For [tex]f(x)=x^2+12[/tex] and [tex]g(x)=x^2-12[/tex] find the following.
a. [tex](f \circ g)(x)[/tex]
b. [tex](g \circ f)(x)[/tex]
c. [tex](f \circ g)(2)[/tex]
a. What is [tex](f \circ g)(x)[/tex]?
[tex](f \circ g)(x)=[/tex] $\square$ (Simplify your answer.)
Answer (1)
Find ( f c i rc g ) ( x ) by substituting g ( x ) into f ( x ) and simplifying: ( f c i rc g ) ( x ) = ( x 2 − 12 ) 2 + 12 = x 4 − 24 x 2 + 156 .
Find ( g c i rc f ) ( x ) by substituting f ( x ) into g ( x ) and simplifying: ( g c i rc f ) ( x ) = ( x 2 + 12 ) 2 − 12 = x 4 + 24 x 2 + 132 .
Evaluate ( f c i rc g ) ( 2 ) by substituting x = 2 into ( f c i rc g ) ( x ) : ( f c i rc g ) ( 2 ) = 2 4 − 24 ( 2 2 ) + 156 = 76 .
The composite function ( f c i rc g ) ( x ) is x 4 − 24 x 2 + 156 , so the answer is x 4 − 24 x 2 + 156 .
Explanation
Understanding the Problem We are given two functions, f ( x ) = x 2 + 12 and g ( x ) = x 2 − 12 . We need to find the composite functions ( f ∘ g ) ( x ) and ( g ∘ f ) ( x ) , and then evaluate ( f ∘ g ) ( 2 ) .
Finding ( f c i rc g ) ( x ) First, we find ( f ∘ g ) ( x ) , which means f ( g ( x )) . We substitute g ( x ) into f ( x ) : f ( g ( x )) = f ( x 2 − 12 ) = ( x 2 − 12 ) 2 + 12
Simplifying ( f c i rc g ) ( x ) Now, we expand and simplify the expression: ( x 2 − 12 ) 2 + 12 = ( x 4 − 24 x 2 + 144 ) + 12 = x 4 − 24 x 2 + 156
Finding ( g c i rc f ) ( x ) Next, we find ( g ∘ f ) ( x ) , which means g ( f ( x )) . We substitute f ( x ) into g ( x ) :
g ( f ( x )) = g ( x 2 + 12 ) = ( x 2 + 12 ) 2 − 12
Simplifying ( g c i rc f ) ( x ) Now, we expand and simplify the expression: ( x 2 + 12 ) 2 − 12 = ( x 4 + 24 x 2 + 144 ) − 12 = x 4 + 24 x 2 + 132
Evaluating ( f c i rc g ) ( 2 ) Finally, we find ( f ∘ g ) ( 2 ) . We substitute x = 2 into the expression we obtained for ( f ∘ g ) ( x ) :
( f ∘ g ) ( 2 ) = ( 2 ) 4 − 24 ( 2 ) 2 + 156 = 16 − 24 ( 4 ) + 156 = 16 − 96 + 156 = 76
Final Answer Therefore, ( f ∘ g ) ( x ) = x 4 − 24 x 2 + 156 , ( g ∘ f ) ( x ) = x 4 + 24 x 2 + 132 , and ( f ∘ g ) ( 2 ) = 76 .
Examples
Composite functions are useful in many real-world scenarios. For example, consider a store that marks down all items by 10% for a sale, and then applies a 5% discount to all purchases for customers who have a loyalty card. If x is the original price, the sale price can be represented by the function f ( x ) = 0.9 x , and the loyalty discount can be represented by g ( x ) = 0.95 x . Applying both discounts is a composite function, either f ( g ( x )) or g ( f ( x )) , depending on the order in which the discounts are applied. In this case, f ( g ( x )) = 0.9 ( 0.95 x ) = 0.855 x and g ( f ( x )) = 0.95 ( 0.9 x ) = 0.855 x , so the order doesn't matter, and the final price is 85.5% of the original price.
