Show that if f:R - {7/5} → R - {3/5} is defined by f(x) = (3x + 4

1.	Show that if f:R - {7/5} → R - {3/5} is defined by f(x) = (3x + 4)/(5x - 7) and g: R - {3/5} → R - {7/5} is defined by g(x) = (7x + 4)/(5x - 3), then fog = IA and gof = IB, where, A = R - {3/5}, B = R - {7/5}; IA(x) = x, ∀ x ∈ A, IB(x) = x, ∀ x ∈ B are called identity functions on sets A and B, respectively.
Answer» We have gof(x) = g((3x + 4)/(5x - 7)) = (7((3x + 4)/(5x - 7)) + 4)/(5((7x + 4)/(5x - 7)) - 3) = (21x + 28 + 20x - 28)/(15x + 20 - 15x + 21) = 41x/41 = x Similarly, fog(x) = f((7x + 4)/(5x - 3)) = (3(7x + 4)/(5x - 3)) + 4)/(5(7x + 4)/(5x - 3)) - 7) = (21x + 12 + 20x - 12)/(35x + 20 - 35x + 21) = 41x/41 = x Thus, gof (x) = x, ∀ x ∈ B and fog (x) = x, ∀ x ∈ A, which implies that gof = I_B and fog = I_A.

Show that if f:R - {7/5} → R - {3/5} is defined by f(x) = (3x + 4)/(5x - 7) and g: R - {3/5} → R - {7/5} is defined by g(x) = (7x + 4)/(5x - 3), then fog = IA and gof = IB, where, A = R - {3/5}, B = R - {7/5}; IA(x) = x, ∀ x ∈ A, IB(x) = x, ∀ x ∈ B are called identity functions on sets A and B, respectively.

Answer»

We have gof(x) = g((3x + 4)/(5x - 7)) = (7((3x + 4)/(5x - 7)) + 4)/(5((7x + 4)/(5x - 7)) - 3)

= (21x + 28 + 20x - 28)/(15x + 20 - 15x + 21) = 41x/41 = x

Similarly, fog(x) = f((7x + 4)/(5x - 3)) = (3(7x + 4)/(5x - 3)) + 4)/(5(7x + 4)/(5x - 3)) - 7)

= (21x + 12 + 20x - 12)/(35x + 20 - 35x + 21) = 41x/41 = x

Thus, gof (x) = x, ∀ x ∈ B and fog (x) = x, ∀ x ∈ A, which implies that gof = I_B and fog = I_A.