Composition of functions:

The composition of the function f with the function g, denoted f ◦ g, is defined by (f ◦ g)(x) = f(g(x)).

The domain of f ◦ g consists of those in x in the domain of g for which g(x) is in the domain of f.

Ex 1: Find (f ◦ g)(2) and (g ◦ f)(2) for f(x) = x – 2 and g(x) = x2 – 1.

(f ◦ g)(2) = f(g(2)) = f(3) = 1

(g ◦ f)(2) = g(f(2)) = g(0) = -1

Notice that f(g(2)) does not equal g(f(2)).

Ex 2: Let 2

1)(1)(x

xgandxxf

a. Find (f ◦ g)(x) and the domain of f ◦ g.b. Find (g ◦ f)(x) and the domain of g ◦ f.

Solution:

111

))(())((.)22

xxfxgfxgfa

),0()0,(

),1[

g

f

D

D]1,0()0,1[ gfD

11

)1(1

)1())(())((.)2

xxxgxfgxfgb

),1( fgD

Note: examine all steps along the way to your final answer. If there are any restrictions on the domain along the way do not forget to eliminate them from your domain.

Ex 3: Write the function 1)( 2 xxf as the composition of two functions.

Solution: We see that we have two different functions here.•Quadratic•Square root

Start with x, and add parts too it up to the next function

x 12 x )(12 xfx

Now, we have to define a pair of functions that match the above.

1)( 2 xxgLet xxhLet )(

1)())(()(

...2 xxgxghxf

Therefore

Ex 4: Write the function1

2)(

2

xxk as the composition of three functions.

Solution: x 12 x 12 x1

22 x

1)( 2 xxfLet

xxgLet )(

xxhLet

2)(

1

2

))((

2)))((()(

...

2

xxghxfghxk

therefore

Do Now: Write 14

1)(

2

xxxf as the composition of functions, and then sketch

the graph of f.

Solution: Since there is no single way to write functions as the compositionof functions, there is however a natural way.

Lets complete the square so that we can see the pieces easierand it will give us the vertex for when we graph.

3)2(

1)4(1)44(

1)(1)4(

2

2

2

x

xx

xx

3)2(

1)(:

2 x

xfhavenowWe

x 2)2( x 3)2( 2 x 3)2( 2 x3)2(

12 x

21 )2()( xxgLet

3)(2 xxgLet

xxgLet )(3

xxgLet

1)(4 ))))(((()( 234 xggggxf

Sketch graph on board.

Ex 6: Oil is leaking from a tanker into a lake. Suppose the shape of the oil spill is approximately circular, and at any time t minutes after the leak began, the radius of the circle is 3)( tttr

Find the area of the spill at any time t after the leak has begun.

Solution: The area of a circle of a radius r is A(r) = πr2

The area of the spill depends on the radius, and the radiusdepends on the time.

233 )()())(())(( ttttAtrAtrA