Let f = {(1, – 1), (4, – 2), (9, – 3), (16, 4)} and g = {(– 1,

1.	Let f = {(1, – 1), (4, – 2), (9, – 3), (16, 4)} and g = {(– 1, – 2), (– 2, – 4), (– 3, – 6), (4, 8)}. Show that gof is defined while fog is not defined. Also, find gof.
Answer» We have, f = {(1, – 1), (4, – 2) , (9, – 3), (16,4)} and g = {(– 1, – 2), (– 2, – 4), (– 3, – 6), (4,8)} Now, Domain of f = {1,4,9,16} Range of f = {– 1, – 2, – 3, 4} Domain of g = (– 1, – 2, – 3,4} Range of g = (– 2, – 4, – 6, 8} Clearly range of f = domain of g ∴ gof is defined. but, range of g ≠ domain of f So, fog is not defined. Now, gof(1) = g(– 1)= – 2 gof(4) = g(– 2) = – 4 gof(9) = g (– 3) = – 6 gof(16) = g(4)= 8 So, gof = {(1, – 2), (4, – 4), (9 , – 6), (16,8)}

Let f = {(1, – 1), (4, – 2), (9, – 3), (16, 4)} and g = {(– 1, – 2), (– 2, – 4), (– 3, – 6), (4, 8)}. Show that gof is defined while fog is not defined. Also, find gof.

Answer»

We have,

f = {(1, – 1), (4, – 2) , (9, – 3), (16,4)} and

g = {(– 1, – 2), (– 2, – 4), (– 3, – 6), (4,8)}

Now,

Domain of f = {1,4,9,16}

Range of f = {– 1, – 2, – 3, 4}

Domain of g = (– 1, – 2, – 3,4}

Range of g = (– 2, – 4, – 6, 8}

Clearly range of f = domain of g

∴ gof is defined.

but, range of g ≠ domain of f
So, fog is not defined.

Now,

gof(1) = g(– 1)= – 2

gof(4) = g(– 2) = – 4

gof(9) = g (– 3) = – 6

gof(16) = g(4)= 8

So, gof = {(1, – 2), (4, – 4), (9 , – 6), (16,8)}