Composition of functions

1. Pronounce (f∘g)(x)

2. (f∘g)(x) =?

1. The composition of F and G of X

2. f( g(x) )

f( g(5) ) and f( g(x) )

Do you work your way outside to inside, or inside to outside?

(answer is the same for both)

You start with the inside, and work your way to the outside, while substituting along the way.

g(x) = x + 2

f(x) = x - 4

f( g(5) ) vs f( g(x) ) process of what you are doing to substitute at the start

With f( g(5) ), you are getting the equation simplified as the y value which you then plug into f( - - ) and repeat with that new value with x

With f( g(x) ), you are plugging in the whole equation into f( - - ) without simplifying anything

g(x) = x + 2

f(x) = x - 4

f( g(5) ) and f( g(x) )

For the g(5) and the g(x) only, substitute

f(7) and f(x+2)

g(x) = x + 2

f(x) = x - 4

f(7) vs f(x+2) process of what you are doing to substitute 7 and x + 2

For f(7) you plug 7 into the f(x) equation f(7) = (7) - 4

For f(x+2) you plug x+2 into the f(x) equation f(x+2) = (x+2) - 4

g(x) = x + 2

f(x) = x - 4

f( g(5) ) and f( g(x) ) —> f(7) and f(x+2)

For f(7) solve, and for f(x+2) substitute only

f(7) = 3

f(x+2) = (x+2) -4

since (x+2) is not being multiplied in this case, you can take off the parenthesis and get f(x+2) = x+2 - 4 —> f(x+2) = x - 2

g(x) = 2x

f(x) = x - 3

g( f(x) ) = ? How do you plug in the f(x)? and do you keep or take off the parenthesis this time after substituting?

You plug in like g( 2(x - 3) )

You can keep the parenthesis, since the 2 of the 2x was already going to multiply with the values of x

(h∘c)(x) =?

(h (c(x) )