What Does \(f(a)\) Mean?
In the previous tutorial, a function \(f:A\to B\) was described by its domain \(A\), its codomain \(B\), and the outputs it attains. Function notation gives a concise way to name the output assigned to a particular input. If \(a\in A\), then \(f(a)\) means the value of the function \(f\) at the input \(a\). It is an element of \(B\), because a function from \(A\) to \(B\) assigns every input in \(A\) an output in \(B\).
The letter \(f\) is the name of the function. The expression inside the parentheses indicates the input at which we are evaluating it. Thus \(f(a)\) is read “\(f\) of \(a\)” or “the value of \(f\) at \(a\).” The parentheses do not indicate multiplication: in general, \(f(a)\) is not the product of \(f\) and \(a\). A function is a mapping, not a number to be multiplied by its input.
The symbol inside the parentheses need not be the letter \(a\). If \(x\in A\), then \(f(x)\) denotes the output at input \(x\); if \(r\in A\), then \(f(r)\) denotes the output at input \(r\). The letter used for an input is a placeholder. What matters is which object is supplied as the argument and whether it belongs to the domain.
Function Notation and Ordered Pairs
Earlier in this course, a function was also treated as a relation consisting of ordered pairs. Its graph records an input together with the output assigned to that input. Function notation and ordered-pair notation describe the same assignment in two different forms: saying \(f(a)=b\) says that the graph of \(f\) contains the ordered pair \((a,b)\).
Proposition (Function Values and Graph Pairs). Let \(f:A\to B\), let \(a\in A\), and let \(b\in B\). Then $$ f(a)=b\quad\Longleftrightarrow\quad (a,b)\in f, $$ where \(f\) is identified with its graph, as in the definition of a function as a mapping.
Proof. Suppose first that \(f(a)=b\). The graph of \(f\) consists of the ordered pairs \((x,f(x))\) for inputs \(x\in A\). In particular, since \(a\in A\), the pair \((a,f(a))\) belongs to the graph. Because \(f(a)=b\), this pair is \((a,b)\), so \((a,b)\in f\).
Conversely, suppose that \((a,b)\in f\). Since \(f\) is the graph of a function with domain \(A\), membership of \((a,b)\) in the graph means that the output assigned to the input \(a\) is \(b\). By the meaning of function notation, that statement is \(f(a)=b\). This proves both implications. \(\square\)
The proposition gives a practical translation. An assignment written \(f(a)=b\) can be recorded as the ordered pair \((a,b)\); a pair in the graph can be read as a function value. This equivalence does not change the domain or codomain. It only changes how one records a particular input-output assignment.
Worked Example: Reading Assignments from a Graph
Let \(A=\{p,q,r\}\), \(B=\{4,7,9\}\), and let the graph of \(f:A\to B\) be $$ f=\{(p,7),(q,4),(r,9)\}. $$ The pair \((q,4)\) belongs to the graph, so the Function Values and Graph Pairs Proposition gives \(f(q)=4\). Similarly, \((p,7)\in f\) gives \(f(p)=7\), and \((r,9)\in f\) gives \(f(r)=9\). In particular, \(f(q)\) is \(4\), not \(q\), and the pair \((q,4)\) places the input first and its output second.
Evaluating a Rule at an Input
A function may be described by a formula together with its domain and codomain. To evaluate the function at a particular input, replace the input variable in the formula with the stated argument, then simplify the resulting expression. Parentheses are important when the argument itself is an expression: they make clear that the entire argument is substituted.
For example, if \(g(x)=x^2-3x\), then evaluating at \(x=2\) means $$ g(2)=2^2-3(2)=4-6=-2. $$ The notation \(g(2)\) is the output, and the calculation determines that output. The expression \(x^2-3x\) is the rule used in the calculation; it is not itself a function value until an input has been specified.
Worked Example: Evaluating at a Negative Input
Let \(h:\mathbb R\to\mathbb R\) be defined by \(h(x)=2x^2+x-1\). To find \(h(-3)\), substitute the whole input \(-3\) for \(x\): $$ h(-3)=2(-3)^2+(-3)-1 =2(9)-3-1 =18-4 =14. $$ The parentheses show that the input is \(-3\), so its square is \((-3)^2=9\). The value \(h(-3)\) is therefore \(14\), which belongs to the stated codomain \(\mathbb R\).
Worked Example: Substituting an Expression as the Input
Let \(p:\mathbb R\to\mathbb R\) be defined by \(p(t)=3t+5\). Find \(p(u-2)\), where \(u\in\mathbb R\). The argument is the entire expression \(u-2\), so replace each occurrence of \(t\) with \((u-2)\): $$ p(u-2)=3(u-2)+5 =3u-6+5 =3u-1. $$ Thus the output at input \(u-2\) is \(3u-1\). The result is an expression because the input \(u-2\) depends on the variable \(u\); for each real \(u\), that expression gives the corresponding value of \(p\).
It is useful to distinguish an input variable from a specific input. In a rule such as \(p(t)=3t+5\), the letter \(t\) stands for an arbitrary element of the domain. When we write \(p(4)\), we have chosen one particular input and substitute \(4\) for \(t\). The variable name can be changed without changing the assignment rule: for example, \(p(x)=3x+5\) describes the same formula and inputs as \(p(t)=3t+5\), provided the domain and codomain remain the same.
Finite Tables as Function Notation
For a finite domain, a table can specify all the function values. Each row states one assignment: the first entry is an input, and the entry beside it is the value of the function there. To describe a function from \(A\) to \(B\), every input in \(A\) must appear exactly once, and each listed output must belong to \(B\). If any input is omitted, the table has not assigned a value to every member of the domain. If an input is assigned two different outputs, the function condition fails.
Theorem (Finite Table Criterion). Let \(A=\{a_1,\ldots,a_n\}\), where \(n\geq1\) and the \(a_i\) are distinct, and let \(B\) be a set. A list of assignments $$ a_1\mapsto b_1,\quad a_2\mapsto b_2,\quad\ldots,\quad a_n\mapsto b_n $$ defines a function \(f:A\to B\) if and only if \(b_i\in B\) for every \(i\). When it defines a function, that function is unique.
Proof. First suppose the assignments define a function \(f:A\to B\). By the definition of a function from \(A\) to \(B\), the output assigned to each input in \(A\) belongs to \(B\). Since the output assigned to \(a_i\) is \(b_i\), we have \(b_i\in B\) for every \(i\).
Conversely, suppose \(b_i\in B\) for every \(i\). Form the set of ordered pairs $$ R=\{(a_1,b_1),\ldots,(a_n,b_n)\}. $$ Each first coordinate belongs to \(A\), and each second coordinate belongs to \(B\). Every element of \(A\) is one of the listed, distinct inputs, so it appears as a first coordinate. Because the inputs are distinct, each appears in exactly one listed pair, and that pair gives it exactly one output. Thus \(R\) assigns every member of \(A\) exactly one element of \(B\). By the definition of a function, \(R\) is the graph of a function \(f:A\to B\) with the stated assignments.
To verify uniqueness, suppose \(g:A\to B\) is another function with the same listed assignments. Each element of \(A\) is some \(a_i\), and both functions assign that input the value \(b_i\). Therefore \(f(a)=g(a)\) for every \(a\in A\). The Pointwise Equality Criterion established earlier in this course gives \(f=g\). This proves existence and uniqueness when every \(b_i\in B\). \(\square\)
The theorem makes explicit what a table must contain to specify a mapping. It also shows that the table's outputs need not be distinct. Different inputs may be assigned the same output; the function condition requires one output for each input, not a different output for every input.
Worked Example: Checking a Finite Table
Let \(A=\{m,n,s\}\) and \(B=\{\alpha,\beta\}\). Consider the assignments $$ m\mapsto\alpha,\qquad n\mapsto\beta,\qquad s\mapsto\alpha. $$ The inputs \(m,n,s\) are exactly the distinct elements of \(A\), and each assigned output is in \(B\). The Finite Table Criterion therefore gives a unique function \(f:A\to B\). In function notation, $$ f(m)=\alpha,\qquad f(n)=\beta,\qquad f(s)=\alpha. $$ The repeated output \(\alpha\) causes no difficulty: it is assigned to two different inputs, and each input still has exactly one assigned value.
Notation Does Not Replace the Mapping
A formula by itself does not always tell the entire story of a function. The same formula can be used with different domains or codomains, and these sets are part of the stated mapping. For example, the formula \(x\mapsto x^2\) can be used on the domain \(\{0,1,2\}\) or on the domain \(\{1,2\}\). The first mapping has the value at \(0\) as well as the values at \(1\) and \(2\); the second does not have an input \(0\). Therefore, before evaluating \(f(a)\), read the mapping specification and confirm that \(a\) belongs to its domain.
The rule and the output are also different kinds of information. In \(f(x)=x^2+1\), the right-hand side describes how to find the output for an input \(x\). In \(f(3)\), the argument is fixed, and the expression denotes the output at that input. Substituting gives \(f(3)=3^2+1=10\). The formula is not the number \(10\); \(10\) is the function value for the particular input \(3\).
When a mapping has been defined only on a specific domain, do not use \(f(a)\) for an object outside that domain unless a separate definition extends the function. For instance, if \(f:\{1,2,3\}\to\mathbb R\) is given by \(f(k)=k+4\), then \(f(2)=6\) is defined. Although the formula \(k+4\) can be evaluated at \(k=10\), \(f(10)\) is not a value of this specified function, because \(10\notin\{1,2,3\}\).
Check Your Understanding
- In \(g(r)\), what does \(r\) represent, and what does the whole expression \(g(r)\) denote?
- Suppose \(h(x)=x^2-2x\). Evaluate \(h(5)\), showing the substitution and simplification.
- If \((c,8)\) belongs to the graph of a function \(f\), what function-notation statement follows?
- Let \(p:\{a,b\}\to\{0,1\}\) be specified by \(p(a)=1\) and \(p(b)=1\). Does the repeated output prevent these assignments from defining a function? Explain.
- Let \(q:\{1,2\}\to\mathbb R\) be given by \(q(x)=x+3\). Is \(q(4)\) a value of this specified function? Explain.