Introduction to Combinations of Functions
Most of the functions studied in calculus are some combination of only a few families of functions, most of the combinations are arithmetic. We can add two functions, f + g(x), subtract them, f − g (x), multiply them, fg(x), and divide them 
. The domain of f + g ( x ), f − g(x), and fg(x), is the intersection of the domain of f(x) and g(x). In other words, their domain is where the domain of f(x) overlaps the domain of g(x). The domain of is the same, except we need to remove any x that makes g(x) = 0.
Examples
Find f + g(x), f − g(x), fg(x), and 
and their domain.
-
f ( x ) = x 2 − 2 x + 5 and g (x) = 6 x −10
f + g (x) = f(x) + g (x) = (x2 − 2 x + 5) + (6 x − 10) = x 2 + 4 x − 5
f - g (x) = f(x) − g (x) = (x 2 − 2 x + 5) − (6 x − 10) = x 2 − 8x + 15
fg (x) = f(x) g (x) = (x2 − 2 x + 5)(6 x − 10) = 6 x 3 − 10 x2 − 12 x 2 + 20 x + 30 x − 50
= 6 x 3 − 22 x2 + 50 x − 50
The domain of f + g(x), f − g (x), and fg(x) is (−∞, ∞). The domain of is 
(from 6 x − 10 = 0), or 
.
-
f ( x ) = x − 3 and 
The domain for f + g(x), f − g (x), and fg(x) is [−2, ∞) (from x + 2 ≥ 0). The domain for is (−2, ∞) because we need 
.
Function Composition
An important combination of two functions is function composition . This involves evaluating one function at the other. The notation for composing f with g is f o g(x). By definition, f o g(x) = f (g(x)), this means that we substitute g(x) for x in f(x).
Examples
Find f o g(x) and g o f(x).
-
f (x) = x 2 + 1 and g (x) = 3 x + 2
f o g(x) = f(g(x))
= f (3 x + 2) Replace g (x) with 3 x + 2.
= (3 x + 2) 2 + 1 Substitute 3 x + 2 for x in f(x).
= (3x + 2)(3x + 2) + 1 = 9 x2 + 12 x + 5
g o f ( x ) = g (f (x))
= g(x2 + 1) Replace f(x) with x2 + 1.
= 3( x 2 + 1) + 2 Substitute x 2 + 1 for x in g(x).
= 3 x 2 + 3 + 2 = 3 x 2 + 5
At times, we only need to find f o g(x) for a particular value of x. The y-value for g(x) becomes the x -value for f(x).
Example
- Find f o g (−1), f o g (0), and g o f (1) for f ( x ) = 4 x + 3 and g ( x ) = 2−x2.
-
f o g (−1) = f ( g (−1)) Compute g (−1).
= f (1) g (−1) = 2−(−1) 2 = 1
= 4(1) + 3 = 7 Evaluate f ( x ) at x = 1.
f o g (0) = f ( g (0)) Compute g (0).
= f (2) g (0) = 2 − 0 2 = 2
= 4(2) + 3 = 11 Evaluate f(x) at x = 2.
g o f (1) = g ( f (1)) Compute f (1).
= g (7) f (1) =4(1) + 3 = 7
= 2 − 7 2 = −47 Evaluate g ( x ) at x = 7.
We can compose two functions at a single x -value by looking at the graphs of the individual functions. To find f o g(a), we will look at the graph of g(x) to find the point whose x -coordinate is a. The y -coordinate of this point will be g(a). Then we will look at the graph of f(x) to find the point whose x -coordinate is g(a). The y -coordinate of this point will be f (g(a)) = f o g (a).
Example
Refer to Figure 4.1. The solid graph is the graph of f ( x ), and the dashed graph is the graph of g ( x ).
- Find f o g (−1), f f g (3), f o g (5), and g o f (0).
-
f o g (−1) = f ( g (−1)) Look for x = −1 on g ( x ).
= f (−2) (−1, −2) is on the graph of g ( x ), so g (−1) = −2.
= 0 (−2, 0) is on the graph of f ( x ), so f (−2) = 0.
Fig. 4.1 .
f o g (3) = f ( g (3)) Look for x = 3 on g ( x ).
= f (−2) (3, −2) is on the graph of g ( x ), so g (3) = −2.
= 0 (−2, 0) is on the graph of f ( x ), so f (−2) = 0.
f o g (5) = f ( g (5)) Look for x = 5 on g ( x ).
= f (0) (5, 0) is on the graph of g (x), so g (5) = 0.
= −1 (0, −1) is on the graph of f (x), so f (0) = −1.
g o f (0) = g ( f (0)) Look for x = 0 on f (x).
= g (−1) (0, −1) is on the graph of f ( x ), so f (0) = −1.
= −2 (−1, −2) is on the graph of g ( x ), so g (−1)= −2.
Unfortunately, finding the domain for the composition of two functions is not straightforward. The definition for the domain of f o g ( x ) is the set of all real numbers x such that g ( x ) is in the domain of f ( x ). When finding the domain f o g ( x ), begin with the domain with g ( x ). Then remove any x -value whose y -value is not in the domain for f ( x ). For example if 
, the y -values for g ( x ) are x + 3. We need for x + 3 to be nonzero for 
.
Examples
Find the domain for f o g ( x ).
The domain for g ( x ) is x ≥ 3 (from 2 x − 6 ≥ 0). Are there any x -values in [3, ∞) we cannot put into 
We cannot allow 
to be zero, so we cannot allow x = 3. The domain for f o g ( x ) is (3, ∞).
The domain for g ( x ) is x ≠ −1. Are there any x -values we need to remove from x ≠ −1? We need to find any real numbers that are not in the domain for
The denominator of this fraction is 
, so we cannot allow to be zero. A fraction equals zero only when the numerator is zero, so we cannot allow x − 1 to be zero. We must remove x = 1 from the domain of g(x).
The domain of f o g (x) is x ≠ −1, 1, or (−∞, −1) ∪ (−1, 1) ∪ (1, ∞). This function simplifies to 
, which hides the fact that we cannot let x = − 1.
Composing Three or More Functions Together
Any number of functions can be composed together. Functions can even be composed with themselves. When composing three or more functions together, we will work from the right to the left, performing one composition at a time.
Examples
Find f o f ( x ) and f o g o h ( x ).
- f ( x ) = x 3 , g ( x ) = 2 x − 5, and h ( x ) = x 2 + 1.
f o f ( x ) = f ( f ( x )) = f ( x 3 ) = ( x 3 ) 3 = x 9
For f o g o h ( x ), we will begin with g o h ( x ) = g ( h ( x )) = g ( x 2 + 1) = 2 ( x 2 + 1) − 5 = 2 x 2 − 3. Now we need to evaluate f ( x ) at 2 x 2 − 3.
f o g o h ( x ) = f ( g ( h ( x ))
= f (2 x 2 − 3) = (2 x 2 − 3) 3
f o f ( x ) = f ( f ( x )) = f (3 x + 7) = 3(3 x + 7) + 7 = 9 x + 28
f o g o h ( x ) = f o g ( h ( x ))
g ( h ( x )) = g ( x 4 − 5) = |( x 4 − 5) − 2| = | x 4 −7|
f o g ( h ( x )) = f ( g ( h ( x ))) = f (| x 4 − 7|)
= 3| x 4 − 7| + 7
Recognizing Complicated Functions
In order for calculus students to use some formulas, they need to recognize complicated functions as a combination of simpler functions. Sums, differences, products, and quotients are easy to see, but some compositions of functions are less obvious.
Examples
Find functions f ( x ) and g ( x ) so that h ( x ) = f o g ( x ).
-
Although there are many possibilities for f ( x ) and g ( x ), there is usually one pair of functions that is obvious. Usually we want g ( x ) to be the computation that is done first and f ( x ), the computation to be done last. Here, when computing the y -value for h ( x ), we would calculate x + 16. This will be g ( x ). The last calculation will be to take the square root. This will be f ( x ). If we let 
= x + 16, we have f o g ( x ) = f ( g ( x )) = 
.
-
When computing a y -value for h ( x ), we would first find x 2 + 1. This will be g ( x ). This number will be the denominator of a fraction whose numerator is 2. This will be f ( x ), a fraction whose numerator is 2 and whose denominator is x . If 
and g ( x ) = x 2 + 1,
Combinations of Functions Practice Problems
Practice
-
f ( x ) = 3 x 2 + x and g ( x ) = x −4
(a) Find f + g ( x ), f − g ( x ), fg ( x ), 
.
(b) What is the domain for ?
(c) Find f o g ( x ) and g o f ( x ).
(d) What is the domain for f o g ( x )
(e) Find f o g (1) and g o f (0).
(f) Find f o f ( x ).
-
Find f o g ( x ), g o f ( x ), and the domain for f o g ( x ).
-
Refer to the graphs in Figure 4.2. The solid graph is the graph of f ( x ), and the dashed graph is the graph of g ( x ). Find f o g (1), f o g (4), and g o f (-2).
-
Find f o g o h ( x ) for 
, g ( x ) = 4 x + 9, and h ( x ) = 5 x 2 − 1.
-
Find functions f ( x ) and g ( x ) so that h ( x ) = f o g ( x ), where h ( x ) = ( x - 5) 3 + 2.
Fig. 4.2 .
Solutions
-
(a)
(b) The domain is x ≠ 4, (from x − 4 = 0), or (−∞, 4) ∪ (4, ∞). (c)
(c)
f o g ( x ) = f ( g ( x )) = f ( x − 4) = 3( x − 4) 2 + ( x − 4)
= 3( x − 4)( x − 4) + x − 4 = 3 x 2 − 23 x + 44
g o f ( x ) = g ( f ( x )) = g (3 x 2 + x ) = 3 x 2 + x − 4
(d) The domain for g ( x ) is all real numbers. We can let x be any real number for f ( x ), so we do not need to remove anything from the domain of g ( x ). The domain of f o g ( x ) is all real numbers, or (−∞, ∞).
(e)
f o g (1) = f ( g (1))
= f (−3) g (1) = 1 − 4 = − 3
= 24 f (−3) = 3(−3) 2 + (−3) = 24
g o f (0) = g ( f (0))
= g (0) f (0) = 3(0) 2 + 0 = 0
= −4 g (0) = 0−4 = −4
(f)
f o f ( x ) = f ( f ( x )) = f (3 x 2 + x ) = 3(3 x 2 + x ) 2 + (3 x 2 + x )
= 3(3 x 2 + x )(3 x 2 + x ) + 3 x 2 + x = 27 x 4 + 18 x 3 + 6 x 2 + x
-
The domain of g ( x ) is x ≠ 1. Now we need to see if there is anything we need to remove from x ≠ 1. Before simplifying f o g ( x ), we have
The denominator of this fraction cannot be zero, so we must have 
.
The domain is 
.
While it seems that x = − 4 might not be allowed in the domain of f o g ( x ), x = −4 is in the domain.
-
f o g (1) = f ( g (1)) Look for x = 1 on g ( x ).
= f (4) (1, 4) is on the graph of g ( x ), so g (1) = 4.
= 1 (4, 1) is on the graph of f ( x ), so f (4) = 1.
f o g (4) = f ( g (4)) Look for x = 4 on g ( x ).
= f (0) (4, 0) is on the graph of g ( x ), so g (4) = 0.
= −2 (0, −2) is on the graph of f ( x ), so f (0) = −2.
g o f (−2) = g ( f (−2)) Look for x = −2 on the graph of f ( x ).
= g (1) (−2, 1) is on the graph of f ( x ), so f (−2) = 1.
= 4 (1, 4) is on the graph of g ( x ), so g (1) = 4.
-
-
One possibility is g ( x ) = x − 5 and f ( x ) = x 3 + 2.
f o g ( x ) = f ( g ( x )) = f ( x − 5) = ( x − 5) 3 + 2 = h ( x )
Find practice problems and solutions for these concepts at Combinations of Functions and Inverse Functions Practice Test.
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