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# f(x) =X-2     g(X)=x^2+1

f(x) = X-2     (f+g) 2       ( f g )-4            (fog )2            (fof) 3           (fogof)X

f( X) = X - 2     g(x) = X^2 +1

(f + g)(X) = X^2 +1 +X -2 = X^2 + X -1

( f + g ) ( 2) = 2^2 + 2 - 1 = 5

(f . g)(X)  = ( X- 2) ( X^2 + 1)
=  X^3 - 2 X^2 +X - 2

( f . g )(- 4 )  = (-4 ^ 3  ) - 2 ( -4^2) + (-4) - 2
= - 64  - 32 -4 -2
= - 102

( f . f ) = ( X -2 ) ^2 = X^2 - 4X +4

( f.f ) ( 3 ) = f( f(3)) =f(1) = 1 -2 = -1

( f . g . f ) =f(g(f(x)) =  f ( g( x-2) = f ( ( X-2)^2 +1) = ( X^2 -4x +5 -2) = X^2 -4X +3

Given: f(x) = x - 2,  g(x) = x^2 + 1
Find: (f + g)(2),  (f • g)(-4),  (f o g)(2),  (f o f)(3),  (f o g o f)(x)

(f + g)(2) = f(2) + g(2) = (2 - 2) + (2^2 + 1) = 0 + 5 = 5

(f • g)(-4) = ( f(-4) ) ( g(-4) ) = (-4 - 2) ( (-4)^2 + 1 ) = -6(17) = -102

(f o g)(2) = f( g(2) ) = f( 2^2 + 1 ) = f(5) = 5 - 2 = 3

(f o f)(3) = f( f(3) ) = f( 3 - 2 ) = f(1) = 1 - 2 = -1

(f o g o f)(x) = f( g( f(x) ) ) = f( g( x - 2 ) ) = f( (x - 2)^2 + 1 )
= ( (x - 2)^2 + 1 ) - 2 = (x - 2)^2 - 1 = x^2 - 4x + 4 - 1 = x^2 - 4x + 3

This is an exercise in Functional Notation; learn it well. You will use all these operations in calculus. Composition of functions is used in the very important and key Chain Rule of Derivatives.
Hi Carli.  You can think of these as substitution problems.

f(x) = x-2
g(x)=x^2+1

So for (fog)-4, you will start with your f(x) equation, f(x)=x-2.  However, wherever you see an x in your f(x) equation, you will substitute it with whatever your g(x) equation is equal to:

so since f(x)= (x)-2
(fog) = (x^2+1) -2
and to solve (fog)-4, you will plug -4 into your (fog) equation:

(fog)-4 = ((-4^2)+1))-2 = 15