Question

If f : Q → Q, g : Q → Q are two functions defined by f(x) = 2 x and g(x) = x + 2, show that f and g are bijective maps. Verify that (gof)⁻¹ = f⁻¹ og ⁻¹.

Answer 1

Injectivity of f:
Let x and y be two elements of domain (Q), such that

f(x) = f(y)

⇒">⇒ 2x= 2y
⇒">⇒ x = y

So, f is one-one.
Surjectivity of f:
Let y be in the co-domain (Q), such that f(x) = y.

⇒ 2x = y 

⇒ `x = y/2 in Q` (domain)

⇒ 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 = 2y

⇒ `y = x/2`

So, ` f^1 (x) = x/2`    (from (1))

njectivity of g:
Let x and y be two elements of domain (Q), such that
g (x) = g (y)

⇒">⇒  x + 2 = y + 2

⇒">⇒ x = y

So, g is one-one.

Surjectivity of g:
Let y be in the co domain (Q), such that g(x) = y.

⇒ x +2 =y

⇒ x= 2 -y ∈ Q (domain)

 ⇒ g is onto.
So, g is a bijection and, hence, it is invertible.

Finding g -1:

Let g−1(x) = y             ...(2)

⇒ x = g (y)

⇒ x = y+2

⇒ y = x − 2

So, g−1 (x) = x − 2        (From (2)

Verification of (gof)−1 = f−1 og −1:

f(x) = 2x ; g (x) = x + 2

and `f^-1 (x) = x/2 ; g^-1 (x)= x-2`

`Now, (f^-1 o  g^-1) (x) = f^-1 (g^-1)(x))  `

⇒ `(f^-1 o  g ^-1)(x) = f^-1 (x-2) `

⇒ `(f ^-1 o   g^-1) (x) = (x-2)/2 .......... (3)`

(gof) (x) = g (f(x))

= g (2x)

= 2x + 2

Let (gof)-1 (x) = y  ............ (4)

x = (gof) (y)

⇒ x = 2y +2

⇒ 2y = x - 2 

⇒ `y= (x-2)/2`

⇒` (gof)^-1 (x) = (x-2)/2`       [form (4) ....... (5) ]

from (3) and (5)

⇒ `(gof)^-1  = f^-1  o  g^-1`