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One to one function is essentially the mapping of two sets. If every element of g corresponds to exactly one element of g's domain, the function g is one-to-one. A function f() is a procedure that connects the elements/values of one variable to the elements/values of another variable in such a way that the elements of the first variable determine the elements of the second variable exactly.
Read more: Linear Equation
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KeyTerms: One to One function, Inverse function, Domain, Range, Element, Variables, Function, Set, One-to-one relationship, Component
Definition of One to One function
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A one to one function is a function that maps every element of the range to exactly one element of its domain, ensuring that the outputs never repeat. For example, the function g(x) = x - 4 is a one to one function because it produces a different result for each input. Also, the function g(x) = x2 is NOT a one-to-one function because it gives 4 when the inputs are 2 and -2. A many-to-one function is one that does not have a one-to-one relationship.
One to one function can be defined algebraically as:
If the function g: D → F is one-to-one, it is said to be one-to-one.
g(x1) = g(x2) ⇒ x1 = x2
for all x1 and x2 components D. A one-to-one function is also regarded as an injection; that is, a function is only injective if it is one-to-one. This definition's inverse is a function g: D → F is one-to-one if x1 ≠ x2 ⇒ g(x1) ≠ g (x2).
One to One function
Examples of One to One function
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The following are some examples of One to One function –
- The identity function X → X is always injective.
- If the function f: R→ R, then f(x) = 5x is injective.
- If the function f: R→ R, then f(x) = 7x+1 is injective.
- If function f: R→ R, then f(x) = x4 is not an injective function, because here if x = -1, then f(-1) = 1 = f(1). Hence, the element of codomain is not discrete here.
- If the function f: R→ R, then f(x) = x/6 is injective.
- If the function f: R→ R, then f(x) = x11 is injective.
- If the function f: R→ R, then f(x) = 5x+4 is injective.
Horizontal Line Test
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When a function's graph is provided, the horizontal line test is performed to establish whether it is one-one. We have already seen how to use the criterion (g(x1) = g(x2) ⇒ x1 = x2) to check whether a function g(x) is algebraically one-one. To determine whether the function is one-one based on its graph,
- Simply draw a horizontal line (think of it as a horizontal stick) and let it run through the graph.
- The function is one-one if the horizontal line does not cross through more than one point on the graph at any point in time.
- If the horizontal line crosses more than one point on the graph at any time, the function is NOT one-one.
Horizontal Line Test
- The horizontal line test revealed that f(x) = x2 (a quadratic function) is NOT one-one (as the horizontal line passes through more than one point of the graph)
- Because it passed the horizontal line test, f(x) = x3 (a cubic function) is one-one (as the horizontal line passes through only one point of the graph every time)
Inverse of One to One function
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If f is a function defined as y = f(x), then x = f -1(y), i.e. f-1 defined from y to x, is the inverse function of f. The co-domain of f is the domain of f -1 in the inverse function, and the domain of f is the domain of f -1.
Only one-to-one functions have an inverse because they have one-to-one correspondences, i.e. each range element relates to one and only one domain element.
If a function f: A → B is specified, then f is said to be invertible if there exists a function g: B → A in such a way that we may retrieve the initial point or value by operating f{g(x)} or g{f(x)}.
Solved Example: Show that the function f : X → Y, such that f(x)= 7x + 5,
For all x, y ∈ N is invertible.
Ans: Let y ∈ N → y = f(x) = 7x + 5 for x ∈ N
x = (y-5)/7
If we define h : Y → X by h(y) = (y-5)/ 7
Again h: → f(x) = h[ f(x) ] = h{ 7x + 5 } = 7(y-5) / 7 + 5 = x
And f: → h(y) = f [ h(y) ] = f( (y-5) / 7) = 7(y-5) / 7 + 5 = y
Hence f is an invertible function and h is the inverse of f.
Read more:
| Relevant Concepts | ||
|---|---|---|
| Polynomial Functions | Types of functions | Cube root |
| Parabola graph | Quadratic equation | Cardioid |
Steps to Find the Inverse of One to One Function
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The following is a step-by-step technique for calculating the inverse function g-1(x) for a one-to-one function g(x):
- Make g(x) equal to y.
- Replace x with y since every (x, y) has a (y, x) mate.
- Determine y
- Rename y as g-1 in the equation you just discovered (x).
Solved Example: Find the inverse function g-1(x) of the function g(x) = 2 x + 5.
Ans:
| Set g(x) = y | y = 2x + 5 |
| Switch x with y | x = 2y + 5 |
| Solve for y | y = (x - 5)/2 |
| Rename y as g-1(x). This is the inverse. | g-1(x) = (x - 5)/2 |
Properties of One to One function
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In order to return to the original y-value, the inverse of a one to one function undoes what the original function did to a value in its domain.
Properties of One to One function
The inverse of a one to one function has the following properties:
- If f and g are both one to one, then f o g is determined by injectivity.
- If g o f is one to one, function f must be one to one as well, but function g may not be.
- If and only if, given any functions g, h: P → X whenever f o g = f o h, then g = h, then f: X → Y is one-one. In other words, one-one functions are the category of sets' monomorphisms.
- If f: X → Y is a one-one relationship and P is a subset of X, then f-1 (f(A)) = P. As a result, P can be recovered from its image f. (P).
- If f: X → Y is a one-one function and P and Q are both subsets of X, then f(P ∩ Q) = f(P) ∩ f (Q).
- If X and Y have the same number of elements, then f: X → Y is one-one if and only if f is a surjective or onto function.
Things to Remember
- In mathematics, one to one functions are functions with an equal number of items in the domain and range, or where one item may only be paired with another item.
- Understanding the concept of one to one functions is required in order to understand the concept of inverse functions and solve certain sorts of equations.
- Geometrically and algebraically, it is simple to determine whether a function is one to one.
- The range of f–1 is the same as the domain of f–1, and the range of f is the same as the domain of f–1.
- Every time x is in the domain of f, f–1(f(x))=x, and every time x is in the domain of f–1, f(f–1(x))=x.
- The graphs of a function and its inverse are symmetric about the line y=x.
- The horizontal line test is a useful technique for determining whether the inverse of a function is also a function.
Sample Questions
Ques. Let A = {1, 2, 3} and B = {a, b, c, d}. Which of the following is a one-to-one function? (1 mark)
(a) {(1, c), (2, c)(2, c)}
(b) {(1, a),(2, b),(3, c)}
(c) {(1, b)(1, c)}
Ans: Option b. fulfills the one-to-one criteria because items of set B(range) are uniquely mapped with elements of set A. (domain).
Ques. Which function is not said to be one to one? (1 mark)
Ans: If a horizontal line can intersect the function's graph more than once, the function is not one-to-one mapped.
Ques. What is one to one function? (2 marks)
Ans: Each element of one set, say Set (A), is mapped with a unique element of another set, say Set (B), where A and B are two separate sets. It can also be written 1-1. In terms of function, if f(x) = f(y) means that x = y, then f is one to one.
Ques. Show that f: R→ R defined as f(a) = 3a3 – 4 is one to one function? (2 marks)
Ans: Let f ( a1 ) = f ( a2 ) for all a1 , a2 ∈ R
so 3a13 – 4 = 3a23 – 4
a13 = a23
a13 – a23 = 0
(a1 – a2) (a1 + a1a2 + a22) = 0
a1 = a2 and (a12 + a1a2 + a22) = 0
(a12 + a1a2 + a22) = 0 is not considered because there are no real values of a1 and a2.
Therefore, the given function f is one-one.
Ques. Let D = {3, 4, 8, 10} and C = {w, x, y, z}. Which of the following relations represent a one to one function? (3 marks)
(a) {(3, w), (3, x), (3, y), (3, z)}
(b) {(4, w), (3, x), (10, z), (8, y)}
(c) {(4, w), (3, x), (8, x), (10, y)}
Ans. To be a one-one function, each element from D must be paired with a unique element from C.
- Because the identical value of x is mapped to each value of y in the first relation, it cannot be regarded as a function and so is not a one-to-one function.
- In the third relationship, 3 and 8 have the same x range. As a result, it is not a one-to-one function.
- The second relation represents a one-to-one function by mapping a unique element from D to every unique element from C.
Thus, {(4, w), (3, x), (10, z), (8, y)} represents a one to one function.
Ques. Determine if g(x) = – 3x3 – 1 is a one-to-one function using the algebraic approach. (3 marks)
Ans: In order for a function to be a one to one function, g( x1 ) = g( x2 ) if and only if x1 = x2 .
g( x1 ) = -3 x13 – 1
g( x2 ) = -3 x23 – 1
start with g( x1 ) = g( x2 ). Then
- 3 x13 – 1 = – 3 x23 – 1
- 3 x13 = – 3 x23
( x1 )3 = ( x2 )3
Taking the cube root on both sides of the equation will lead us to x1 = x2.
Hence, g(x) = – 3x3 – 1 is a one to one function.
Ques. Is function f defined by a one to one function? (1 mark)
f = {(1 , 2),(3 , 4),(5 , 6),(8 , 6),(10 , -1)}
Ans: Since two different domain values, 5 and 6, produce the same output, function f is not a one to one function.
Ques. Show that all linear functions of the form where a and b are real numbers such that a not equal to zero, are one to one functions. (2 marks)
f(x) = ax + b
Ans: Begin with f(A) = f(B) and demonstrate that this leads to a = b.
a(A) + b equals a(B) + b
Add -b to both sides of the equation to obtain
a(A) = a(B)
Because an is not equal to zero, divide both sides by it to get
A = B
Given that f(A) = f(B) leads to A = B,
Hence concluded that all linear functions of the form
f(x) = an x + b are one-to-one functions.
A = B
Ques. Show that all functions of the form where a, h and k are real numbers such that a not equal to zero, are one to one functions. (3 marks)
f(x) = a (x - h) 2 + k , for x >= h
Ans: Start with f(A) = f(B)
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)
The first equation leads
A = B
The domain of f includes all x values such that x >= h. This results in x - h >= 0, which leads to A - h >= 0 and B - h >= 0, indicating that the second equation (A - h) = - (B - h) has no solution.
Ques. Show that all the rational functions of the form where a, and b are real numbers such that a not equal to zero, are one to one functions. (3 marks)
f(x) = 1 / (ax + b)
Ans. Write an equation beginning 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”
B = A
The given functions are one to one functions.
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