Question

Find f^-1 (if exists), where f: A → B, such that

(i) A = {0,-1,- 3, 2}, B = {-9, – 3,0, 6}, f(x) = 3x

(ii) A = {1, 3, 5, 7,9}, B = {0, 1, 9, 25, 49, 81), f(x) = x²

(iii) A = B = R, f(x) = x³

Answer 1

(i) Give function

f : A → B, f(x) = 3x where,

A = {0,-1,-3, 2}

B = {-9, – 3,0, 6}

Now, f(0)= 3 x 0 = 0

f(-1) = 3 x – 1 = -3

f(-3) = 3 x (- 3) = -9

f(2)= 3 x 2 = 6

So, f= {(0,0), (- 1, -3), (- 3,-9), (2, 6)}

Since, different element of A have different image in B under f.

So ‘f is one-one function.

Here, range of f = {-9, -3,0, 6} = B (co-domain).

So, f is onto function.

It is clear that ‘f’ is one-one onto function.

So, f^-1 : B → A exist.

f^-1 = {(0,0), (-3,- 1), (-9, – 3), (6, 2)}

(ii) Given function

f : A → B, f(x) = x²

where, A = {1, 3, 5, 7, 9}

B = {0, 1, 9, 25, 49, 81}

Now, f(1) = 1² = 1

f(3) = 3² = 9

f(5)= 5² = 25

f (7) = 7² = 49

f(9) = 9² = 81

So, f= {(1, 1), (3,9), (5, 25), (7,49), (9,81)}

Since, different element of A have different image in B under f.

So ‘f’ one-one function.

Here, range of R = {1,9, 25, 49, 81} ≠ B (co-domain)

∴ ‘f’ is not onto function.

Hence, f^-1 does not exist.

(iii) Given, A = B = R, f(x) = x³

f(x)= x³

Let a, b ∈ R

and f(a) = f(b)

⇒ a³ = b³

a = b, ∀ a, b ∈ R

Let f(x)= y and y ∈ B or y ∈ R

∴ x³ = y

⇒ x = y^1/3

Since, different element of A have different image of B under f.

So ‘f’ is one-one function.

and range of f= co-domain of f.

So, ‘f’ is onto function.

Hence, it is clear that f is one-one function f^-1 exists.

If y is image of x under ‘f’, then

f^-1 : B → A is defined

as f^-1(y) = x ⇒ f(x) = y

⇒ x³ = y

⇒ x = y^1/3

⇒ f^-1(y) = y^1/3

⇒ f^-1(x) = x^1/3

So, f^-1: B → A, f^-1(x) = x^1/3, ∀ x ∈ B.