Find ( g ∘ f ) ( x ) ( g \circ f ) ( x ) ( g ∘ f ) ( x ) where f ( x ) = 4 x + 7 f ( x ) = 4 x + 7 f ( x ) = 4 x + 7 and g ( x ) = x 2 − 1 g ( x ) = x ^ { 2 } - 1 g ( x ) = x 2 − 1 .A) 4 x 3 28 x 2 − 4 x − 7 4 x ^ { 3 } 28 x ^ { 2 } - 4 x - 7 4 x 3 28 x 2 − 4 x − 7 B) 4 x 2 + 3 4 x ^ { 2 } + 3 4 x 2 + 3 C) 16 x 2 + 56 x + 48 16 x ^ { 2 } + 56 x + 48 16 x 2 + 56 x + 48 D) 16 x 2 + 6 16 x ^ { 2 } + 6 16 x 2 + 6 E) 4 x 2 + 6 4 x ^ { 2 } + 6 4 x 2 + 6

Question

Find   ( g ∘ f )  ( x )  ( g \circ f )  ( x )  ( g ∘ f )  ( x )    where   f ( x )  = 4 x + 7 f ( x )  = 4 x + 7 f ( x )  = 4 x + 7   and   g ( x )  = x 2 − 1 g ( x )  = x ^ { 2 } - 1 g ( x )  = x 2 − 1   .A)    4 x 3 28 x 2 − 4 x − 7 4 x ^ { 3 } 28 x ^ { 2 } - 4 x - 7 4 x 3 28 x 2 − 4 x − 7 B)    4 x 2 + 3 4 x ^ { 2 } + 3 4 x 2 + 3 C)    16 x 2 + 56 x + 48 16 x ^ { 2 } + 56 x + 48 16 x 2 + 56 x + 48 D)    16 x 2 + 6 16 x ^ { 2 } + 6 16 x 2 + 6 E)    4 x 2 + 6 4 x ^ { 2 } + 6 4 x 2 + 6

SUSHANT BISTA · Accepted Answer

Final Answer : C, Explanation :   To find   ( g ∘ f ) ( x ) (g \circ f)(x) ( g ∘ f ) ( x )  , first apply   f ( x ) f(x) f ( x )   and then apply   g ( x ) g(x) g ( x )   to the result. Substituting   f ( x ) = 4 x + 7 f(x) = 4x + 7 f ( x ) = 4 x + 7   into   g ( x ) = x 2 − 1 g(x) = x^2 - 1 g ( x ) = x 2 − 1  , we get   g ( f ( x ) ) = ( 4 x + 7 ) 2 − 1 g(f(x)) = (4x + 7)^2 - 1 g ( f ( x )) = ( 4 x + 7 ) 2 − 1  . Expanding this gives   16 x 2 + 56 x + 49 − 1 = 16 x 2 + 56 x + 48 16x^2 + 56x + 49 - 1 = 16x^2 + 56x + 48 16 x 2 + 56 x + 49 − 1 = 16 x 2 + 56 x + 48  .

Find $\circ f ) ( x )$ where $f (x) = 4 x + 7$ and $g ( x ) = x ^ { 2 } - 1$ .

A) $4 x ^ { 3 } 28 x ^ { 2 } - 4 x - 7$
B) $4 x ^ { 2 } + 3$
C) $16 x ^ { 2 } + 56 x + 48$
D) $16 x ^ { 2 } + 6$
E) $4 x ^ { 2 } + 6$

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