Find f o g and g o f if : f ( x ) = c , c ∈ R , g ( x ) = sin x 2 .

f ( x ) = c , g ( x ) = sin x 2 f : R → c ; g : R → [ - 1 , 1 ] Computing f o g : Clearly, the range of g is a subset of the domain of f . ⇒ f o g : R → R So, ( f o g ) ( x ) = f ( g ( x ) ) = f ( sin x 2 ) = c Computing g o f : Clearly, the range of f is a subset of the domain of g . ⇒ f o g : R → R So, ( g o f ) ( x ) = g ( f ( x ) ) = g ( c ) = sin c 2

Find f o g and g o f if : f ( x ) = x 2 + 2 , g ( x ) = 1 − 1 ( 1 - x )

We have, f x = x 2 + 2 , g x = 1 - 1 1 - x = - x 1 - x Domain of f = R = Domain of g Range of f = ( 2 , ∞ ) Range of g = R - 1 Therefore, Range of f ⊂ Domain of g ⇒ g o f exist And, Range of g ⊂ Domain of f ⇒ f o g exist. Now, g ∘ f ( x ) = g ( f ( x ) ) = g x 2 + 2 = - x 2 - 2 ( 1 - x 2 - 2 ) = x 2 + 2 x 2 + 1 and, f ∘ g ( x ) = f ( g ( x ) ) = f - x 1 - x = - x 1 - x 2 + 2

Let f ( x ) = x 2 + x + 1 and g ( x ) = sin x . Show that f o g ≠ g o f .

We have, f x = x 2 + x + 1 and g x = sin x So, ( f o g ) ( x ) = f ( g ( x ) ) = f ( sin x ) = sin 2 x + sin x + 1 and ( g o f ) ( x ) = g ( f ( x ) ) = g ( x 2 + x + 1 ) = sin ( x 2 + x + 1 ) Clearly, So, f o g ≠ g o f . Hence Proved