Function Notation and Evaluation, precalculus lesson

A weather forecast lists the temperature hour by hour. If $T$ is that function and $t$ counts hours after midnight, then $T(14)$ is the temperature at two in the afternoon. Function notation records which input you used and which output you got in a few symbols. Calculus uses it on every page, often with inputs like $a + h$ that are expressions rather than numbers.

Key terms

Function notation

Definition Let $f$ be a function with domain $D$ . For each $x \in D$ , the symbol $f(x)$ , read ” $f$ of $x$ ”, denotes the unique output that $f$ assigns to $x$ . A formula such as

f(x) = 3x^2 - 1

defines a function by saying that for every $x$ in its domain the output is $3x^2 - 1$ . The letter used for the input is a placeholder, so $f(t) = 3t^2 - 1$ defines the same function. When no domain is stated, the domain is the natural domain of the formula. Writing $y = f(x)$ names the output $y$ .

In plain words The name of the function goes outside the parentheses and the input goes inside. The whole symbol stands for the output.

Example The rule “square the input, then add $1$ ” is written $f(x) = x^2 + 1$ . Then $f(3) = 3^2 + 1 = 10$ , so the point $(3, 10)$ is on the graph of $f$ . Writing $g(u) = u^2 + 1$ describes the same rule with a different letter, and indeed $g(3) = 10$ as well.

Function value as an output

Definition For an input $a$ in the domain of $f$ , the function value $f(a)$ is a single number, the output at $a$ . The letter $f$ names the function itself, the whole rule or set of pairs, and is not a number. The equation $f(a) = b$ says exactly that $(a, b)$ belongs to $f$ , which means that the point $(a, b)$ lies on the graph of $f$ .

In plain words The letter $f$ is the machine. The symbol $f(a)$ is what comes out when you feed it $a$ .

Example A ball thrown upward is at height

s(t) = 20t - 5t^2 \text{ meters}

after $t$ seconds. The value $s(2) = 40 - 20 = 20$ says that two seconds after the throw the ball is $20$ meters high, and the point $(2, 20)$ is on the graph of $s$ . A statement like " $s = 20$ " makes no sense, since $s$ is the whole height function and not one height.

Evaluation

Definition To evaluate $f$ at $a$ means to find the value $f(a)$ . This is possible exactly when $a$ is in the domain of $f$ . If $a$ is not in the domain, then $f(a)$ is undefined, and no algebra can produce a value for it.

In plain words Evaluating is asking the function for its answer at one input. Some inputs simply have no answer.

Example Let

g(x) = \frac{1}{x - 1}.

Then $g(3) = \frac{1}{2}$ and $g(0) = -1$ . At $x = 1$ the denominator is $0$ , so $1$ is not in the domain and $g(1)$ is undefined.

Substitution

Definition When $f$ is given by a formula in $x$ , the value $f(a)$ is found by replacing every occurrence of $x$ in the formula by $a$ , written in parentheses, and then simplifying with the usual order of operations.

In plain words Swap each $x$ for the input, keep the input in parentheses, and then simplify.

Example Let $f(x) = x^2 - 3x + 2$ . Then

\begin{aligned} f(-2) &= (-2)^2 - 3(-2) + 2 \\[0.4em] &= 4 + 6 + 2 \\[0.4em] &= 12. \end{aligned}

The parentheses matter. If the first term were written as $-2^2$ , it would be read as $-(2^2) = -4$ , and the result would come out as $4$ instead of $12$ .

Evaluating from a table or a graph

Definition If $f$ is given completely by a table, then $f(a)$ is the output listed with the input $a$ , and $f(a)$ is undefined if $a$ is not among the listed inputs. If $f$ is given by its graph, then $f(a)$ is the second coordinate of the point where the vertical line $x = a$ meets the graph. So

\begin{aligned} &f(a) = b \\[0.4em] &\iff (a, b) \text{ is on the graph of } f. \end{aligned}

If the vertical line $x = a$ misses the graph, then $a$ is not in the domain.

In plain words In a table, find the input and read the output next to it. On a graph, go to the input on the horizontal axis, move up or down to the graph, and read the height there.

Example A function $f$ is given by this table.

$x$	$-1$	$0$	$1$	$2$	$3$
$f(x)$	$4$	$1$	$0$	$1$	$4$

Then $f(2) = 1$ , while $f(5)$ is undefined because $5$ is not listed. Reading the table backwards answers a different question. The inputs with $f(x) = 1$ are $x = 0$ and $x = 2$ . If the graph of a function $g$ passes through $(-1, 3)$ and $(2, 0)$ , then $g(-1) = 3$ and $g(2) = 0$ .

Arbitrary inputs

Definition The input of $f$ may be any expression whose value lies in the domain of $f$ , such as a letter $a$ , or $2x$ , $x + 1$ or $x^2$ . To find $f$ of an expression, substitute the whole expression, in parentheses, for every occurrence of the variable in the formula, and then simplify.

In plain words The input slot does not care whether it receives a number or an expression. Whatever goes in replaces every $x$ , all of it at once.

Example Let $f(x) = x^2 - 3x$ . Then

f(2a) = (2a)^2 - 3(2a) = 4a^2 - 6a

and

\begin{aligned} f(x + 1) &= (x + 1)^2 - 3(x + 1) \\[0.4em] &= x^2 + 2x + 1 - 3x - 3 \\[0.4em] &= x^2 - x - 2. \end{aligned}

A quick numerical check helps. For $x = 2$ the last formula gives $4 - 2 - 2 = 0$ , and indeed $f(3) = 9 - 9 = 0$ .

Evaluating at a plus h

Definition For numbers $a$ and $h$ such that $a + h$ is in the domain of $f$ , the value $f(a + h)$ is the output of $f$ at the single input $a + h$ . It is computed by substituting $a + h$ for the variable. In general $f(a + h)$ is neither $f(a) + h$ nor $f(a) + f(h)$ .

In plain words Think of $a + h$ as one input, a step of size $h$ away from $a$ . Put the whole sum in the input slot before doing anything else.

Example Let $f(x) = x^2 + 1$ . Then

\begin{aligned} f(a + h) &= (a + h)^2 + 1 \\[0.4em] &= a^2 + 2ah + h^2 + 1. \end{aligned}

Compare this with the two tempting wrong answers, $f(a) + h = a^2 + 1 + h$ and $f(a) + f(h) = a^2 + h^2 + 2$ . With $a = 1$ and $h = 2$ the correct value is $f(3) = 10$ , while the wrong ones give $4$ and $7$ . Subtracting $f(a)$ and dividing by $h \ne 0$ gives

\begin{aligned} \frac{f(a + h) - f(a)}{h} &= \frac{2ah + h^2}{h} \\[0.8em] &= 2a + h, \end{aligned}

which is the difference quotient behind derivatives.

Function value versus multiplication

Definition In $f(x)$ the parentheses mean “apply $f$ to $x$ ”, not “multiply $f$ by $x$ ”. Function evaluation has no distributive law. The equations

f(a + b) = f(a) + f(b)

and

f(cx) = c\,f(x)

hold for all real inputs when $f(x) = mx$ for a constant $m$ , but they fail for most other functions, for example for $f(x) = x^2$ and for $f(x) = x + 1$ . So neither may be used unless it has been checked for the function at hand. Whether a letter followed by parentheses means multiplication or evaluation depends on whether the letter names a number or a function.

In plain words A function is not a number you can hand out across a sum. To find $f(a + b)$ , add first and then apply $f$ .

Example Take $f(x) = x^2$ . Then $f(3 + 4) = f(7) = 49$ , while $f(3) + f(4) = 9 + 16 = 25$ . Similarly $f(2 \cdot 3) = 36$ , while $2f(3) = 18$ . By contrast, if $k$ is the number $5$ , then $k(x + 1) = 5x + 5$ really is a product.

Check yourself

Let $g(x) = 2x^2 - x$ . Find $g(-3)$ and $g(a + 1)$ .

Show the answer

Substituting gives

\begin{aligned} g(-3) &= 2(-3)^2 - (-3) \\[0.4em] &= 18 + 3 \\[0.4em] &= 21. \end{aligned}

For the second value,

\begin{aligned} g(a + 1) &= 2(a + 1)^2 - (a + 1) \\[0.4em] &= 2a^2 + 4a + 2 - a - 1 \\[0.4em] &= 2a^2 + 3a + 1. \end{aligned}

Let $f(x) = x^2 - 4x$ . Simplify the following expression for $h \ne 0$ .

\frac{f(3 + h) - f(3)}{h}

Show the answer

First

\begin{aligned} f(3 + h) &= (3 + h)^2 - 4(3 + h) \\[0.4em] &= 9 + 6h + h^2 - 12 - 4h \\[0.4em] &= h^2 + 2h - 3, \end{aligned}

and $f(3) = 9 - 12 = -3$ . So

\begin{aligned} \frac{f(3 + h) - f(3)}{h} &= \frac{h^2 + 2h}{h} \\[0.8em] &= h + 2. \end{aligned}

For $f(x) = x^2$ , is it true that $f(2x) = 2f(x)$ for every $x$ ?

Show the answer

No. Substituting gives $f(2x) = (2x)^2 = 4x^2$ , while $2f(x) = 2x^2$ . The two agree only at $x = 0$ . For example, $f(2 \cdot 1) = 4$ but $2f(1) = 2$ .

Function Notation and Evaluation

Theory, part 5

Practice, part 6

Key terms