# Questions on one to one Functions

Several questions with detailed solutions as well as exercises with answers on one to one functions are presented.

## One to One Function

From the definition of one-to-one functions we can write that a given function f(x) is one-to-one
if A is not equal to B then f(A) is not equal f(B)

where A and B are any values of the variable x in the domain of function f.
The contrapositive of the above definition is as follows:
if f(A) = f(B) then A = B

where A and B are any values of x included in the domain of f. We will use this contrapositive of the definition of one to one functions to find out whether a given function is a one to one.

## Questions with Solutions

### Question 1

Is 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 6, have the same output, hence function f is not a one to one function.

### Question 2

Is function g defined by
g = {(-1 , 2),(0 , 4),(2 , -4),(5 , 6),(10 , 0)}
,
a one to one function?

Solution to Question 2

• Consider any two different values in the domain of function g and check that their corresponding output are different. Hence function g is a one to one function.

### Question 3

Is function f given by
f(x) = -x 3 + 3 x 2 - 2
,
a one to one function?

Solution to Question 3:

• A graph and the horizontal line test can help to answer the above question. • Since a horizontal line cuts the graph of f at 3 different points, that means that they are at least 3 different inputs x1, x2 and x3 with the same output Y and therefore f is not a one to one function.

### Question 4

Show that all linear functions of the form
f(x) = a x + b
,
where a and b are real numbers such that a not equal to zero, are one to one functions.

Solution to Question 4

a(A) + b = a(B) + b
• Add -b to both sides of the equation to obtain
a(A) = a(B)
• Divide both sides by a since it is not equal to zero to obtain
A = B
• Since we have proved that f(A) = f(B) leads to A = B then all linear functions of the form f(x) = a x + b are one-to-one functions.
A = B

### Question 5

Show that all functions of the form
f(x) = a (x - h) 2 + k , for x >= h
,
where a, h and k are real numbers such that a not equal to zero, are one to one functions.

Solution to Question 5

a (A - h) 2 + k = a (B - h) 2 + k
• Add -k to both sides of the equation to obtain
a (A - h) 2 = a (B - h) 2
• Divide both sides by a since it not equal to 0
(A - h) 2 = (B - h) 2
• The above equation leads to two other equations
(A - h) = (B - h) or (A - h) = - (B - h)
A = B
• Let us examine the second equation. The domain of f is all values of x such that x >= h. This leads to x - h >= 0 which in turn leads to A - h >= 0 and B - h >= 0 which means the second equation (A - h) = - (B - h) does not have a solution.

### Question 6

Is function f given by
f(x) = 1 / (x - 2) 2
,
a one to one function?

Solution to Question 6

• It is easy to find two values of x that correspond to the same value of the function.
f(0) = 1 / 4 and f(4) = 1 / 4. For two different values in the domain of f correspond one same value of the range and therefore function f is not a one to one.

### Question 7

Show that all the rational functions of the form
f(x) = 1 / (a x + b)

where a, and b are real numbers such that a not equal to zero, are one to one functions.

Solution to Question 7

• Let us write an equation starting with f(A) = f(B)
1 / (a A + b) = 1 / (a B + b)
• Multiply both sides of the equation by (a A + b)(a B + b) and simplify
a B + b = a A + b
• Add -b to both sides
a B = a A
• Divide both sides by a to obtain
B = A
• The given functions are one to one functions

## Exercises

For each of these functions, state whether it is a one to one function.
1. f = {(12 , 2),(15 , 4),(19 , -4),(25 , 6),(78 , 0)}
2. g = {(-1 , 2),(0 , 4),(9 , -4),(18 , 6),(23 , -4)}
3. h(x) = x 2 + 2
4. i(x) = 1 / (2x - 4)
5. j(x) = -5x + 1/2
6. k(x) = 1 / |x - 4|