2a) = -20/(2 \times -4.9) \approx 2.041\) s, and

So the range is about \(\lbrack 0, 21.9]\) metres. Outside the domain the formula still produces numbers (\(h(6) \approx -55\)), but they mean nothing: the ball does not burrow 55 m underground.

A mathematical function is only an input-to-output assignment. It does not "run", take time, print anything, or remember previous calls. A programming function (a subroutine, procedure or method) is a piece of code that may do all of those things.

add_impure fails the defining property of a function: the input 5 produced two different outputs. Mathematically, it is a function of two inputs (the argument and the hidden state), and making that hidden input explicit, as add_pure does, is the standard repair. Functional programming is largely the discipline of making every input explicit.

Other differences worth knowing:

In mathematics, functions are objects you can pass around: composition takes two functions and returns a function; the derivative (Chapter 17) takes a function and returns a function. Languages with first-class functions allow the same: functions can be stored in variables, passed as arguments, and returned as results. A function that takes or returns functions is a higher-order function. compose, map, filter, sorted(key=…​), and decorators are all higher-order. The Python lambda x: x ** 2 is a direct transcription of \(x \mapsto x^2\).

Mathematically, a function on a finite domain is nothing more than its table of input–output pairs. Computing can trade time for space by storing that table:

Memoisation (here via lru_cache) builds the table lazily: six calls, but only three distinct inputs, so only three real computations. This is only safe for pure functions; memoising add_impure would silently change its behaviour. Precomputed tables appear in CRC checksums (a 256-entry table per byte), gamma correction for images, sine tables in old games and embedded systems, and dynamic programming generally (Chapter 24).

A hash function maps data of any size (strings, files, objects) to a fixed-size value, typically an integer in a fixed range. A hash table uses it to choose a bucket: \(\text{bucket}(k) = \text{hash}(k) \bmod n\).

(We used zlib.crc32 because it is deterministic; Python’s built-in hash for strings is deliberately randomised per process, so its output would differ every run.)

A hash must be a true function: the same key must always land in the same bucket, or lookups fail. It is never one-to-one when the domain is bigger than the codomain: ten words into eight buckets must produce at least one collision (two keys, one bucket). This is the pigeonhole principle (Chapter 24). Hash-table design is the art of making collisions rare and cheap, not impossible. Cryptographic hashes (SHA-256) go further: collisions certainly exist, but finding one should be computationally infeasible.

The one-to-one idea from Section 11.7 has two partners, and the three together classify functions precisely.

Surjectivity depends on the codomain you declare. \(x \mapsto x^2\) is not surjective onto \(\mathbb{R}\) (−1 is never hit) but is surjective onto \(\lbrack 0, \infty)\). Injectivity depends on the domain: \(x^2\) is not injective on \(\mathbb{R}\) but is on \(\lbrack 0, \infty)\). So \(x \mapsto x^2\) as a function \(\lbrack 0, \infty) \to \lbrack 0, \infty)\) is a bijection, and its inverse is the square root, matching what we found earlier.

Worked example. Classify, with reasons:

On a finite domain these checks are complete proofs: the program examines every input. The injectivity check is the same algorithm as "detect duplicate values with a dictionary".