OBJECTIVES

When you finish this section, you should be able to:

1. Determine whether a function is one-to-one (p. 32)
2. Determine the inverse of a function defined by a set of ordered pairs (p. 34)
3. Obtain the graph of the inverse function from the graph of a one-to-one function (p. 35)
4. Find the inverse of a one-to-one function defined by an equation (p. 35)

spanDEFINITIONspan One-to-One Function

A function \(f\) is a one-to-one function if any two different inputs in the domain correspond to two different outputs in the range. That is, if \(x_{1} \neq x_2\), then \(f (x_{1}) \neq f (x_{2})\).

IN WORDS

A function is not one-to-one if there are two different inputs in the domain corresponding to the same output.

THEOREM Horizontal-line Test

The graph of a function in the \(xy\)-plane is the graph of a one-to-one function if and only if every horizontal line intersects the graph in at most one point.

Determining Whether a Function Is One-to-One

Determine whether each of these functions is one-to-one:

1. \(f(x) = x^{2}\)
2. \(g(x) = x^{3}\)

Solution (a) Figure 49 illustrates the Horizontal-line Test for the graph of \(f(x) = x^{2}\). The horizontal line \(y = 1\) intersects the graph of \(f\) twice, at \((1, 1)\) and at \((-1, 1)\), so \(f\) is not one-to-one.

Figure 49

(b) Figure 50 illustrates the horizontal-line test for the graph of \(g(x) = x^{3}\). Because every horizontal line intersects the graph of \(g\) exactly once, it follows that \(g\) is one-to-one.

Figure 50

NOW WORK

Problem 9.

THEOREM One-to-One Function

• A function that is increasing on an interval \(I\) is a one-to-one function on \(I\).
• A function that is decreasing on an interval \(I\) is a one-to-one function on \(I\).

NOTE

\({f}^{-1}\) is not the reciprocal function. That is, \(f^{-1} (x) \neq \dfrac{1}{f(x)}\). The reciprocal function \(\dfrac{1}{f(x)}\) is written \([f(x)]^{-1}\).

spanDEFINITIONspan Inverse Function

Let \(f\) be a one-to-one function. The inverse of \(f\), denoted by \(f^{-1}\), is the function defined on the range of \(f\) for which \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{x = f^{ - 1} (y)\quad {\rm{if}}\;{\rm{and}}\;{\rm{only}}\;{\rm{if}}\quad y = f{\rm{(}}x{\rm{)}}}} \]

Finding the Inverse of a Function Defined by a Set of Ordered Pairs

Find the inverse of the one-to-one function: \[ \{(-3,-5), (-1,1), (0,2), (1,3)\} \]

State the domain and the range of the function and its inverse.

Solution The inverse of the function is found by interchanging the entries in each ordered pair. The inverse is \[ \{(-5,-3), (1,-1), (2,0), ( 3,1)\} \]

The domain of the function is \(\{-3, -1,0, 1\}\); the range of the function is \(\{-5, 1,2,3\}\). The domain of the inverse function is \(\{-5, 1,2,3\}\); the range of the inverse function is \(\{ -3, -1,0,1\}\).

NOW WORK

Problem 19.

THEOREM

Given a one-to-one function \(f\) and its inverse function \(f^{-1}\), then \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{\begin{array}{*{20}c} { \bullet \;(f^{ - 1} \circ \;f)(x)\; = \;f^{ - 1} (f(x))\; = \;x} & {{\rm{where}}\;x\;{\rm{is}}\;{\mathop{\rm in}\nolimits} \;{\rm{the}}\;{\rm{domain}}\;{\rm{of}}\;f}\\ { \bullet \;(f \circ \;f^{ - 1} )(x)\; = \;f(f^{ - 1} (x))\; = \;x} & \;\;\;{{\rm{where}}\;x\;{\rm{is}}\;{\mathop{\rm in}\nolimits} \;{\rm{the}}\;{\rm{domain}}\;{\rm{of}}\;f^{ - 1} } \\ \end{array}}} \]

Verifying Inverse Functions

Verify that the inverse of \(f(x) = \dfrac{1}{x-1}\) is \(f^{-1}(x) = \dfrac{1}{x}+1\). For what values of \(x\) is \(f^{-1} (f( x)) = x?\) For what values of \(x\) is \(f (f^{-1}(x)) = x?\)

Solution The domain of \(f\) is \(\{x|x\neq 1\}\) and the domain of \(f^{\,-1}\) is \(\{ x|x\neq 0\}\). Now \[ \begin{array}{*{20}c} \;\;\;\;\;\;\;\;\;\;{f^{ - 1} (f(x)) = f^{ - 1} \left( {\frac{1}{{x - 1}}} \right) = \frac{1}{{\left( {\frac{1}{{x - 1}}} \right)}} + 1 = x - 1 + 1 = x,} & {{\rm{provided }}\;\;x \ne \;1} \\ {f(f^{ - 1} (x)) = f\left( {\frac{1}{x} + 1} \right) = \frac{1}{{\left( {\frac{1}{x} + 1} \right) - 1}} = \frac{1}{{\frac{1}{x}}} = x,} & {{\rm{provided}}\;x\; \ne 0} \\ \end{array} \]

NOW WORK

Problem 15.

THEOREM Symmetry of Inverse Functions

The graph of a one-to-one function \(f\) and the graph of its inverse function \(f^{-1}\) are symmetric with respect to the line \(y = x\).

Graphing the Inverse Function from the Graph of a Function

The graph in Figure 55 is that of a one-to-one function \(y = f(x)\). Draw the graph of its inverse function.

Figure 55

Solution Since the points \((-2,-1)\), \((-1,0)\), and \(( 2,1)\) are on the graph of \(f\), the points \((-1,-2)\), \(( 0,-1)\), and \(( 1,2)\) are on the graph of \(f^{\,-1}\). Using the points and the fact that the graph of \(f^{-1}\) is the reflection about the line \(y = x\) of the graph of \(f\), draw the graph of \(f^{-1}\), as shown in Figure 56.

Figure 56

NOW WORK

Problem 27.

Steps for Finding the Inverse of a One-to-One Function

• Step 1  Write \(f\) in the form \(y = f(x)\).
• Step 2  Interchange the variables \(x\) and \(y\) to obtain \(x = f(y)\) This equation defines the inverse function \(f^{-1}\) implicitly.
• Step 3  If possible, solve the implicit equation for \(y\) in terms of \(x\) to obtain the explicit form of \(f^{-1}\): \(y = f^{-1}( x)\).
• Step 4  Check the result by showing that \(f^{-1} (f(x)) = x\) and \(f( f^{-1}(x)) = x\).

Finding the Inverse Function

The function \(f(x) = 2x^{3}-1\) is one-to-one. Find its inverse.

Solution We follow the steps given above.

Step 1  Write \(f\) as \(y = 2x^{3}-1\).

Step 2  Interchange the variables \(x\) and \(y\). \[ \ x = 2y^{3}-1 \]

This equation defines \(f^{-1}\) implicitly.

Step 3 Solve the implicit form of the inverse function for \(y\). \begin{eqnarray*} x+1 &=&2y^{3} \\[5pt] y^{3} &=&\dfrac{x+1}{2} \\[5pt] y &=&\sqrt[3]{\dfrac{x+1}{2}}=f^{-1}( x) \end{eqnarray*}

Figure 57

Step 4  Check the result. \[ \begin{array}{c} \;\;\;\;\;\;\;\;\;\;f^{ - 1} (f(x)) = f^{ - 1} (2x^3 - 1) = \sqrt[3]{{\frac{{(2x^3 - 1) + 1}}{2}}} = \sqrt[3]{{\frac{{2x^3 }}{2}}} = \sqrt[3]{{x^3 }} = x \\ f(f^{ - 1} (x)) = f\left( {\sqrt[3]{{\frac{{x + 1}}{2}}}} \right) = 2\left( {\sqrt[3]{{\frac{{x + 1}}{2}}}} \right)^3 - 1 = 2\left( {\frac{{x + 1}}{2}} \right) - 1 \\ \;\;\;\;\;\;\;\;\;= x + 1 - 1 = x \\ \end{array} \]

NOW WORK

Problem 31.

Finding the Inverse of a Domain-Restricted Function

Find the inverse of \(f(x) = x^{2}\) if \(x\geq 0\).

Figure 58

Solution The function \(f(x) = x^{2}\) is not one-to-one (see Example 1(a)). However, by restricting the domain of \(f\) to \(x\geq 0\), the new function \(f\) is one-to-one, so \(f^{-1}\) exists. To find \(f^{-1}\), follow the steps.

• Step 1  \(y=x^{2}\), where \(x\geq 0\).
• Step 2  Interchange the variables \(x\) and \(y\): \(x = y^{2}\), where \(y\geq 0\). This is the inverse function written implicitly.
• Step 3  Solve for \(y\): \(y=\sqrt{x}=f^{-1}(x)\). (Since \(y\geq 0\), only the principal square root is obtained.)
• Step 4  Check that \(f^{-1}(x) = \sqrt{x}\) is the inverse function of \(f\). \[ \begin{array}{rcl@{\qquad}l} f^{-1}( f( x))~&=&\sqrt{f( x) }=\sqrt{x^{2}}=\vert x\vert =x, & \hbox{where }x\geq 0 \\[2pt] f( f^{-1}( x))~&=&[ f^{-1}( x) ] ^{2}=[ \sqrt{x}] ^{2}=x, & \hbox{where }x\geq 0 \end{array} \]

NEED TO REVIEW?

Principal roots are discussed in Appendix A.1, p. A-9.

NOW WORK

Problem 37.