Functions: Chapter Review

Vocabulary and notation

Function

A rule that assigns to each input exactly one output, so one input can never yield two different results.

Function notation $f(x)$

$f$ names the rule, $x$ records the input, and the whole symbol is the output: $f(3) = 7$ says the input $3$ gave the output $7$.

Domain

The set of inputs a function is allowed to take.

Range

The set of outputs it actually produces: what comes out, where the domain is what may go in.

Composition $f \circ g$

Feeding one function's output straight into another, read "$f$ after $g$". The ring is neither multiplication nor the letter o.

Inner and outer function

In $f(g(x))$, $g$ is inner and runs first on the raw input; $f$ is outer and runs last, so it decides what kind of answer comes out.

Inverse function $f^{-1}$

The function that reverses $f$, sending each output back to the input it came from. It swaps every pair: $(5, 8)$ becomes $(8, 5)$.

One-to-one function

A function whose different inputs always give different outputs, so no two inputs share an output.

Binary operation

A rule taking two inputs to one output: ordinary $+$, or an invented symbol such as $\star$, $\diamond$, or $\oplus$ given a defining formula.

Commutative operation

One with $a \star b = b \star a$ for every $a$ and $b$, so the two inputs may trade places freely.

Associative operation

One where the grouping never matters: $(a \star b) \star c = a \star (b \star c)$ always.

Formulas and theorems

• Vertical line test

  A graph represents a function exactly when no vertical line crosses it more than once.

  Text description

  The curve x equals y squared, with a dashed vertical line at x equals 4 crossing it twice, at the outputs 2 and negative 2.

  Use when Input on the horizontal axis. A line that misses the graph is fine, since an input outside the domain has no point above it.

  e.g. $x = y^2$ fails: at $x = 4$ both $2$ and $-2$ appear.

• Arithmetic combinations of two functions

  $$\begin{gathered} (f + g)(x) = f(x) + g(x) \\ (f - g)(x) = f(x) - g(x) \\ (fg)(x) = f(x) \cdot g(x) \\ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} \end{gathered}$$

  Use when Defined wherever both $f$ and $g$ are. The quotient also needs $g(x) \neq 0$, even where $f$ is happy.

• Composition

  $$(f \circ g)(x) = f(g(x))$$

  Text description

  A chain: x enters the inner function g first, and the output g of x then enters the outer function f, giving f of g of x.

  Use when The inner function $g$ runs first on the raw input; $f$ runs second on its output, so $f \circ g$ is read right to left, not left to right.

  e.g. $f(x) = x^2$, $g(x) = x + 4$: $(f \circ g)(x) = (x + 4)^2 = x^2 + 8x + 16$.

• Composition is not commutative

  In general $f \circ g$ and $g \circ f$ are different functions.

  Use when "In general" is exact: particular pairs can agree, as a function and its inverse do. Never assume the swap is free.

  e.g. $f(x) = x^2$, $g(x) = x + 4$: $f(g(3)) = 49$ but $g(f(3)) = 13$.

• Domain of a composition

  An input is in the domain of $f \circ g$ exactly when it is allowed into $g$ AND the output $g(x)$ is allowed into $f$.

  Use when Both tests, always. The second can bar an input the inner function accepts, so the domain is neither function's by itself.

• Defining property of an inverse

  $$\begin{gathered} f^{-1}(f(x)) = x \\ f(f^{-1}(x)) = x \end{gathered}$$

  Text description

  Two ovals: an arrow labelled f carries x across to f of x, and a return arrow labelled f inverse carries it back to x.

  Use when The first holds on the domain of $f$, the second on the domain of $f^{-1}$. The raised $-1$ names the inverse under composition, never a reciprocal.

• Domain and range trade places

  The domain of $f^{-1}$ is the range of $f$, and its range is the domain of $f$.

  Use when Whenever $f^{-1}$ exists. It follows from swapping every pair, so the forbidden inputs swap as well.

  e.g. $f(x) = \dfrac{1}{x - 3}$ bans $3$; $f^{-1}(x) = \dfrac{1}{x} + 3$ bans $0$.

• One-to-one, and the horizontal line test

  A function has an inverse function exactly when it is one-to-one, which on a graph means no horizontal line crosses it more than once.

  Use when Algebraically, set $f(a) = f(b)$ and check that $a = b$ follows. A rule that fails can often be repaired by shrinking the domain to a one-to-one piece, and that restriction travels with the inverse.

  e.g. $f(x) = x^2$ fails ($3$ and $-3$ both give $9$); on $x \ge 0$ its inverse is $\sqrt{x}$.

• Evaluating a defined operation

  Substitute the first input for every $a$ and the second for every $b$ in the defining rule, then simplify.

  Use when The symbol carries no meaning of its own; only the rule does.

  e.g. With $a \star b = 2a + b - 1$: $3 \star 5 = 2(3) + 5 - 1 = 10$.

• Identity element of an operation

  $$\begin{gathered} a \star e = a \\ e \star a = a \end{gathered}$$

  Use when One fixed number $e$ must satisfy both, for EVERY $a$. Both sides, since the operation need not be commutative; some operations have no identity.

  e.g. $a \oplus b = a + b + ab$ has identity $0$; $a \star b = 2a + b - 1$ has none.

Problem types, step by step

Exam traps

• Trap Reading $f^{-1}$ as a reciprocal: for $f(x) = x + 3$, calling $f^{-1}(5)$ the value $\dfrac{1}{f(5)} = \dfrac{1}{8}$.

  Fix The raised $-1$ names the inverse under composition, never $\dfrac{1}{f}$. Here $f^{-1}(x) = x - 3$, so $f^{-1}(5) = 2$, the input that $f$ sends to $5$.

• Trap Reading $f(3)$ as "$f$ times $3$": for $f(x) = 2x + 1$, writing $f(3) = 3(2x + 1) = 6x + 3$.

  Fix The parentheses are an instruction to apply the rule, not a multiplication sign. Substitute instead: $f(3) = 2(3) + 1 = 7$.

• Trap Substituting only the variable: for $f(x) = x^2$ and $g(x) = x + 4$, writing $f(g(x)) = x^2 + 4$.

  Fix The entire inner rule goes in: $f(g(x)) = (x + 4)^2 = x^2 + 8x + 16$. The answer $x^2 + 4$ is really $g(f(x))$, the other order.

• Trap Answering "which input gives an output of $7$?" by computing $f(7)$.

  Fix $f(7)$ evaluates AT the input $7$; finding the input that yields $7$ means solving $f(x) = 7$. For $f(x) = 2x + 1$ that is $x = 3$, not $f(7) = 15$.

• Trap Calling $\{(1, 5), (2, 5), (3, 5)\}$ not a function because the output $5$ repeats, or rejecting $f(x) = x^2$ because a horizontal line meets its graph twice.

  Fix Repeated outputs are legal; only one input with two outputs is banned. Vertical crossings decide whether a graph is a function, horizontal crossings only whether it is one-to-one.

• Trap Reading the domain of $f \circ g$ off the simplified formula: with $f(x) = x^2$ and $g(x) = \sqrt{x}$, $\left(\sqrt{x}\right)^2 = x$ looks defined everywhere.

  Fix The inner function must accept the input first. Since $g$ rejects every negative, the domain is $x \ge 0$, though the simplified rule shows no trace of the root.

• Trap Declaring two functions inverses after checking one direction only.

  Fix One direction can hold while the other fails. For $f(x) = x^2$ on all numbers and $g(x) = \sqrt{x}$: $f(g(x)) = x$, but $g(f(-3)) = \sqrt{9} = 3$, not $-3$.

• Trap Letting the invented symbol override the order of operations: reading $3 \star 5$ with $a \star b = 2a + b - 1$ as $2(3 + 5) - 1 = 15$.

  Fix Substitution changes nothing about how the expression is read. Only $a$ is doubled: $2(3) + 5 - 1 = 10$.

• Trap Substituting only part of an expression input: with $a \diamond b = a^2 - b$, writing $(k + 1) \diamond 4 = k^2 + 1 - 4 = k^2 - 3$.

  Fix The whole expression drops into the slot as one block: $(k + 1)^2 - 4 = k^2 + 2k - 3$. The middle term $2k$ is exactly what dropping the parentheses loses.