Section 4.1

Definition 43 (Relation)

A relation is a set of ordered pairs.

Definition 44 (Function)

A function is a relation in which, for each distinct value of the first component of the ordered pair, there is exactly one value of the second component.

Functions are a special type of relation. The following is a special condition on functions.

Definition 45 (One-to-one Functions)

A function \(f\) is one-to-one function if, for each element \(a\) and \(b\) in the domain of \(f\)

\[ a\ne b \implies f(a)\ne f(b)\]

That is, different values in the domain correspond to different range values.

  • The following set \(\{(1,4),(1,5),(2,6),(3,7)\}\) is a relation but not a function since \(1\to4\) and \(1\to 5\).

  • The following set \(\{(1,4),(2,4),(3,4)\}\) is a relation and a function, but not a one-to-one function since \(4\) corresponds to \(1\), \(2\), and \(3\).

  • The following set \(\{(1,4),(2,5),(3,6)\}\) is a relation, a function, and a one-to-one function.

Example 66

  • \(f(x)=x\) is a one-to-one function.

  • \(f(x)=mx+b\) is a one-to-one function.

  • \(f(x)=x^2\) is not a one-to-one function since \(1\ne-1\), \(f(-1)=(-1)^2=1\), and \(f(1)=(1)^2=1\).

  • \(f(x)=x^2\) with a domain of \([0,\infty)\) is a one-to-one function.

  • \(f(x)=x^3\) is a one-to-one function.

We have the following, like the vertical line test for determining if a graph is for a function.

Property 8

Horizontal line test for one-to-one function. A function is one-to-one if every horizontal line intersects the graph of the function at most once.

A function must be one-to-one in order for the function to have an inverse.

Definition 46

Let \(f\) be a one-to-one function. Then \(g\) is the inverse of \(f\) if \(f(g(x))=x\) for all \(x\) in the domain of \(g\) and \(g(f(x))=x\) for all \(x\) in the domain of \(f\).

Property 9

Let \(f\) and \(f^{-1}\) exists.

  • The domain of \(f\) is equal to the range of \(f^{-1}\).

  • The randge of \(f\) is equal to the domain of \(f^{-1}\).

  • The graph of \(y=f^{-1}(x)\) is the graph of \(y=f(x)\) but reflected about the line \(y=x\).

Example 67

Let \(f(x)=8x+5\) and \(g(x)=\frac{1}{8}x-\frac{5}{8}\). Show \(g\) is the inverse of \(f\).

Scratch work:

Solution:
\[\begin{align*} f(g(x)) & = 8\left(\frac{1}{8}x-\frac{5}{8}\right)+5\\ & = x-5+5\\ & = x \end{align*}\]
\[\begin{align*} g(f(x)) & = \frac{1}{8}\left(8x+5\right)-\frac{5}{8}\\ & = x+\frac{5}{8}-\frac{5}{8}\\ & = x \end{align*}\]
“Proof:”

Since \(f\) is a linear function we say \(f\) is a one-to-one function. Furthermore, since \(f(g(x))=x\) and \(g(f(x))=x\) we say \(g\) is the inverse of \(f\).

When \(g(x)\) is the inverse of \(f(x)\), we say, \(g(x)=f^{-1}(x)\).

Example 68

Let \(f(x)=2x+5\) and \(g(x)=\frac{1}{2}x-5\). Show \(g\) is not the inverse of \(f\).

Solution:
\[\begin{align*} f(g(x)) & = 2\left(\frac{1}{2}x-5\right)+5\\ & = x-\frac{5}{2}+5\\ & = \ne x \end{align*}\]

Since \(f(g(x))\ne x\) we know \(g\) is not the inverse of \(f\).