Questions on One-to-One Functions

This page presents several questions with detailed solutions, as well as exercises with answers, on one-to-one (injective) functions.

Definition of a One-to-One Function

A function \(f(x)\) is said to be one-to-one if different inputs always produce different outputs. Formally,

\[ A \neq B \; \Rightarrow \; f(A) \neq f(B) \]

where \(A\) and \(B\) are any values in the domain of the function \(f\).

The contrapositive form of this definition, which is often more convenient to use, is:

\[ f(A) = f(B) \; \Rightarrow \; A = B \]

We will use this contrapositive to determine whether a given function is one-to-one.

Questions with Solutions

Question 1

Is the function \(f\) defined by

\[ f = {(1,2),(3,4),(5,6),(8,6),(10,-1)} \]

a one-to-one function?

Solution to Question 1

Two different values in the domain, namely \(5\) and \(8\), have the same output \(6\). Therefore, the function \(f\) is not a one-to-one function.

Question 2

Is the function \(g\) defined by

\[ g = {(-1,2),(0,4),(2,-4),(5,6),(10,0)} \]

a one-to-one function?

Solution to Question 2

Each value in the domain corresponds to a distinct output. Hence, the function \(g\) is a one-to-one function.

Question 3

Is the function \(f\) given by

\[ f(x) = -x^3 + 3x^2 - 2 \]

a one-to-one function?

Solution to Question 3

A graph of the function and the horizontal line test can be used to answer this question.

Graph of the function f(x) = -x^3 + 3x^2 - 2

Since a horizontal line can intersect the graph at three different points, there exist distinct inputs \(x_1, x_2, x_3\) that produce the same output. Therefore, the function \(f\) is not one-to-one.

Question 4

Show that all linear functions of the form

\[ f(x) = ax + b \]

where \(a\) and \(b\) are real numbers and \(a \neq 0\), are one-to-one functions.

Solution to Question 4

Assume \(f(A) = f(B)\). Then \[ aA + b = aB + b. \]
Subtract \(b\) from both sides: \[ aA = aB. \]
Divide both sides by \(a \neq 0\): \[ A = B. \]
Since \(f(A) = f(B)\) implies \(A = B\), all linear functions of this form are one-to-one.

Question 5

Show that all functions of the form

\[ f(x) = a(x - h)^2 + k, \quad x \ge h \]

where \(a, h, k\) are real numbers and \(a \neq 0\), are one-to-one functions.

Solution to Question 5

Start with \(f(A) = f(B)\): \[ a(A - h)^2 + k = a(B - h)^2 + k. \]
Subtract \(k\) and divide by \(a\): \[ (A - h)^2 = (B - h)^2. \]
This implies \[ A - h = B - h \quad \text{or} \quad A - h = -(B - h). \]
The first equation gives \(A = B\).
Since the domain satisfies \(x \ge h\), we have \(A - h \ge 0\) and \(B - h \ge 0\). Thus the second equation has no solution.

Question 6

Is the function \(f\) given by

\[ f(x) = \frac{1}{(x - 2)^2} \]

a one-to-one function?

Solution to Question 6

We find that \(f(0) = \frac{1}{4}\) and \(f(4) = \frac{1}{4}\). Since two different inputs give the same output, the function \(f\) is not one-to-one.

Question 7

Show that all rational functions of the form

\[ f(x) = \frac{1}{ax + b} \]

where \(a\) and \(b\) are real numbers with \(a \neq 0\), are one-to-one functions.

Solution to Question 7

Assume \(f(A) = f(B)\): \[ \frac{1}{aA + b} = \frac{1}{aB + b}. \]
Multiply both sides by \((aA + b)(aB + b)\): \[ aB + b = aA + b. \]
Subtract \(b\) and divide by \(a\): \[ B = A. \]
Therefore, all functions of this form are one-to-one.

Exercises

For each of the following functions, state whether it is a one-to-one function.

\(f = \{(12,2),(15,4),(19,-4),(25,6),(78,0)\}\)
\(g = \{(-1,2),(0,4),(9,-4),(18,6),(23,-4)\}\)
\(h(x) = x^2 + 2\)
\(i(x) = \frac{1}{2x - 4}\)
\(j(x) = -5x + \frac{1}{2}\)
\(k(x) = \frac{1}{|x - 4|}\)

Answers to the Exercises

\(f\) is a one-to-one function
\(g\) is not a one-to-one function
\(h\) is not a one-to-one function
\(i\) is a one-to-one function
\(j\) is a one-to-one function
\(k\) is not a one-to-one function

Explore more with this interactive tutorial on One-to-One Functions.