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) = x 2 – 2. Express (gof) -1 and f -1 og -1 as the sets of ordered pairs and verify (gof) -1 = f -1 og -1 .

Question

Philoid · Accepted Answer

We have f : A → B & f(x) = 2x + 1

⇒ f = {(1, 2×1 + 1), (2, 2×2 + 1), (3, 2×3 + 1), (4, 2×4 + 1)}

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

Function f is clearly one-one and onto.

Thus, f^-1 exists and f^-1 = {(3, 1), (5, 2), (7, 3), (9, 4)}

We have g : B → C & g(x) = x² – 2

⇒ g = {(3, 3² – 2), (5, 5² – 2), (7, 7² – 2), (9, 9² – 2)}

∴ g = {(3, 7), (5, 23), (7, 47), (9, 79)}

Function g is clearly one-one and onto.

Thus, g^-1 exists and g^-1 = {(7, 3), (23, 5), (47, 5), (79, 9)}

We know (gof)(x) = g(f(x))

Thus, gof : A → C and

(gof)(1) = g(f(1)) = g(3) = 7

(gof)(2) = g(f(2)) = g(5) = 23

(gof)(3) = g(f(3)) = g(7) = 47

(gof)(4) = g(f(4)) = g(9) = 79

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

Clearly, gof is also both one-one and onto.

Thus, the function gof has an inverse.

We have (gof)^-1 = {(7, 1), (23, 2), (47, 3), (79, 4)}

Now, let us consider f^-1og^-1.

We know (f^-1og^-1)(x) = f^-1(g^-1(x))

Thus, f^-1og^-1 : C → A and

(f^-1og^-1)(7) = f^-1(g^-1(7)) = f^-1(3) = 1

(f^-1og^-1)(23) = f^-1(g^-1(23)) = f^-1(5) = 2

(f^-1og^-1)(47) = f^-1(g^-1(47)) = f^-1(7) = 3

(f^-1og^-1)(79) = f^-1(g^-1(79)) = f^-1(9) = 4

⇒ f^-1og^-1 = {(7, 1), (23, 2), (47, 3), (79, 4)}

Therefore, we have (gof)^-1 = f^-1og^-1.

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 (gof)-1 = f-1og-1.

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 (gof)^-1 = f^-1og^-1.