Functions and invented operations

A function is a rule

A function takes an input and gives exactly one output. Writing $f(x) = 2x^2 - 3x$ says: whatever goes in the brackets, square it, double that, and subtract three times the input. The letter $x$ is only a placeholder. $f(5)$, $f(-2)$, $f(a + 1)$ and $f(2x)$ all mean "replace every $x$ in the rule with what is in the brackets".

Substitute with brackets

The safest habit is to rewrite the rule with empty brackets where each $x$ was, then fill them:

$$f(\;) = 2(\;)^2 - 3(\;)$$

For $f(-2)$: $2(-2)^2 - 3(-2) = 8 + 6 = 14$. Without the brackets it is easy to write $-2^2 = -4$, or to lose the sign in $-3 \times (-2)$.

For $f(x + 1)$, the whole of $x + 1$ goes into each bracket:

$$f(x + 1) = 2(x + 1)^2 - 3(x + 1) = 2x^2 + 4x + 2 - 3x - 3 = 2x^2 + x - 1$$

Four expressions that look alike

Expression	Meaning	For $f(x) = 3x + 2$
$f(x + 1)$	change the input, then apply $f$	$3x + 5$
$f(x) + 1$	apply $f$, then change the output	$3x + 3$
$f(2x)$	double the input	$6x + 2$
$2f(x)$	double the output	$6x + 4$

These are different unless $f$ happens to be special. In particular $f(2x)$ and $2f(x)$ are equal only for functions like $f(x) = kx$ with no constant term. A comparison that sets one against the other is testing exactly this.

Composition: inside out

$f(g(x))$ means: apply $g$ first, then feed its output to $f$. Work from the innermost brackets outwards. With $f(x) = 3x - 1$ and $g(x) = x^2 + 2$:

• $f(g(2))$: $g(2) = 6$, then $f(6) = 17$.
• $g(f(2))$: $f(2) = 5$, then $g(5) = 27$.

The order matters, and the two results are usually different. Read the expression from the inside, not from left to right.

Solving for the input

Some questions give the output and ask for the input. "If $f(k) = 10$" means: substitute $k$ into the rule and set the result equal to 10, then solve the equation that results. If the rule is quadratic, expect two inputs, and check whether the stem limits which is allowed.

Where a function is defined

A rule can fail for some inputs. At this level, two things cause it: a denominator that would be zero, and an even root of a negative number. For $h(x) = \frac{1}{x - 4}$, the input 4 is not allowed. For $r(x) = \sqrt{x + 3}$, inputs must satisfy $x \ge -3$.

Operations with invented symbols

Questions often define a new operation with an unfamiliar symbol, for example

$$a \star b = \frac{a + b}{a - b} \quad (a \neq b)$$

There is nothing to know in advance; the definition is the whole rule. Treat it exactly like a function of two inputs:

1. Match the inputs by position: the first number replaces $a$, the second replaces $b$.
2. Work innermost brackets first, as with composition.
3. Assume nothing the definition does not say. An invented operation need not be commutative ($a \star b$ may differ from $b \star a$) or associative ($(a \star b) \star c$ may differ from $a \star (b \star c)$).

With the definition above, $3 \star 1 = \frac{4}{2} = 2$, but $1 \star 3 = \frac{4}{-2} = -2$. Swapping the inputs changed the answer.

Functions on a graph

When a function is drawn, $f(a)$ is the height of the graph at $x = a$, and "$f(x) = 0$" asks where the graph crosses the $x$-axis. The same substitution rules apply; the graph is just the rule drawn as points $(x, f(x))$.