9.5 The Algebra of Functions
In this section, we will take a look at all our basic operations as they apply to functions.
Operations on Functions
If f and g are functions and x is and element of the domain of each function then,
( f + g )(x ) = f (x ) + g (x )
( fg )(x ) = f (x )g (x )
(f
⎛f⎞
f (x )
⎜⎜ ⎟⎟( x ) =
, g (x ) ≠ 0
g (x )
⎝g⎠
− g )( x ) = f ( x ) − g ( x )
This just means in order to perform a basic operation on two functions, you simply need to do that
particular operation pointwise.
Example 1:
Let f ( x ) = x − x and
2
a.
(f
− g )(3)
e.
( fg )(x )
g ( x ) = 3x − 2 . Find the following.
b. ( f + g )(2)
c. ( fg )(− 1)
⎛ f ⎞⎛ 2 ⎞
⎛f ⎞
f. ⎜⎜ ⎟⎟(0 )
g. ⎜⎜ ⎟⎟⎜ ⎟
⎝ g ⎠⎝ 3 ⎠
⎝g⎠
d.
(f
+ g )( x )
Solution:
(
)( )
()
()
a. First we recall the rule given above f − g x = f x − g x .
There are two primary ways in which we can approach this problem. One way is to just
replace the x in the rule with 3 and simplify as follows:
(f
− g )(3) = f (3) − g (3)
( )
= 3 2 − 3 − (3(3) − 2)
= 9−3−9+ 2
= −1
The alternate way to do this problem is to just find ( f − g )( x ) in general and then substitute
the value in the end.
(f
− g )( x ) = f ( x ) − g ( x )
= x 2 − x − (3 x − 2 )
= x 2 − 4x + 2
(f
− g )(3) = 3 2 − 4(3) + 2 = −1
We can see they both give us –1. We prefer the latter method since once the expression for
f − g x has been found we can evaluate it at as many values as we would like.
(
)( )
b. Again we will start by finding
(f
+ g )( x ) and then evaluate at 2.
(f
+ g )( x ) = f ( x ) + g ( x )
= x 2 − x + (3 x − 2 )
= x 2 + 2x − 2
(f
c.
+ g )(2 ) = 2 2 + 2(2 ) − 2 = −2
( fg )(x ) = f (x )g (x )
= (x 2 − x )(3 x − 2 )
= 3x 3 − 5x 2 + 2 x
( fg )(− 1) = 3(− 1)3 − 5(− 1)2 + 2(− 1)
= −10
c. Since we have already found ( f + g )( x ) for part b. we simply need to refer back.
( f + g )(x ) = x 2 + 2 x − 2
d. We have also already found
e. We will again start by finding
( fg )(x ) = 3x 3 − 5 x 2 + 2 x
in part c above.
⎛f⎞
⎜⎜ ⎟⎟( x ) , then evaluate at 0.
⎝g⎠
⎛f ⎞
f (x )
⎜⎜ ⎟⎟( x ) =
g (x )
⎝g⎠
x2 − x
3x − 2
⎛f⎞
02 − 0
⎜⎜ ⎟⎟(0 ) =
3(0) − 2
⎝g⎠
=
0
−2
=0
=
f.
In part e. we found
⎛f ⎞
x2 − x
⎜⎜ ⎟⎟( x ) =
. Evaluating at
3x − 2
⎝g⎠
2
3
we get
⎛ f ⎞⎛ 2 ⎞ ( 23 ) − ( 23 )
⎜⎜ ⎟⎟⎜ ⎟ = 2
⎝ g ⎠⎝ 3 ⎠ 3( 3 ) − 2
4
−2
= 9 3
0
2
But we recall that we cannot have a zero in the denominator. Also, from the rules above,
we know that
g (x ) ≠ 0 . Thus, we can see that
Thus, we can say
2
3
⎛ f ⎞⎛ 2 ⎞
⎜⎜ ⎟⎟⎜ ⎟ is not a real number.
⎝ g ⎠⎝ 3 ⎠
is not in the domain of
⎛f⎞
⎜⎜ ⎟⎟( x ) .
⎝g⎠
We now turn our attention to another very important operation on functions, the composition.
Definition: Composition of two functions
Let f and g be two functions such that g ( x ) is in the domain of f for all x in the domain of g .
Then the composition of f and g , written f o g , is the function given by ( f o g )( x ) = f ( g ( x )) .
It usually helps to visualize the composition as
f ( g (x ) ).
The composition idea is one of putting an entire function into another function.
Example 2:
Let
a.
e.
f ( x ) = 2 x 2 − x + 1 , g ( x ) = 2 x − 3 and h( x ) =
( f o g )(0)
(g o h )(x )
b.
f.
1
. Find the following.
x +1
c. (h o f )(2)
d. ( f o g )( x )
(h o g )(2)
( f o f )(x )
Solution:
a. First, by the definition we know
( f o g )(x ) = f (g (x )) .
So we really want
example above, we can do this by finding the formula for
at 0. So, if we insert
( f o g )(x ) first and then evaluate it
g (x ) = 2 x − 3 into f ( g ( x )) and simplify we get
( f o g )(x ) = f (g (x ))
= f (2 x − 3)
2
= 2(2 x − 3) − (2 x − 3) + 1
= 2(4 x 2 − 12 x + 9 ) − 2 x + 3 + 1
= 8 x 2 − 24 x + 18 − 2 x + 4
= 8 x 2 − 24 x + 22
Now we evaluate it for x=0
( f o g )(0) = 8(0)2 − 24(0) + 22
= 22
b. In a similar fashion we start by finding
f ( g (0)) . Like the
(h o g )(x ) , the evaluate as follows
(h o g )(x ) = h(g (x ))
= h(2 x − 3)
1
(2 x − 3) + 1
1
=
2x − 2
(h o g )(2) = 1
2(2) − 2
1
=
4
=
c.
(h o f )(x ) = h( f (x ))
(
)
= h 2x 2 − x + 1
1
2x − x + 1 + 1
1
= 2
2x − x + 2
1
(h o f )(2) =
2
2(2 ) − (2 ) + 2
1
=
8
2
d. In part a. above we already found ( f o g )( x ) = 8 x − 24 x + 22 .
=
e.
(
2
)
(g o h )(x ) = g (h(x ))
⎛ 1 ⎞
= g⎜
⎟
⎝ x + 1⎠
⎛ 1 ⎞
= 2⎜
⎟−3
⎝ x + 1⎠
2
=
−3
x +1
e. Lastly, we want to take the composition of a function with itself. We do this in the same
manner as all other compositions. We simply insert f into f and simplify.
( f o f )(x ) = f ( f (x ))
(
= 2(2 x
= 2(4 x
)
− x + 1) − (2 x
= f 2x 2 − x + 1
2
4
2
2
)
− x +1 +1
)
− 4x + 5x − 2x + 1 − 2x 2 + x − 1 + 1
3
2
= 8 x − 8 x + 10 x − 4 x + 2 − 2 x 2 + x
4
3
2
= 8 x 4 − 8 x 3 + 8 x 2 − 3x + 2
9.5 Exercises
f (x ) = 2 x − 1 and g ( x ) = x 2 − 2 x + 1 . Find the following.
2. ( f − g )(2)
3. ( g − f )( x )
1. ( f + g )(− 1)
⎛g⎞
⎛f ⎞
5. ⎜⎜ ⎟⎟(0 )
6. ⎜⎜ ⎟⎟( x )
7. ( f − g )(a + h )
⎝f⎠
⎝g⎠
⎛f⎞
9. ⎜⎜ ⎟⎟(ab )
10. ( fg )( x )
⎝g⎠
Let
Let g ( x ) =
(g + h )(1)
12.
(g − h)(− 2)
13.
(gh)(x )
15.
(h − g )(x )
16.
(g + h )(x 2 )
17.
(h − g )(a − 1)
19.
⎛h⎞
⎜⎜ ⎟⎟(x )
⎝g⎠
20.
(h − g )(a + b)
Let f ( x ) =
x − 1 , g ( x ) = x 2 + 1 and h( x ) =
22.
(g − g )(x )
2
. Find the following.
x
23. (h − g )( x )
(gh)(x )
26.
( fh)(x )
27.
29. ⎜
⎛g⎞
⎟( x )
⎝h⎠
30. ⎜
f ( x ) = 3x + 1 and
31. ( f o g )(− 1)
35. ( f o g )(− 4)
39. ( f o f )( x )
g (x ) = 2 x 2 − x + 4 . Find the following.
32. ( f o g )(0)
33. ( g o f )(2)
36. ( g o f )(6)
37. ( f o g )( x )
40. ( g o g )( x )
(f
25.
Let
( fg )(− 2)
8.
(f
+ g )(2 + h )
1
x
and h( x ) =
. Find the following.
x +1
x +1
11.
21.
4.
+ g )(x )
( fg )(x )
⎛g⎞
⎟(− 2 )
⎝h⎠
18. ( gh)(0)
14. ⎜
24.
(g − h)(x )
28.
⎛g⎞
⎜⎜ ⎟⎟( x )
⎝f⎠
⎛f ⎞
⎟( x )
⎝h⎠
34.
38.
(g o f )(− 3)
(g o f )(x )
1
2
and h( x ) = . Find the following.
x +1
x
42. (h o g )(− 1)
43. (h o g )(0)
41. ( g o h )(2 )
45. (h o h )(2)
46. ( g o g )(0 )
47. ( g o h )( x )
50. ( g o g )( x )
49. (h o h )( x )
Let g ( x ) =
Let f ( x ) =
51.
55.
59.
63.
67.
44.
48.
x + 1 , g ( x ) = x 2 − 2 and h( x ) = 2 x − 1 . Find the following.
( f o g )(2)
( f o h)(t )
(g o g )(a − b )
(g o f )(x )
( f o g )(x )
52.
56.
60.
64.
68.
(g o f )(0)
(g o f )(c )
(h o g )(a + b)
(h o g )(x )
(h o h)(x )
53.
57.
61.
65.
69.
(h o f )(3)
(g o f )(8 + a )
(g o f )(x + h)
(g o h )(x )
(g o g )(x )
54.
58.
62.
66.
70.
(g o h)(0)
(h o g )(x )
(g o h )(− 1)
( f o g )(∆ )
(h o f )(x )
( f o h)(x )
( f o f )(x )