Question

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−1og−1 as the sets of ordered pairs and verify that (gof)−1 = f−1og−1.

Answer 1

Given 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 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)}

So, f and g are bijections and,

Hence, f−1: B → A and g−1: C → B exist.

Therefore, f−1= {(3, 1), (5, 2), (7, 3), (9, 4)}

And g−1= {(7, 3), (23, 5), (47, 7), (79, 9)}

Now, (f−1og−1): C → A

f−1og−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

Therefore, f−1og−1and (gof)−1have 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

Now, (gof)(1) = g(f(1))

= 4 + 4 − 1

= 7,

(gof)(2) = g(f(2))

= 4 + 4 – 1 = 23,

(gof)(3) = g(f(3))

= 4 + 4 – 1 = 47 and

(gof)(4) = g(f(4))

= 4 + 4 − 1 = 79

Therefore, gof = {(1, 7), (2, 23), (3, 47), (4, 79)}

⇒ (gof) – 1 = {(7, 1), (23, 2), (47, 3), (79, 4)} …(2)

From equations (1) and (2), we get:

(gof)−1= f−1og−1