Answer:
Explanation:
Like terms, functions may be combined by addition, subtraction, multiplication or division.
Example 1. Given f ( x ) = 2x + 1 and g ( x ) = x2
+ 2x – 1 find ( f + g ) ( x ) and
( f + g ) ( 2 )
Solution
Step 1. Find ( f + g ) ( x )
Since ( f + g ) ( x ) = f ( x ) + g ( x ) then;
( f + g ) ( x ) = ( 2x + 1 ) + (x2
+ 2x – 1 )
= 2x + 1 + x2
+ 2x – 1
= x
2
+ 4x
Step 2. Find ( f + g ) ( 2 )
To find the solution for ( f + g ) ( 2 ), evaluate the solution above for 2.
Since ( f + g ) ( x ) = x2
+ 4x then;
( f + g ) ( 2 ) = 22
+ 4(2)
= 4 + 8
= 12
Example 2. Given f ( x ) = 2x – 5 and g ( x ) = 1 – x find ( f – g ) ( x ) and ( f – g ) ( 2 ).
Solution
Step 1. Find ( f – g ) ( x ).
( f – g ) ( x ) = f ( x ) – g ( x )
= ( 2x – 5 ) – ( 1 – x )
= 2x – 5 – 1 + x
= 3x – 6
Step 2. Find ( f – g ) ( 2 ).
( f – g ) ( x ) = 3x – 6
( f – g ) ( 2 ) = 3 (2) – 6
= 6 – 6
= 0
Example 2
Given f ( x ) = x2
+ 1 and g ( x ) = x – 4 , find ( f g ) ( x ) and ( f g ) ( 3 ).
Solution
Step 1. Solve for ( f g ) ( x ).
Since ( f g ) ( x ) = f ( x ) * g ( x ) , then
= (x2
+ 1 ) ( x – 4 )
= x
3
– 4 x2
+ x – 4 .
Step 2. Find ( f g ) ( 3 ).
Since ( f g ) ( x ) = x3
– 4 x2
+ x – 4, then
( f g ) ( 3 ) = (3)3
– 4 (3)2
+ (3) – 4
= 27 – 36 + 3 – 4
= -10
Example 4. Given f ( x ) = x + 1 and g ( x ) = x – 1 , find ( x ) and ( 3 ). f
g
⎛ ⎞ ⎜
⎝ ⎠
f
g
⎛ ⎞ ⎜
⎝ ⎠ ⎟ ⎟
Solution
Step 1. Solve for ( x ). f
g
⎛
⎜
⎝ ⎠
⎞
⎟
Since ( x ) = , then ( )
( )
f x
g x
f
g
⎛
⎜
⎝ ⎠
⎞
⎟
= ; x ≠ 1 1
1
x
x
+
−
Step 2 Find . ( ) 3 f
g
⎛ ⎞ ⎜ ⎟ ⎝ ⎠
Since = , then 1
1
x
x
+
− ( ) f x
g
⎛ ⎞ ⎜ ⎟ ⎝ ⎠
=
3 1
3 1
+
− ( ) 3 f
g
⎛ ⎞ ⎜ ⎟ ⎝ ⎠
=
4
2
= 2
Composition is another operation that may be performed among functions. Simply stated, it is
evaluating one function in terms of another. The format for composition is: (f B g)(x) = f(g(x)).
Example 5. Given f ( x ) = x2 and g ( x ) = x + 1 , find (f B g)(x) and (g B f)(x).
Solution
Step 1. Find (f B g)(x)
Since (f B g)(x) = f( g(x) ), then
= f( x + 1 )
= ( x + 1 )2
Step 2. Find (g B f)(x)
Since (g B f)(x) = g( f(x) ), then
= g ( x2
)
= ( x2
) + 1
Note that (f B g)(x) ≠ (g B f)(x). This means that, unlike multiplication or addition,
composition of functions is not a commutative operation.
The following example will demonstrate how to evaluate a composition for a given value.
Example 6. Find (f B g)(3) and (g B f)(3) if f ( x ) = x + 2 and g ( x ) = 4 – x2
.
Solution
Step 1. Find (f B g)(x) then evaluate for 3.
Since (f B g)(x) = f( g(x) ), then
= f(4 – x2
)
= (4 – x2
) + 2
= 6 – x2
Evaluating for 3
(f B g)(3) = 6 – (3)2
= 6 – 9
= -3