Functions and Function Notation

A function is a rule that assigns exactly one output to every allowed input. The rule may be expressed with a formula, table, graph, mapping, or description.

Learning objectives

You will identify functions, use function notation, determine domains and ranges, evaluate and interpret outputs, and distinguish input restrictions from output behavior.

The defining condition

A relation $f:A\to B$ is a function if every input in domain $A$ is paired with exactly one output in codomain $B$. Different inputs may share an output. One input may not have two outputs.

The notation

$$f(x)$$

means “the output of $f$ at input $x$.” It does not mean $f$ multiplied by $x$.

If

$$f(x)=2x^2-3,$$

then

$$f(-2)=2(-2)^2-3=5.$$

Domain and range

The domain is the set of permitted inputs. The range is the set of outputs actually produced.

For

$$g(x)=\frac1{x-4},$$

$x=4$ is excluded because it makes the denominator zero. For

$$h(x)=\sqrt{x+2},$$

real-valued inputs require $x\ge-2$.

Context can impose additional restrictions. A formula for cost may algebraically accept negative item counts while the physical model does not.

Representations

A graph represents a function when each vertical line intersects it at most once. This vertical-line test checks the single-output condition visually.

A table is a function if no input appears with conflicting outputs. Repeated outputs are allowed.

Inputs can be expressions

If $f(x)=x^2+1$, then

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

not $f(x)+h$. Function notation instructs us to replace every occurrence of the input variable.

A function is a dependency statement

Calling $y$ a function of $x$ asserts that choosing an allowed $x$ determines one and only one $y$. It does not assert that the rule has a simple formula. A temperature record, a lookup table, and a piecewise rule can all define functions.

The independent/dependent language is useful in applications, but domain and output are more precise. Time may be the input in a motion model because position is recorded at each time. In a different question, position might be the input used to infer time. The mathematical role follows the mapping being studied.

Determine the domain before calculating

For a real-valued formula, denominators cannot equal zero and even roots cannot receive negative radicands. Consider

$$p(x)=\frac{\sqrt{x+1}}{x-3}.$$

The radical requires $x\ge-1$, while the denominator requires $x\ne3$. The domain is

$$[-1,3)\cup(3,\infty).$$

These restrictions must be combined: satisfying only one condition is insufficient. A contextual domain can be narrower. If $x$ counts manufactured parts, negative and noninteger inputs may be excluded even when the formula accepts them.

Composition tracks a chain of dependence

If $g$ transforms an input and $f$ acts on the result, the composite is

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

Order matters. Let $f(x)=x^2$ and $g(x)=x+1$. Then

$$(f\circ g)(x)=(x+1)^2, \qquad (g\circ f)(x)=x^2+1.$$

The expressions differ because the operations occur in a different sequence. The domain of a composite contains inputs allowed by $g$ whose outputs are allowed by $f$.

Inverse functions reverse a mapping

An inverse function exists only when each output in the relevant range identifies exactly one input. The function $f(x)=x^2$ on all real numbers is not one-to-one because $f(2)=f(-2)$. Restricting its domain to $x\ge0$ produces an inverse, $f^{-1}(x)=\sqrt{x}$.

Do not confuse $f^{-1}(x)$ with $\frac{1}{f(x)}$. The superscript $-1$ names the inverse mapping, not a reciprocal. A correct inverse satisfies both $f^{-1}(f(x))=x$ on the original domain and $f(f^{-1}(x))=x$ on the original range.

Common mistakes

• Reading $f(x)$ as multiplication.
• Assuming every relation is a function.
• Confusing the codomain with the range.
• Ignoring domain restrictions.
• Substituting into only one occurrence of a variable.

Test Your Knowledge

1. For $f(x)=3x-7$, find $f(5)$ and solve $f(x)=8$.
2. State the real domain of $r(x)=\sqrt{5-x}$.
3. Is $\{(1,2),(2,2),(1,3)\}$ a function?
4. Find the domain of $q(x)=\frac{1}{\sqrt{x-2}}$.
5. If $f(x)=2x-1$ and $g(x)=x^2$, find $(f\circ g)(3)$ and $(g\circ f)(3)$.

Solutions

1. $f(5)=8$; solving $3x-7=8$ gives $x=5$.
2. $5-x\ge0$, so $x\le5$.
3. No. Input $1$ has outputs $2$ and $3$.
4. The square root requires $x-2\ge0$, but the denominator may not be zero. Therefore $x>2$, or $(2,\infty)$.
5. $(f\circ g)(3)=f(9)=17$, while $(g\circ f)(3)=g(5)=25$. The unequal results show that composition is order-dependent.

Connection forward

Graphs reveal how outputs change with inputs. Calculus will later formalize that change through limits and derivatives.

Sources

• OpenStax, College Algebra 2e, chapters on functions, composition, and inverse functions.
• National Council of Teachers of Mathematics, Principles and Standards for School Mathematics, algebra standard.