- Books Name
- Mathmatics Book Based on NCERT
- Publication
- KRISHNA PUBLICATIONS
- Course
- CBSE Class 12
- Subject
- Mathmatics
Invertible Functions
A function f : X → Y is defined to be invertible if there exists a function g : Y → X such that gof = IX and fog = IY. The function g is called the inverse of f and is denoted by f–1.
An important note is that, if f is invertible, then f must be one-one and onto and conversely if f is one-one and onto, then f must be invertible.
Example 1:. Find f −1 if it exists: f: A → B, where
(i) A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3 x.
(ii) A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x2
Solution:
(i) Given A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3 x.
So, f = {(0, 0), (-1, -3), (-3, -9), (2, 6)}
Here, different elements of the domain have different images in the co-domain.
Clearly, this is one-one.
Range of f = Range of f = B
so, f is a bijection and,
Thus, f -1 exists.
Hence, f -1= {(0, 0), (-3, -1), (-9, -3), (6, 2)}
(ii) Given A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x2
So, f = {(1, 1), (3, 9), (5, 25), (7, 49), (9, 81)}
Here, different elements of the domain have different images in the co-domain.
Clearly, f is one-one.
But this is not onto because the element 0 in the co-domain (B) has no pre-image in the domain (A)
⇒ f is not a bijection.
So, f -1does not exist.
Example 2:. Consider f: {1, 2, 3} → {a, b, c} and g: {a, b, c} → {apple, ball, cat} defined as f (1) = a, f (2) = b, f (3) = c, g (a) = apple, g (b) = ball and g (c) = cat. Show that f, g and gof are invertible. Find f−1, g−1 and gof−1and show that (gof)−1 = f −1o g−1
Solution:
Given f = {(1, a), (2, b), (c , 3)} and g = {(a , apple) , (b , ball) , (c , cat)} Clearly , f and g are bijections.
So, f and g are invertible.
Now,
f -1 = {(a ,1) , (b , 2) , (3,c)} and g-1 = {(apple, a), (ball , b), (cat , c)}
So, f-1 o g-1= {apple, 1), (ball, 2), (cat, 3)}……… (1)
f: {1,2,3,} → {a, b, c} and g: {a, b, c} → {apple, ball, cat}
So, gof: {1, 2, 3} → {apple, ball, cat}
⇒ (gof) (1) = g (f (1)) = g (a) = apple
(gof) (2) = g (f (2))
= g (b)
= ball,
And (gof) (3) = g (f (3))
= g (c)
= cat
∴ gof = {(1, apple), (2, ball), (3, cat)}
Clearly, gof is a bijection.
So, gof is invertible.
(gof)-1 = {(apple, 1), (ball, 2), (cat, 3)}……. (2)
Form (1) and (2), we get
(gof)-1 = f-1 o g -1
Example 3:. Let A = {1, 2, 3, 4}; B = {3, 5, 7, 9}; C = {7, 23, 47, 79} and f: A → B, g: B → C be defined as f(x) = 2x + 1 and g(x) = x2 − 2. Express (gof)−1 and f−1 og−1 as the sets of ordered pairs and verify that (gof)−1 = f−1 og−1.
Solution:
Given that f (x) = 2x + 1
⇒ f= {(1, 2(1) + 1), (2, 2(2) + 1), (3, 2(3) + 1), (4, 2(4) + 1)}
= {(1, 3), (2, 5), (3, 7), (4, 9)}
Also given that g(x) = x2−2
⇒ g = {(3, 32−2), (5, 52−2), (7, 72−2), (9, 92−2)}
= {(3, 7), (5, 23), (7, 47), (9, 79)}
Clearly f and g are bijections and, hence, f−1: B→ A and g−1: C→ B exist.
So, f−1= {(3, 1), (5, 2), (7, 3), (9, 4)}
And g−1= {(7, 3), (23, 5), (47, 7), (79, 9)}
Now, (f−1 o g−1): C→ A
f−1 o g−1 = {(7, 1), (23, 2), (47, 3), (79, 4)}……….(1)
Also, f: A→B and g: B → C,
⇒ gof: A → C, (gof) −1 : C→ A
So, f−1 o g−1and (gof)−1 have same domains.
(gof) (x) = g (f (x))
=g (2x + 1)
=(2x +1 )2− 2
⇒ (gof) (x) = 4x2 + 4x + 1 − 2
⇒ (gof) (x) = 4x2+ 4x −1
Then, (gof) (1) = g (f (1))
= 4 + 4 − 1
=7,
(gof) (2) = g (f (2))
= 4(2)2 + 4(2) – 1 = 23,
(gof) (3) = g (f (3))
= 4(3)2 + 4(3) – 1 = 47 and
(gof) (4) = g (f (4))
= 4(4)2 + 4(4) − 1 = 79
So, gof = {(1, 7), (2, 23), (3, 47), (4, 79)}
⇒ (gof)– 1 = {(7, 1), (23, 2), (47, 3), (79, 4)}…… (2)
From (1) and (2), we get:
(gof)−1 = f−1 o g−1
Example 4:. Show that the function f: Q → Q, defined by f(x) = 3x + 5, is invertible. Also, find f−1
Solution:
Given function f: Q → Q, defined by f(x) = 3x + 5
Now we have to show that the given function is invertible.
Injection of f:
Let x and y be two elements of the domain (Q),
Such that f(x) = f(y)
⇒ 3x + 5 = 3y + 5
⇒ 3x = 3y
⇒ x = y
so, f is one-one.
Surjection of f:
Let y be in the co-domain (Q),
Such that f(x) = y
⇒ 3x +5 = y
⇒ 3x = y – 5
⇒ x = (y -5)/3 belongs to Q domain
⇒ f is onto.
So, f is a bijection and, hence, it is invertible.
Now we have to find f-1:
Let f-1(x) = y…… (1)
⇒ x = f(y)
⇒ x = 3y + 5
⇒ x −5 = 3y
⇒ y = (x – 5)/3
Now substituting this value in (1) we get
So, f-1(x) = (x – 5)/3
Example 5: Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
Solution:
Given f: R → R given by f(x) = 4x + 3
Now we have to show that the given function is invertible.
Consider injection of f:
Let x and y be two elements of domain (R),
Such that f(x) = f(y)
⇒ 4x + 3 = 4y + 3
⇒ 4x = 4y
⇒ x = y
So, f is one-one.
Now surjection of f:
Let y be in the co-domain (R),
Such that f(x) = y.
⇒ 4x + 3 = y
⇒ 4x = y -3
⇒ x = (y-3)/4 in R (domain)
⇒ f is onto.
So, f is a bijection and, hence, it is invertible.
Now we have to find f -1
Let f-1(x) = y……. (1)
⇒ x = f (y)
⇒ x = 4y + 3
⇒ x − 3 = 4y
⇒ y = (x -3)/4
Now substituting this value in (1) we get
So, f-1(x) = (x-3)/4
Example 6: Consider f: R → R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with inverse f−1 of f given by f−1(x) = √ (x-4) where R+ is the set of all non-negative real numbers.
Solution:
Given f: R → R+ → [4, ∞) given by f(x) = x2 + 4.
Now we have to show that f is invertible,
Consider injection of f:
Let x and y be two elements of the domain (Q),
Such that f(x) = f(y)
⇒ x2 + 4 = y2 + 4
⇒ x2 = y2
⇒ x = y (as co-domain as R+)
So, f is one-one
Now surjection of f:
Let y be in the co-domain (Q),
Such that f(x) = y
⇒ x2 + 4 = y
⇒ x2 = y – 4
⇒ x = √ (y-4) in R
⇒ f is onto.
So, f is a bijection and, hence, it is invertible.
Now we have to find f-1:
Let f−1 (x) = y…… (1)
⇒ x = f (y)
⇒ x = y2 + 4
⇒ x − 4 = y2
⇒ y = √ (x-4)
So, f-1(x) = √ (x-4)
Now substituting this value in (1) we get,
So, f-1(x) = √ (x-4)
Example 7:. If f(x) = (4x + 3)/ (6x – 4), x ≠ (2/3) show that fof(x) = x, for all x ≠ (2/3). What is the inverse of f?
Solution:
It is given that f(x) = (4x + 3)/ (6x – 4), x ≠ 2/3
Now we have to show fof(x) = x
(fof)(x) = f (f(x))
= f ((4x+ 3)/ (6x – 4))
= (4((4x + 3)/ (6x -4)) + 3)/ (6 ((4x +3)/ (6x – 4)) – 4)
= (16x + 12 + 18x – 12)/ (24x + 18 – 24x + 16)
= (34x)/ (34)
= x
Therefore, fof(x) = x for all x ≠ 2/3
=> fof = 1
Hence, the given function f is invertible and the inverse of f is f itself.
Example 8:. Consider f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with
f-1(x) = (√(x +6)-1)/3
Solution:
Given f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x – 5
We have to show that f is invertible.
Injectivity of f:
Let x and y be two elements of domain (R+),
Such that f(x) = f(y)
⇒ 9x2 + 6x – 5 = 9y2 + 6y − 5
⇒ 9x2 + 6x = 9y2 + 6y
⇒ x = y (As, x, y ∈ R+)
So, f is one-one.
Surjectivity of f:
Let y is in the co domain (Q)
Such that f(x) = y
⇒ 9x2 + 6x – 5 = y
⇒ 9x2 + 6x = y + 5
⇒ 9x2 + 6x +1 = y + 6 (By adding 1 on both sides)
⇒ (3x + 1)2 = y + 6
⇒ 3x + 1 = √(y + 6)
⇒ 3x = √ (y + 6) – 1
⇒ x = (√ (y + 6)-1)/3 in R+ (domain)
f is onto.
So, f is a bijection and hence, it is invertible.
Now we have to find f-1
Let f−1(x) = y….. (1)
⇒ x = f (y)
⇒ x = 9y2 + 6y − 5
⇒ x + 5 = 9y2 + 6y
⇒ x + 6 = 9y2+ 6y + 1 (adding 1 on both sides)
⇒ x + 6 = (3y + 1)2
⇒ 3y + 1 = √ (x + 6)
⇒ 3y =√(x +6) -1
⇒ y = (√ (x+6)-1)/3
Now substituting this value in (1) we get,
So, f-1(x) = (√ (x+6)-1)/3
Example 9:. If f: R → R be defined by f(x) = x3 −3, then prove that f−1 exists and find a formula for f−1. Hence, find f−1 (24) and f−1 (5).
Solution:
Given f: R → R be defined by f(x) = x3 −3
Now we have to prove that f−1 exists
Injectivity of f:
Let x and y be two elements in domain (R),
Such that, x3 − 3 = y3 − 3
⇒ x3 = y3
⇒ x = y
So, f is one-one.
Surjectivity of f:
Let y be in the co-domain (R)
Such that f(x) = y
⇒ x3 – 3 = y
⇒ x3 = y + 3
⇒ x = ∛(y+3) in R
⇒ f is onto.
So, f is a bijection and, hence, it is invertible.
Finding f -1:
Let f-1(x) = y…….. (1)
⇒ x= f(y)
⇒ x = y3 − 3
⇒ x + 3 = y3
⇒ y = ∛(x + 3) = f-1(x) [from (1)]
So, f-1(x) = ∛(x + 3)
Now, f-1(24) = ∛ (24 + 3)
= ∛27
= ∛33
= 3
And f-1(5) =∛ (5 + 3)
= ∛8
= ∛23
= 2
Binary Operations
A binary operation ∗ on a set A is a function ∗ : A × A → A.
We denote * (a, b) by a * b.
Example Problems
Example 1: Show that subtraction and division are not binary operations on R.
Solution: N × N → N, given by (a, b) → a – b, is not binary operation, as the image of (2, 5) under ‘–’ is 2 – 5 = – 3 ∉ N.
Similarly, ÷: N × N → N, given by (a, b) → a ÷ b is not a binary operation, as the image of (2, 5) under ÷ is 2 ÷ 5 = 2/5 ∉ N.
Example 2: Let f : {2, 3, 4, 5} → {3, 4, 5, 9} and g : {3, 4, 5, 9} → {7, 11, 15} be functions defined as f(2) = 3, f(3) = 4, f(4) = f(5) = 5 and g (3) = g (4) = 7 and g (5) = g (9) = 11. Find gof.
Solution: From the given, we have:
gof(2) = g (f(2)) = g (3) = 7
gof (3) = g (f(3)) = g (4) = 7
gof(4) = g (f(4)) = g (5) = 11
gof(5) = g (5) = 11
Example 3: Show that the relation R in the set Z of integers given by R = {(a, b) : 2 divides a – b} is an equivalence relation.
Solution: R is reflexive, as 2 divides (a – a) for all a ∈ Z.
Further, if (a, b) ∈ R, then 2 divides a – b.
Therefore, 2 divides b – a.
Hence, (b, a) ∈ R, which shows that R is symmetric.
Similarly, if (a, b) ∈ R and (b, c) ∈ R, then (a – b) and (b – c) are divisible by 2.
Now, a – c = (a – b) + (b – c) is even. (from the above statements)
From this,
(a – c) is divisible by 2.
This shows that R is transitive.
Thus, R is an equivalence relation in Z.