2.8 One-to-One Functions and Their Inverses
2.8.1 Learning Objectives
By the end of this section, you should be able to:
- determine whether a function is one-to-one;
- apply the horizontal line test;
- restrict a function’s domain so that it has an inverse;
- find and verify an inverse function algebraically; and
- relate the graphs, domains, and ranges of inverse functions.
2.8.2 One-to-One Functions
A function is one-to-one if different inputs always produce different outputs. Equivalently, if \(f(x_1)=f(x_2)\), then \(x_1=x_2\). A one-to-one function pairs each output with exactly one input, allowing the input-output process to be reversed.
Example 2.39 Testing a Linear Function Algebraically
Show that \(f(x)=4x-7\) is one-to-one.
Solution
Assume \(f(x_1)=f(x_2)\). Then
\[ 4x_1-7=4x_2-7. \]
Adding \(7\) and dividing by \(4\) gives \(x_1=x_2\). Therefore,
\[ \boxed{f\text{ is one-to-one}}. \]
Exercise 2.38 Is \(f(x)=x^2\), with domain all real numbers, one-to-one?
Show answer
Answer: No. For example, \(f(2)=f(-2)=4\), although \(2\ne-2\).
2.8.3 The Horizontal Line Test
A graph represents a one-to-one function exactly when every horizontal line intersects it at most once. This is the horizontal line test. The vertical line test decides whether a graph is a function; the horizontal line test decides whether that function is one-to-one.
Example 2.40 Using a Graph to Test One-to-One Behaviour
Is \(f(x)=|x|\) one-to-one on the real numbers?
Solution
Any horizontal line \(y=c\), where \(c>0\), intersects the graph twice: at \(x=c\) and \(x=-c\). Therefore,
\[ \boxed{f(x)=|x|\text{ is not one-to-one on }\mathbb R}. \]
Exercise 2.39 Use the horizontal line test to decide whether \(f(x)=x^3\) is one-to-one.
Show answer
Answer: Yes. Every horizontal line intersects the graph at most once.
2.8.4 Restricting a Domain
A function that is not one-to-one on its full domain may become one-to-one when its domain is restricted. For example, \(f(x)=x^2\) is one-to-one on \(x\ge0\) and also on \(x\le0\), but the two restrictions lead to different inverse functions.
Example 2.41 Restricting a Quadratic
Restrict \(f(x)=(x-3)^2+1\) so that it is one-to-one and increasing.
Solution
The vertex is \((3,1)\). To keep the increasing half of the parabola, restrict the domain to \(x\ge3\).
\[ \boxed{\text{Restricted domain }[3,\infty)} \]
Exercise 2.40 Restrict \(f(x)=(x+2)^2-4\) so that it is one-to-one and decreasing.
Show answer
Answer: Use the domain \(\boxed{(-\infty,-2]}\).
2.8.5 Inverse Functions
If \(f\) is one-to-one, its inverse function, written \(f^{-1}\), reverses the action of \(f\):
\[ f^{-1}(f(x))=x \]
for inputs in the domain of \(f\), and
\[ f(f^{-1}(x))=x \]
for inputs in the domain of \(f^{-1}\). The notation \(f^{-1}(x)\) does not mean \(1/f(x)\).
Example 2.42 Reversing a Linear Rule
If \(f(x)=3x-5\), explain how its inverse reverses the operations.
Solution
The function multiplies by \(3\) and then subtracts \(5\). To reverse it, add \(5\) and divide by \(3\):
\[ \boxed{f^{-1}(x)=\frac{x+5}{3}}. \]
Exercise 2.41 Find the inverse of \(f(x)=5x+2\).
Show answer
Answer: \(\boxed{f^{-1}(x)=\dfrac{x-2}{5}}\)
2.8.6 Finding an Inverse Algebraically
To find an inverse:
- write \(y=f(x)\);
- interchange \(x\) and \(y\);
- solve for \(y\); and
- write \(y=f^{-1}(x)\).
Example 2.43 Finding the Inverse of a Cubic Transformation
Find the inverse of \(f(x)=(x-2)^3+1\).
Solution
\[ y=(x-2)^3+1. \]
Interchange \(x\) and \(y\):
\[ x=(y-2)^3+1. \]
Now solve for \(y\):
\[ \begin{aligned} x-1&=(y-2)^3,\\ \sqrt[3]{x-1}&=y-2,\\ y&=\sqrt[3]{x-1}+2. \end{aligned} \]
\[ \boxed{f^{-1}(x)=\sqrt[3]{x-1}+2} \]
Exercise 2.42 Find the inverse of \(f(x)=\dfrac{x+4}{2}\).
Show answer
Answer: \(\boxed{f^{-1}(x)=2x-4}\)
2.8.7 Verifying Inverses
Two functions are inverses if both compositions simplify to the identity function \(x\), on the appropriate domains.
Example 2.44 Checking Both Compositions
Verify that \(f(x)=2x+7\) and \(g(x)=(x-7)/2\) are inverses.
Solution
\[ f(g(x))=2\left(\frac{x-7}{2}\right)+7=x \]
and
\[ g(f(x))=\frac{(2x+7)-7}{2}=x. \]
Therefore,
\[ \boxed{g=f^{-1}}. \]
Exercise 2.43 Verify one composition for \(f(x)=x-6\) and \(g(x)=x+6\).
Show answer
Answer: \(f(g(x))=(x+6)-6=x\), so this composition confirms the inverse relationship.
2.8.8 Graphs, Domains, and Ranges of Inverses
The graph of \(f^{-1}\) is the reflection of the graph of \(f\) across the line \(y=x\). Consequently:
- if \((a,b)\) lies on \(f\), then \((b,a)\) lies on \(f^{-1}\);
- the domain of \(f\) is the range of \(f^{-1}\); and
- the range of \(f\) is the domain of \(f^{-1}\).
Example 2.45 Inverting a Restricted Quadratic
Let \(f(x)=x^2+4\) with domain \(x\ge0\). Find its inverse and state both domains and ranges.
Solution
Starting with \(y=x^2+4\), interchange \(x\) and \(y\):
\[ x=y^2+4. \]
Since the original domain requires non-negative outputs from the inverse,
\[ y=\sqrt{x-4}. \]
Thus,
\[ \boxed{f^{-1}(x)=\sqrt{x-4}}. \]
The domain and range swap:
\[ \boxed{D_f=[0,\infty),\ R_f=[4,\infty),\ D_{f^{-1}}=[4,\infty),\ R_{f^{-1}}=[0,\infty)}. \]
Exercise 2.44 If the point \((-2,5)\) lies on a one-to-one function \(f\), what point lies on \(f^{-1}\)?
Show answer
Answer: \(\boxed{(5,-2)}\)
2.8.9 Conceptual Takeaways
- A one-to-one function pairs every output with only one input.
- The horizontal line test is the graphical test for one-to-one behaviour.
- Restricting a domain can make a function invertible.
- An inverse reverses a function; it is not the reciprocal of the function.
- Inverse graphs reflect across \(y=x\), and their domains and ranges exchange roles.
2.8.10 Skills You Should Be Able to Do
- Determine algebraically or graphically whether a function is one-to-one.
- Choose a suitable domain restriction for a non-one-to-one function.
- Find an inverse function algebraically.
- Verify inverse functions by composition.
- Relate points, graphs, domains, and ranges of inverse functions.
2.8.11 Practice Problems with Solutions
Determine whether \(f(x)=7x+1\) is one-to-one.
Show Solution
A non-constant linear function has no repeated outputs. Algebraically, \(7x_1+1=7x_2+1\) implies \(x_1=x_2\). Thus,
\[ \boxed{f\text{ is one-to-one}}. \]
Explain why \(f(x)=x^2-4\) is not one-to-one on \(\mathbb R\).
Show Solution
Opposite inputs give equal outputs. For example,
\[ f(2)=0=f(-2). \]
Since \(2\ne-2\), \(\boxed{f\text{ is not one-to-one}}\).
Restrict the domain of \(f(x)=x^2-4\) so that it is one-to-one and increasing.
Show Solution
The increasing half begins at the vertex \(x=0\). Use
\[ \boxed{[0,\infty)}. \]
Find the inverse of \(f(x)=4x-9\).
Show Solution
Write \(y=4x-9\), exchange \(x\) and \(y\), and solve:
\[ x=4y-9,\qquad y=\frac{x+9}{4}. \]
Therefore, \(\boxed{f^{-1}(x)=\dfrac{x+9}{4}}\).
Find the inverse of \(f(x)=\dfrac{3x+2}{5}\).
Show Solution
\[ x=\frac{3y+2}{5} \]
gives \(5x=3y+2\), so
\[ \boxed{f^{-1}(x)=\frac{5x-2}{3}}. \]
Verify that \(f(x)=x^3-2\) and \(g(x)=\sqrt[3]{x+2}\) are inverses.
Show Solution
\[ f(g(x))=\left(\sqrt[3]{x+2}\right)^3-2=x \]
and
\[ g(f(x))=\sqrt[3]{x^3-2+2}=x. \]
Hence, \(\boxed{g=f^{-1}}\).
Let \(f(x)=(x-1)^2+3\) with domain \(x\ge1\). Find \(f^{-1}\) and state its domain.
Show Solution
Exchange \(x\) and \(y\):
\[ x=(y-1)^2+3. \]
Since \(y\ge1\), choose the positive square root:
\[ y-1=\sqrt{x-3}. \]
Thus,
\[ \boxed{f^{-1}(x)=1+\sqrt{x-3},\quad \text{domain }[3,\infty)}. \]
A temperature conversion is \(F(C)=\frac95C+32\). Find the inverse and explain its meaning.
Show Solution
Begin with \(y=\frac95x+32\). Exchange \(x\) and \(y\):
\[ x=\frac95y+32. \]
Solving for \(y\) gives
\[ y=\frac59(x-32). \]
Using \(x\) as the inverse input,
\[ \boxed{F^{-1}(x)=\frac59(x-32)}. \]
The inverse converts a temperature in degrees Fahrenheit to degrees Celsius.