Function Notation

Let $f(x) = x^2 + 1$. Work out $f(2) + f(3)$, and separately work out $f(2 + 3)$. Before you check, predict whether they will be equal. Then say, in one sentence, what your two answers show about the notation.

The symbol $f(x)$ does three jobs at once: $f$ names the function, the brackets announce that what follows is the input, and the whole expression stands for the output. It is never $f$ multiplied by $x$.

Writing

$$f(x) = 3x - 1$$

does three things at once. The letter $f$ gives the function a name, so it can be referred to later. The bracketed $x$ announces which letter is the variable. And the whole symbol $f(x)$ stands for the output, so the equation says what that output is.

$f(x)$ is not $f$ multiplied by $x$. There is no separate quantity called $f$, so there is nothing to multiply and nothing to cancel. The brackets here mean "applied to", not "times".

Any letter may be used as a name, and questions routinely use several at once:

$$f(x) = 3x - 1, \qquad g(x) = x^2, \qquad h(t) = 5 - t$$

The third of those uses $t$ as its variable, which is allowed and common when the input stands for something like time. The letter chosen for the variable carries no mathematical weight: $h(t) = 5 - t$ and $h(x) = 5 - x$ are the same function.

$f(a)$ means: replace every $x$ in the rule by $a$, then simplify. Take $f(x) = x^2 - 4x + 1$.

$$f(3) = 3^2 - 4(3) + 1 = 9 - 12 + 1 = -2$$

Negatives need brackets, every time:

$$f(-2) = (-2)^2 - 4(-2) + 1 = 4 + 8 + 1 = 13$$

Without them, $-2^2$ evaluates as $-4$ rather than $4$, and the answer is wrong by $8$.

The input need not be a number. Anything may be substituted, provided it goes in as a whole, in brackets:

$$f(2t) = (2t)^2 - 4(2t) + 1 = 4t^2 - 8t + 1$$

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

Notice what the second one is not. $f(x+1)$ is not $f(x) + 1$: the addition happens to the input, before $f$ acts, not to the output afterwards. That distinction reappears in Lesson 7 as the difference between a horizontal and a vertical translation.

Every question involving the notation is one of two kinds, and telling them apart is most of the skill.

The input is given. "Find $f(4)$." Substitute $4$ and evaluate. The answer is a number, and there is exactly one, because $f$ is a function.

The output is given. "Solve $f(x) = 4$." Set the rule equal to $4$ and solve for $x$. The answer may be one value, several, or none, because nothing prevents two inputs from sharing an output.

With $f(x) = x^2 - 3$:

$$f(4) = 16 - 3 = 13 \qquad \text{one substitution, one answer}$$

$$f(x) = 13 \ \Rightarrow \ x^2 = 16 \ \Rightarrow \ x = \pm 4 \qquad \text{one equation, two answers}$$

The asymmetry is the function condition doing its work. One input gives one output; one output may come from many inputs.

Two special cases are worth naming. $f(0)$ is the $y$-intercept of the graph, and the solutions of $f(x) = 0$ are the $x$-intercepts. Those two are the ones most often confused, and they are on opposite axes.

The brackets in $f(x)$ look like the brackets in $3(x)$, and that resemblance causes a specific error: treating $f$ as though it distributes.

$$f(a+b) \neq f(a) + f(b) \qquad \text{in general}$$

One example settles it. With $f(x) = x^2$:

$$f(2+3) = f(5) = 25 \qquad \text{but} \qquad f(2) + f(3) = 4 + 9 = 13$$

Likewise $f(2x) \neq 2f(x)$ in general: for the same $f$, the left side is $4x^2$ and the right is $2x^2$.

The reason is that $f$ is an instruction, not a multiplier. "Square the input" applied to $2+3$ squares the whole of $5$; it does not square the parts and add them.

A small number of functions do happen to satisfy $f(a+b) = f(a) + f(b)$, and they are exactly those of the form $f(x) = mx$. That is a special property of straight lines through the origin, not a rule of notation, and assuming it elsewhere is the single most expensive habit in this topic.

Five items on function notation. Work each one, then reveal the answer.

1. 1 $f(x) = 5x - 2$. Find $f(3)$.

   Show answer
   Substitute $3$: $15 - 2$.$13$

2. 2 $f(x) = x^2 + x$. Find $f(-3)$.

   Show answer
   Brackets: $(-3)^2 + (-3) = 9 - 3$.$6$

3. 3 $f(x) = 2x + 1$. Solve $f(x) = 11$.

   Show answer
   The output is given, so solve $2x + 1 = 11$.$x = 5$

4. 4 $f(x) = x^2$. Is $f(1+4)$ equal to $f(1) + f(4)$?

   Show answer
   $f(5) = 25$ while $1 + 16 = 17$.No

5. 5 $f(x) = 3 - x$. Find $f(x+2)$ in simplest form.

   Show answer
   Substitute the whole bracket: $3 - (x+2)$.$1 - x$