Function Notation and Evaluation

Intuition

In Relations and Functions you saw that a function is more than a formula: it is a rule together with a domain. That lesson also showed that the same curve can be a function of one variable but not another. You also already know how to drop a number into $f(x)$ and read off the answer. This lesson is about the notation $f(x)$ itself, and it makes one move that runs through the rest of the chapter. The letter inside the parentheses is not really the variable $x$ . That letter is a slot, and anything can go in it: a number, another variable, or a whole expression like $x + h$ . Once the slot can hold an expression, three things become possible that a bare $y$ could never express. Those three are writing $f(2x)$ and $f(x + h)$ , asking whether $f(a + b)$ equals $f(a) + f(b)$ , and building the difference quotient that measures how fast $f$ changes. We take $f$ apart until the notation is doing real work.

Learning goals

Read $f$ as the function and $f(x)$ as one value
Copy any expression into every slot the argument marks
Reject $f(a + b) = f(a) + f(b)$ as a general rule
Simplify the difference quotient into a secant slope
Require equal domains before calling two functions equal
Evaluate a piecewise rule by choosing the right case

The function is $f$ ; its value is $f(x)$

Start by separating two things that the phrase “the function $f(x) = x^2$ ” quietly runs together. Strictly, $f$ is the function, the whole rule with its domain, as in Relations and Functions, while $f(x)$ is a number: the output the rule produces at the input $x$ . Writing “the function $f(x)$ ” is a mild abuse of language that everyone uses. But when a step gets subtle, it pays to remember that $f$ is the machine and $f(x)$ is what comes out of it.

This also settles what the parentheses are not. They are not multiplication. $f(x)$ does not mean $f$ times $x$ , and $f(3)$ is not $3f$ ; $f$ is a rule, not a number, so there is nothing to multiply it by. The parentheses do one job only: they hold the input the rule is about to act on. Keep that job in mind, because the next section widens what is allowed to sit inside them.

The argument is a slot, not the letter $x$

Picture the rule with an empty box where the input goes:

f(\,\square\,) = \square^2 + 1.

The box says “whatever you feed me gets squared, then $1$ is added.” The $x$ in $f(x) = x^2 + 1$ is just a name written in that box so we can talk about a general input; it is a placeholder, not a special number. So anything at all can go in the slot, and whatever goes in is copied into every place the box appears:

\begin{aligned} f(3) &= 3^2 + 1, &\qquad f(-x) &= (-x)^2 + 1, \\ f(2x) &= (2x)^2 + 1, &\qquad f(a) &= a^2 + 1. \end{aligned}

The input can even be another expression built from $x$ . Putting $x + h$ in the slot gives

f(x + h) = (x + h)^2 + 1,

and you may drop a function value into the slot too: $f(f(x)) = (f(x))^2 + 1 = (x^2 + 1)^2 + 1$ , treating $f(x)$ as just another expression. Combining functions this way earns its own lesson later in the chapter; here it is nothing more than substitution.

Because the slot only cares about what you put in, not what you call it, the name of the input variable carries no meaning. The rules $f(x) = x^2 + 1$ and $f(t) = t^2 + 1$ have the same domain, the same codomain, and pair every input with the same output. So by the equality test later in this lesson, those two rules are the same function. Renaming $x$ to $t$ changes the writing, not the function.

The slot also reads backward. Because $f(x)$ is a number, an equation like $f(x) = 7$ is an ordinary equation in $x$ . You solve that equation for the input that produces that output, using only the equation solving of Chapter 1.

The one habit the slot demands is substitution with parentheses. Wrap the whole input before applying the rule. Skip the wrapping and you silently change the problem: for $f(x) = x^2$ ,

f(-x) = (-x)^2 = x^2, \qquad \text{not} \qquad -x^2.

The correct result squares the input; the careless $-x^2$ negates a square, and the two disagree at every nonzero input (at $x = 3$ , $(-3)^2 = 9$ but $-3^2 = -9$ ). The parentheses are what hold the whole expression together inside the slot.

The argument of f is a slot. Whatever expression you write inside f( ) is copied into every slot of the rule, so f(x + h) squares the whole expression x + h. Here the rule is f(x) = x squared plus 1.

Worked example 1 Substituting whole expressions into the slot

Let $f(x) = x^2 + 1$ . Evaluate $f(-x)$ , $f(2x)$ , and $f(x + h)$ , keeping every input wrapped in parentheses.

For $f(-x)$ , put $-x$ in the slot and square the whole thing:

f(-x) = (-x)^2 + 1 = x^2 + 1.

For $f(2x)$ , the factor $2$ is inside the slot, so it is squared along with $x$ :

f(2x) = (2x)^2 + 1 = 4x^2 + 1.

For $f(x + h)$ , the entire sum $x + h$ is squared:

f(x + h) = (x + h)^2 + 1 = x^2 + 2xh + h^2 + 1.

Notice that $f(-x)$ came back equal to $f(x)$ here, while $f(2x)$ is $4x^2 + 1$ , not $2x^2 + 1$ . The slot squares whatever it holds, coefficients and all.

Check your understanding

If $f(x) = x^2 + 1$ , which expression equals $f(2x)$ ?

2x^2 + 2

4x^2 + 2

4x^2 + 1

2x^2 + 1

$f(a + b)$ is usually not $f(a) + f(b)$

Because the slot swallows the whole input, $f(a + b)$ squares the sum $a + b$ ; it does not add the separate outputs. With $f(x) = x^2$ ,

f(a + b) = (a + b)^2 = a^2 + 2ab + b^2, \qquad f(a) + f(b) = a^2 + b^2.

These differ by the cross term $2ab$ , which is almost never zero. A single pair of numbers makes the gap concrete: $f(2 + 3) = f(5) = 25$ , while $f(2) + f(3) = 4 + 9 = 13$ . Pulling a constant out fails the same way: $f(2x) = (2x)^2 = 4x^2$ , but $2f(x) = 2x^2$ . So $f$ is a rule, not a multiplier. Unless you have checked that the particular rule allows it, you may not distribute $f$ across a sum or slide a constant through it.

One family of rules does respect both moves: the straight-line rules through the origin, $f(x) = mx$ .

The rules $f(x) = mx$ add and scale, and scaling alone pins them down#

Take any rule of the form $f(x) = mx$ with $m$ a constant. For all inputs $a$ and $b$ , the distributive law gives

f(a + b) = m(a + b) = ma + mb = f(a) + f(b),

so the output of a sum is the sum of the outputs. For any constant $k$ ,

f(kx) = m(kx) = k(mx) = k\,f(x),

so scaling the input by $k$ scales the output by $k$ . Both identities hold for every rule $f(x) = mx$ .

The reverse direction holds too, and it is the scaling identity that delivers it. Suppose $f(kx) = k\,f(x)$ for every real $k$ and every $x$ . Put $x = 1$ : then $f(k) = k\,f(1)$ for all $k$ , which is exactly $f(x) = f(1)\,x$ , a rule of the form $mx$ with slope $m = f(1)$ . So the pair of identities does pin down the through-origin lines, with no extra assumption needed. Additivity by itself is weaker: the identity $f(a + b) = f(a) + f(b)$ alone does not force $f(x) = mx$ without a further hypothesis such as the graph having no breaks. Notice as well that adding a nonzero constant already destroys additivity: for $f(x) = mx + c$ ,

f(a) + f(b) = ma + mb + 2c, \qquad f(a + b) = ma + mb + c,

and these agree only when $c = 0$ . Even the lines that everyday language calls linear are additive only when they pass through the origin.

So the rules $f(x) = mx$ are exactly the rules that respect scaling, and each of them is additive as well. The one caution is that additivity on its own is weaker: it does not single out $f(x) = mx$ without an added assumption. That gap is why the scaling identity is the one that pins the rules down.

The difference quotient

Now that $f(x + h)$ is expressible, we can measure how much $f$ changes as the input moves from $x$ to $x + h$ . The change in output is $f(x + h) - f(x)$ , and dividing by the change in input $h$ gives the difference quotient

\frac{f(x + h) - f(x)}{h}.

Work it out for $f(x) = x^2$ . The numerator expands and the $x^2$ terms cancel:

f(x + h) - f(x) = (x + h)^2 - x^2 = (x^2 + 2xh + h^2) - x^2 = 2xh + h^2.

Every term left has a factor of $h$ , so divide:

\frac{2xh + h^2}{h} = \frac{h(2x + h)}{h} = 2x + h, \qquad h \ne 0.

Hold on to that restriction $h \ne 0$ , because it is the whole point. Cancelling the $h$ is a division, and you can only divide by $h$ when $h \ne 0$ . The original difference quotient $\frac{2xh + h^2}{h}$ is undefined at $h = 0$ , where it reads $0/0$ , while the simplified $2x + h$ is perfectly happy at $h = 0$ . The two expressions agree for every $h \ne 0$ but have different domains, so they are genuinely different expressions that happen to match wherever both are defined. This is exactly the effect you met in Chapter 1, where clearing a denominator could change which inputs are allowed. Here cancelling an $h$ does the same thing, and the next section names why it matters.

The quantity itself has a meaning worth naming. The points $(x, f(x))$ and $(x + h, f(x + h))$ sit on the graph of $f$ , and $\frac{f(x + h) - f(x)}{h}$ is the slope of the straight line through them, the secant line. It is the average rate of change of $f$ between the two inputs. A later course lets $h$ shrink toward $0$ to turn this secant slope into the slope of the curve at a single point, but that is calculus. Here we go only as far as the algebra, and the algebra is exactly the careful substitution of this lesson.

The difference quotient is the slope of the secant line through P = (x, f(x)) and Q = (x + h, f(x + h)). The run is h and the rise is f(x + h) - f(x), so the quotient equals the average rate of change of f from x to x + h.

Worked example 2 A difference quotient with a linear term

Simplify $\dfrac{f(x + h) - f(x)}{h}$ for $f(x) = x^2 + 3x$ , with $h \ne 0$ .

Substitute $x + h$ into the whole rule, wrapping it in parentheses everywhere:

f(x + h) = (x + h)^2 + 3(x + h) = x^2 + 2xh + h^2 + 3x + 3h.

Subtract $f(x) = x^2 + 3x$ . The $x^2$ and $3x$ terms cancel, leaving only terms with a factor of $h$ :

f(x + h) - f(x) = 2xh + h^2 + 3h.

Now divide by $h$ , which is allowed because $h \ne 0$ :

\frac{2xh + h^2 + 3h}{h} = 2x + h + 3.

The linear term contributes the constant $3$ to the answer, and the squared term contributes the $2x + h$ you saw for $x^2$ .

Worked example 3 A difference quotient for $f(x) = 1/x$

Simplify $\dfrac{f(x + h) - f(x)}{h}$ for $f(x) = \dfrac{1}{x}$ , with $h \ne 0$ .

Put $x + h$ in the slot, then combine the two fractions over the common denominator $x(x + h)$ :

f(x + h) - f(x) = \frac{1}{x + h} - \frac{1}{x} = \frac{x - (x + h)}{x(x + h)} = \frac{-h}{x(x + h)}.

Dividing by $h$ means multiplying by $\dfrac{1}{h}$ , and the $h$ in the numerator cancels the one in the denominator:

\frac{1}{h} \cdot \frac{-h}{x(x + h)} = \frac{-1}{x(x + h)}, \qquad h \ne 0.

The equality holds where both sides are defined, and that needs $h \ne 0$ , because you divided by $h$ to reach it. The equality also needs $x \ne 0$ and $x + h \ne 0$ , so that $f$ is defined at both $x$ and $x + h$ . The simplified $\dfrac{-1}{x(x + h)}$ is itself perfectly defined at $h = 0$ , where it equals $-\dfrac{1}{x^2}$ . That value of $h$ is the one the cancellation dropped, the same way clearing a denominator in Chapter 1 changed which inputs were allowed.

The same wrap-the-whole-input rule handles higher powers. For $f(x) = x^3$ , expand the cube $(x + h)^3 = x^3 + 3x^2 h + 3x h^2 + h^3$ , subtract $x^3$ , and divide by $h$ :

\frac{(x + h)^3 - x^3}{h} = \frac{3x^2 h + 3x h^2 + h^3}{h} = 3x^2 + 3xh + h^2, \qquad h \ne 0.

Nothing new happens: the whole input is cubed, and every surviving term carries a factor of $h$ .

Check your understanding

For $f(x) = x^2$ , simplify $\dfrac{f(x + h) - f(x)}{h}$ for $h \ne 0$ .

2x

2x + h

2x + h^2

2xh + h^2

When are two functions equal?

The difference quotient just handed us two expressions that agree everywhere they are both defined yet are not the same, because their domains differ. That is a clue about what equality of functions should mean, and it deserves a precise statement.

Two functions are equal exactly when their domains, codomains, and values all match#

Recall from Relations and Functions that a function is a set of input-output pairs, with no repeated first coordinate, together with a declared codomain. We prove that $f$ and $g$ are the same function if and only if three conditions all hold. The three are: the same domain, the same codomain, and $f(x) = g(x)$ at every input $x$ in that domain.

Suppose first that $f$ and $g$ are the same function. Then they are literally the same set of pairs and the same codomain. The domain of a function is the set of its first coordinates, so the two domains match. For each input $x$ the one pair with first coordinate $x$ is shared by both sets, so $f(x) = g(x)$ at every $x$ . Equal functions therefore agree on domain, codomain, and every value.

Conversely, suppose $f$ and $g$ have the same domain $D$ , the same codomain, and $f(x) = g(x)$ for every $x$ in $D$ . Then for each $x$ in $D$ the pair $(x, f(x))$ is the pair $(x, g(x))$ , so the two sets of pairs are identical, and the codomains match by assumption. Having the same pairs and the same codomain, $f$ and $g$ are the same function.

Both directions hold, so the equality of two functions is precisely the equality of their domains, their codomains, and all their values, no more and no less.

This is why $f(x) = \dfrac{x^2}{x}$ and $g(x) = x$ are different functions, not equal ones. For $x \ne 0$ the fraction cancels, $\dfrac{x^2}{x} = x$ , so the two rules agree wherever both are defined. But $x = 0$ lies in $g$ ‘s domain and not in $f$ ‘s, since $\dfrac{x^2}{x}$ reads $0/0$ there. Their domains differ, so by the test above they are not the same function. It is the same lesson the difference quotient taught: $2x + h$ and $\dfrac{2xh + h^2}{h}$ agree for $h \ne 0$ but part ways at $h = 0$ , and matching values on the overlap is not enough. This is also why every cancelling step in Chapter 1 had to record the inputs it removed.

Worked example 4 Same formula after cancelling, different function

Decide whether $f(x) = \dfrac{x^2 - 1}{x - 1}$ and $g(x) = x + 1$ are the same function.

Factor the numerator with the difference of squares, $x^2 - 1 = (x - 1)(x + 1)$ . For $x \ne 1$ the common factor cancels:

\frac{x^2 - 1}{x - 1} = \frac{(x - 1)(x + 1)}{x - 1} = x + 1.

So the two rules agree at every input except possibly $x = 1$ . Check that input directly. The denominator $x - 1$ is $0$ there, so $f(1)$ is $\dfrac{0}{0}$ , undefined, and $1$ is not in $f$ ‘s domain. But $g(1) = 2$ , so $1$ is in $g$ ‘s domain:

f(1) \ \text{is undefined}, \qquad g(1) = 2.

The domains differ, so $f$ and $g$ are different functions, even though $f$ simplifies to $g$ everywhere $f$ is defined.

Piecewise definitions

A piecewise definition gives a single function by splitting the input into cases and naming a rule on each. For example,

f(x) = \begin{cases} x^2 & \text{if } x < 0 \\ 2x + 1 & \text{if } x \ge 0 \end{cases}

is one function, not two. To evaluate it, first find which case the input falls in, then use that case’s rule. Since $-3 < 0$ , use the top rule: $f(-3) = (-3)^2 = 9$ . Since $4 \ge 0$ , use the bottom rule: $f(4) = 2(4) + 1 = 9$ . The boundary input $x = 0$ is caught by the second case, because $0 \ge 0$ is true while $0 < 0$ is false, so $f(0) = 2(0) + 1 = 1$ .

For a piecewise rule to define a function at all, the cases must do two things: cover the whole domain, and never assign one input two different outputs. If some input were left out of every case, it would get no output. And if two cases overlapped and disagreed on the overlap, that input would be handed two different outputs, which Relations and Functions forbids. Cases are allowed to overlap when they agree there. For instance $\lvert x \rvert$ can be written as $x$ for $x \ge 0$ and $-x$ for $x \le 0$ , which overlap at $0$ yet agree, since both give $0$ . The easy way to rule out any conflict is to make the cases disjoint, which is what this example does. Here $x < 0$ and $x \ge 0$ share no input and together cover every real number, so $f$ is well defined.

Worked example 5 Evaluating a piecewise rule

Evaluate $g(-3)$ , $g(0)$ , $g(1)$ , and $g(4)$ for

g(x) = \begin{cases} -x & \text{if } x < -1 \\ x^2 & \text{if } -1 \le x \le 1 \\ 2x - 1 & \text{if } x > 1 \end{cases}

For each input, pick the case whose condition it satisfies, then apply that rule.

At $x = -3$ , the condition $x < -1$ holds, so use $-x$ : $g(-3) = -(-3) = 3$ . At $x = 0$ , the condition $-1 \le x \le 1$ holds, so use $x^2$ : $g(0) = 0^2 = 0$ . At the boundary $x = 1$ , the middle case still applies, because $-1 \le 1 \le 1$ is true while $x > 1$ is false, so $g(1) = 1^2 = 1$ . At $x = 4$ , the condition $x > 1$ holds, so use $2x - 1$ :

g(-3) = 3, \qquad g(0) = 0, \qquad g(1) = 1, \qquad g(4) = 2(4) - 1 = 7.

The only care needed is at a boundary: read the inequalities to see which single case owns that point.

Check your understanding

Let $f(x) = x + 5$ for $x < 0$ and $f(x) = x^2$ for $x \ge 0$ . What is $f(0)$ ?

5

0

25

D undefined

Common mistakes

Reading $f(x)$ as multiplication. $f$ is a rule, not a number, so $f(x)$ means “apply $f$ to $x$ ,” never ” $f$ times $x$ .” There is nothing to multiply, and $f(3)$ is not $3f$ .
Dropping the parentheses when substituting. For $f(x) = x^2$ , the value $f(-x)$ is $(-x)^2 = x^2$ , not $-x^2$ . Wrap the whole input before applying the rule, or the sign or a coefficient goes wrong.
Distributing $f$ across a sum. In general $f(a + b) \ne f(a) + f(b)$ and $f(2x) \ne 2f(x)$ . Only the through-origin rules $f(x) = mx$ are additive and scale, and even $f(x) = mx + c$ with $c \ne 0$ fails.
Forgetting the $h \ne 0$ restriction. The difference quotient simplifies only after you divide by $h$ , which needs $h \ne 0$ . The simplified form and the original agree just where both are defined, not at $h = 0$ .
Calling two functions equal because the formulas match after cancelling. $\dfrac{x^2}{x}$ and $x$ agree for $x \ne 0$ but differ at $0$ , so they are different functions. Equality needs the same domain, not only the same simplified rule.
Leaving a gap or a conflict in piecewise cases. The cases must cover the whole domain and must never give one input two different outputs. Overlap is fine when the rules agree there; making the cases disjoint is just the easy way to guarantee no conflict.

Key takeaways

$f$ names the function (its rule plus its domain); $f(x)$ is the single output value at the input $x$ . The parentheses mark the input, not a product.
The argument is a slot: $f(2x)$ , $f(x + h)$ , $f(-x)$ , and $f(a)$ all mean copy the whole expression into every $x$ . The name of the input variable does not matter, so $f(t) = t^2 + 1$ and $f(x) = x^2 + 1$ define the same function.
In general $f(a + b) \ne f(a) + f(b)$ and $f(2x) \ne 2f(x)$ . The rules $f(x) = mx$ both add and scale, and the scaling identity $f(kx) = k\,f(x)$ characterizes them (set $x = 1$ to get $f(x) = f(1)\,x$ ); additivity by itself is weaker and does not.
The difference quotient $\dfrac{f(x + h) - f(x)}{h}$ simplifies for $h \ne 0$ to the slope of the secant line through $(x, f(x))$ and $(x + h, f(x + h))$ , the average rate of change of $f$ from $x$ to $x + h$ .
Two functions are equal exactly when they share a domain, a codomain, and every value. So $\dfrac{x^2}{x}$ and $x$ are different functions, because $0$ is in one domain and not the other.
A piecewise rule is one function given by cases. Evaluate it by choosing the case the input falls in, and make sure the cases cover the domain and never give one input two different outputs.

Function Notation and Evaluation

Learning goals

The function is $f$ ; its value is $f(x)$

The argument is a slot, not the letter $x$

$f(a + b)$ is usually not $f(a) + f(b)$

The difference quotient

When are two functions equal?

Piecewise definitions

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)

Function Notation and Evaluation

Learning goals

The function is fff; its value is f(x)f(x)f(x)

The argument is a slot, not the letter xxx

f(a+b)f(a + b)f(a+b) is usually not f(a)+f(b)f(a) + f(b)f(a)+f(b)

The difference quotient

When are two functions equal?

Piecewise definitions

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)

The function is $f$ ; its value is $f(x)$

The argument is a slot, not the letter $x$

$f(a + b)$ is usually not $f(a) + f(b)$