15 Functions

15.1 Fundamental Concepts and Definitions

Much of this chapter is taken from the lectures of Dr. John Brunette, University of Southern Maine

A function is a collection of ordered pairs in which no two pairs have the same first element.

The set of all first members of the pairs is called the domain.

The set of all second members of the pairs is called the range.

Suppose now that for any function \(f\) we have two items \(x\) and \(y\) such that \(x\in dom(f)\) and \(y\in ran(x)\) where \(dom(f)\) and \(ran(f)\) denote the domain and range of \(f\), respectively. It is said that \(f\) maps \(x\) onto \(y\), written

\[f:\ x\mapsto y\]

It is common to write the \(ran(f)\) as some expression of \(x\). For example, \(f: x\mapsto x^2\) takes each element in the domain, and pairs it with it’s square. The common shorthand for this is \(f(x)=x^2\), meaning that whatever appears between the parentheses following the \(f\) is to be squared.

15.1.1 Function Operations

The three basic operations that can be performed on functions are addition, multipilication, and composition. For any two functions \(f\) and \(g\) these operations are defined as:

Addition \(\lbrack f+g\rbrack(x)=:\big\{\big(x,f(x)+g(x)\big)\mid x\in dom(f)\cap dom(g)\big\}\)
Multiplication \(\lbrack f\cdot g\rbrack(x):=\big\{\big(x,f(x)\cdot g(x)\big) \mid x\in dom(f)\cap dom(g)\big\}\)
Composition \(\lbrack f\circ g\rbrack(x)=\big\{\big(x,f\big(g(x)\big) \mid x\in dom(g)\) and \(g(x)\in dom(f)\big\}\)

Notice that the composition \(\lbrack f\circ g\rbrack(x)=f \circ g: g(x)\mapsto f(x)\). In other words, the result of \(g\) is then applied to \(f\) to produce the result of the composition.

15.2 Identities and Inverses

Recall that addition and multiplication have identity properties. Specifically, for any real number \(x\), applying one of these identities returns the value \(x\), i.e. \(x+0=x\) and \(x\cdot 1\)=x. Functions also have an identity, denoted \(id(x)\), that is defined as

\[id:\ x\mapsto x\]

Furthermore, the composition of \(id\) with \(f\) behaves in this way:

\[id\circ f=f\circ id=f\]

Functions may also exhibit the property of inverses that are exhibited by addition and multiplication. In the latter two, combining any real number \(x\) and its inverse returns the identity of that operation, i.e. \(x+-x=0\) and \(x\cdot x^{-1}=1,\ x\neq 0\). Likewise, some functions have an inverse function. If a function \(f\) has an inverse \(f^{-1}\), then

\[f\circ f^{-1}=f^{-1}\circ f=id\]

On closer observation, we see

\[f^{-1}\circ f\big(dom(x)\big)=f^{-1}\Big(f\big(dom(x)\big)\Big)=f^{-1}\big(ran(x)\big)=dom(x)\]

So \(f^{-1}\) must be the set of all ordered pairs \((y,x)\) where \(x\in dom(x)\) and \(y\in ran(x)\), i.e. \(f^{-1}(x)=\{(y,x) \mid x\in dom(x)\) and \(y\in ran(x)\}\). By the definition of functions, no two first elements in \(f^{-1}\) can be the same. But the first elements in \(f^{-1}\) are the second elements in \(f\). So \(f^{-1}\) only exists if no two second elements in \(f\) are the same. We thus make the following definition:

A function \(f\) is called a one-to-one function if it has no two ordered pairs with the same second element.

For any one-to-one function \(f\), no two of the first elements are the same, and no two of the second elements are the same. Thus, \(f^{-1}\) is a function, because no two of its first elements are the same, and because the range of \(f^{-1}\) is the domain of \(f\), no two second elements in \(f^{-1}\) are the same, and \(f^{-1}\) is a one-to-one function. Thus, every one-to-one function has an inverse.

If a function \(f\) is not one-to-one, however, then there exist two pairs in \(f\) that have the same second element. The inverse \(f^{-1}\) therefore has two pairs where the first element is the same. When such is the case, the definition of a function is violated, and \(f^{-1}\) cannot be a function. Thus, if a function is invertible, it must be one-to-one.

15.3 Odd and Even Functions

A function is said to be even if for any real number \(x,\ f(-x)=f(x)\).

A function is said to be odd if for any real number \(x,\ f(-x)=-f(x)\).

If neither of these criteria are met, the function is simply said to be neither odd nor even.

15.4 Theorems

15.4.1 Operations on Even Functions

Let \(f\) and \(g\) both be even functions. Then:

  1. \([f+g](x)\) is an even function
  2. \([f\cdot g](x)\) is an even function
  3. \([f\circ g](x)\) is an even function.

Proof:

  1. \[\begin{aligned} [f+g](-x) &= f(-x)+g(-x) \\ &= f(x)+g(x) \\ &= [f+g](x) \end{aligned}\]

so \([f+g](x)\) is an even function.

  1. \[\begin{aligned} [f\cdot g](-x) &= f(-x)\cdot g(-x) \\ &= f(x)\cdot g(x) \\ &= [f\cdot g](x) \end{aligned}\]

so \([f\cdot g](x)\) is an even function.

  1. \[\begin{aligned} [f\circ g](-x) &= f\big(g(-x)\big) \\ &= f\big(g(x)\big) \\ &= [f\circ g](x) \end{aligned}\]

so \([f\circ g](x)\) is an even function.

15.4.2 Operations on Odd Functions

Let \(f\) and \(g\) both be odd functions. Then:

  1. \([f+g](x)\) is an odd function
  2. \([f\cdot g](x)\) is an even function
  3. \([f\circ g](x)\) is an odd function.

Proof:

  1. \[\begin{aligned} [f+g](-x) &= f(-x) + g(-x) \\ &= -f(x) - g(x) \\ &= -[f+g](x) \end{aligned}\]

so \([f+g](x)\) is an odd function.

  1. \[\begin{aligned} [f\cdot g](-x) &= f(-x)\cdot g(-x) \\ &= -f(x)\cdot -g(x) \\ &= f(x)\cdot g(x) \\ &= [f\cdot g](x) \end{aligned}\]

so \([f\cdot g](x)\) is an even function.

  1. \[\begin{aligned} [f\circ g](-x) &= f\big(g(-x)\big) \\ &= f\big(-g(x)\big) \\ &= -f\big(g(x)\big) \\ &= -[f\circ g](x) \end{aligned}\]

so \([f\circ g](x)\) is an odd function.

15.4.3 Operations on an Odd and Even Function

Let \(f\) be an even function and let \(g\) both be an odd function. Then:

  1. \([f+g](x)\) is neither an odd nor an even function
  2. \([f\cdot g](x)\) is an odd function
  3. \([f\circ g](x)\) is an even function
  4. \([g\circ f](x)\) is an even function.

Proof:

  1. \[\begin{aligned} [f+g](-x) &= f(-x) + g(-x) \\ &= -f(x) - g(x) \end{aligned}\]

so \([f+g](x)\) is neither an odd nor an even function.

  1. \[\begin{aligned} [f\cdot g](-x) &= f(-x)\cdot g(-x) \\ &= f(x)\cdot -g(x) \\ &= -\big(f(x)\cdot g(x)\big) \\ &= -[f\cdot g](x) \end{aligned}\]

so \([f\cdot g](x)\) is an odd function.

  1. \[\begin{aligned} [f\circ g](-x) &= f\big(g(-x)\big) \\ &= f\big(-g(x)\big) \\ &= f\big(g(x)\big) \\ &= [f\circ g](x) \end{aligned}\]

so \(\lbrack f\circ g\rbrack(x)\) is an even function.

  1. \[\begin{aligned} [g\circ f](-x) &= g\big(f(-x)\big) \\ &= g\big(f(x)\big) \\ &= [g\circ f](x) \end{aligned}\]

so \(\lbrack f\circ g\rbrack(x)\) is an even function. \end{itemize}

15.4.4 Derivatives and Anti-derivatives of Odd Functions

Let \(f\) be an odd function and let \(f^\prime\) and \(F\) denote the derivative and anti-derivative of \(f\), respectively. Then \(f^\prime\) and \(F\) are both even functions.

Proof:

\[\begin{aligned} f(-x) &= -f(x)\\ \Rightarrow \frac{d}{dx}\big\lbrack f(-x)\big\rbrack &= \frac{d}{dx}\big\lbrack-f(x)\big\rbrack \\ \Rightarrow f^\prime(-x)\cdot -1 &= -f^\prime(x) \\ \Rightarrow -f^\prime(-x) &= -f^\prime(x) \\ \Rightarrow f^\prime(-x) &= f^\prime(x) \end{aligned}\]

So \(f^\prime\) is an even function.

\[\begin{aligned} f(-x) &= -f(x) \\ \Rightarrow \int f(-x) &= \int-f(x)\\ \Rightarrow F(-x)\cdot-1 &= -F(x)\\ \Rightarrow -F(-x) &= -F(x)\\ \Rightarrow F(-x) &= F(x) \end{aligned}\]

So \(F\) is also an even function.  

15.4.5 Derivatives and Anti-derivatives of Even Functions

Let \(g\) be an even function, and let \(g^\prime\) and \(G\) denote the derivative and anti-derivative of \(g\), respectively. Then \(g^\prime\) and \(G\) are both odd functions.

Proof:

\[\begin{aligned} g(-x) &= g(x) \\ \Rightarrow \frac{d}{dx}\big\lbrack g(-x)\big\rbrack &= \frac{d}{dx}\big\lbrack g(x)\big\rbrack \\ \Rightarrow g^\prime(-x)\cdot -1 &= g^\prime(x) \\ \Rightarrow -g^\prime(-x) &= g^\prime(x) \\ \Rightarrow g^\prime(-x) &= -g^\prime(x) \end{aligned}\]

So \(g^\prime\) is an odd function.

\[\begin{aligned} g(-x) &= g(x)\\ \Rightarrow \int g(-x) &= \int g(x)\\ \Rightarrow G(-x)\cdot-1 &= G(x)\\ \Rightarrow -G(-x) &= G(x)\\ \Rightarrow G(-x) &= -G(x) \end{aligned}\]

So \(G\) is also an odd function.