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2.3 The Composition of Functions

Find the composition of two functions and the domain of the composition.
Decompose a function as a composition of two functions.

The Composition of Functions

In real-world situations, it is not uncommon for the output of a function to depend on some input that is itself an output of another function. For instance, the amount that a person pays as state income tax usually depends on the amount of adjusted gross income on the person’s federal tax return, which, in turn, depends on his or her annual earnings. Such functions are called composite functions.

To see how composite functions work, suppose a chemistry student needs a formula to convert Fahrenheit temperatures to Kelvin units. The formula

$c (t) = \frac{5}{9} (t - 32)$

gives the Celsius temperature $c (t)$ that corresponds to the Fahrenheit temperature t. The formula

$k (c (t)) = c (t) + 273$

gives the Kelvin temperature $k (c (t))$ that corresponds to the Celsius temperature $c (t)$ . Thus, $50 °$ Fahrenheit corresponds to

$c (50) = \frac{5}{9} (50 - 32) = \frac{5}{9} (18) = 10 ° Celsius$

and $10 °$ Celsius corresponds to

$k (c (50)) = k (10) = 10 + 273 = 283 Kelvin units,$

which is usually written 283 K. We see that $50 °$ Fahrenheit is the same as 283 K. This two-step procedure can be used to convert any Fahrenheit temperature to Kelvin units.

A student making numerous conversions might look for a formula that converts directly from Fahrenheit to Kelvin. Such a formula can be found by substitution:

$\begin{array}{rcll} k (c (t)) & = & c (t) + 273 \\ = & \frac{5}{9} (t - 32) + 273 & Substituting \frac{5}{9} (t - 32) for c (t) \\ = & \frac{5}{9} t - \frac{160}{9} + 273 \\ = & \frac{5}{9} t - \frac{160}{9} + \frac{2457}{9} \\ = & \frac{5 t + 2297}{9} . & Simplifying \end{array}$

Since the formula found above expresses the Kelvin temperature as a new function K of the Fahrenheit temperature t, we can write

$K (t) = \frac{5 t + 2297}{9},$

where $K (t)$ is the Kelvin temperature corresponding to the Fahrenheit temperature, t. Here we have $K (t) = k (c (t))$ . The new function K is called the composition of k and c and can be denoted $k \circ c$ (read “k composed with c,” “the composition of k and c,” or “k circle c”).

Example 1

Given that $f (x) = 2 x - 5$ and $g (x) = x^{2} - 3 x + 8$ , find each of the following.

$(f \circ g) (x)$ and $(g \circ f) (x)$
$(f \circ g) (7)$ and $(g \circ f) (7)$
$(g \circ g) (1)$
$(f \circ f) (x)$

Solution

Consider each function separately:

$\begin{array}{l} f (x) = 2 x - 5 & This function multiplies each \\ input by 2 and then subtracts 5. \end{array}$

and

$\begin{array}{l} g (x) = x^{2} - 3 x + 8. & This function squares an input, subtracts three times \\ the input from the result, and then adds 8. \end{array}$

To find $(f \circ g) (x)$ , we substitute $g (x)$ for x in the equation for $f (x)$ :

$\begin{array}{l} (f \circ g) (x) = f (g (x)) & = & f (x^{2} - 3 x + 8) & x^{2} - 3 x + 8 is \\ the input for f . \\ = & 2 (x^{2} - 3 x + 8) - 5 & f multiplies the input by 2 \\ and then subtracts 5. \\ = & 2 x^{2} - 6 x + 16 - 5 \\ = & 2 x^{2} - 6 x + 11. \end{array}$

To find $(g \circ f) (x)$ , we substitute $f (x)$ for x in the equation for $g (x)$ :

$\begin{array}{rcll} (g \circ f) (x) & = & g (f (x)) = g (2 x - 5) & 2 x - 5 is the \\ input for g . \\ = & {(2 x - 5)}^{2} - 3 (2 x - 5) + 8 & g squares the \\ input, subtracts \\ three times the \\ input, and then \\ adds 8 . \\ = & 4 x^{2} - 20 x + 25 - 6 x + 15 + 8 \\ = & 4 x^{2} - 26 x + 48. \end{array}$

Technology Connection

We can check our work in Example 1(b) using a graphing calculator. We enter the following on the equation-editor screen:

$y_{1} = 2 x - 5$

and

$y_{2} = x^{2} - 3 x + 8.$

Then, on the home screen, we find $(f \circ g) (7)$ and $(g \circ f) (7)$ using the function notations $Y 1 (Y 2 (7))$ and $Y 2 (Y (7))$ , respectively.
To find $(f \circ g) (7)$ , we first find $g (7)$ . Then we use $g (7)$ as an input for f:

$\begin{array}{rcl} (f \circ g) (7) & = & f (g (7)) = f (7^{2} - 3 \cdot 7 + 8) \\ = & f (36) = 2 \cdot 36 - 5 \\ = & 72 - 5 = 67. \end{array}$

To find $(g \circ f) (7)$ , we first find $f (7)$ . Then we use $f (7)$ as an input for g:

$\begin{array}{rcl} (g \circ f) (7) & = & g (f (7)) = g (2 \cdot 7 - 5) \\ = & g (9) = 9^{2} - 3 \cdot 9 + 8 \\ = & 81 - 27 + 8 = 62. \end{array}$

We could also find $(f \circ g) (7)$ and $(g \circ f) (7)$ by substituting 7 for x in the equations that we found in part (a):

$\begin{array}{rcl} (f \circ g) (x) & = & 2 x^{2} - 6 x + 11 \\ (f \circ g) (7) & = & 2 \cdot 7^{2} - 6 \cdot 7 + 11 = 67; \\ (g \circ f) (x) & = & 4 x^{2} - 26 x + 48 \\ (g \circ f) (7) & = & 4 \cdot 7^{2} - 26 \cdot 7 + 48 = 62. \end{array}$
$\begin{array}{rcl} (g \circ g) (1) = g (g (1)) & = & g (1^{2} - 3 \cdot 1 + 8) \\ = & g (1 - 3 + 8) = g (6) \\ = & 6^{2} - 3 \cdot 6 + 8 \\ = & 36 - 18 + 8 = 26 \end{array}$
$\begin{array}{rcll} (f \circ f) (x) = f (f (x)) & = & f (2 x - 5) \\ = & 2 (2 x - 5) - 5 \\ = & 4 x - 10 - 5 = 4 x - 15 \end{array}$

Now Try Exercises 1 and 15.

Example 1 illustrates that, as a rule, $(f \circ g) (x) \neq (g \circ f) (x)$ . We can see this graphically, as shown in the graphs at left.

Example 2

Given that $f (x) = \sqrt{x}$ and $g (x) = x - 3$ :

Find $f \circ g$ and $g \circ f$ .
Find the domain of $f \circ g$ and the domain of $g \circ f$ .

Solution

$(f \circ g) (x) = f (g (x)) = f (x - 3) = \sqrt{x - 3}$

$(g \circ f) (x) = g (f (x)) = g (\sqrt{x}) = \sqrt{x} - 3$
Since $f (x)$ is not defined for negative radicands, the domain of $f (x)$ is ${x | x \geq 0}$ , or $[0, \infty)$ . Any real number can be an input for $g (x)$ , so the domain of $g (x)$ is $(- \infty, \infty)$ .

Since the inputs of $f \circ g$ are outputs of g, the domain of $f \circ g$ consists of the values of x in the domain of g, $(- \infty, \infty)$ , for which $g (x)$ is nonnegative. (Recall that the inputs of $f (x)$ must be nonnegative.) Thus we have

$\begin{array}{rcll} g (x) & \geq & 0 \\ x - 3 & \geq & 0 & Substituting x - 3 for g (x) \\ x & \geq & 3. \end{array}$

We see that the domain of $f \circ g$ is ${x | x \geq 3}$ , or $[3, \infty)$ .

We can also find the domain of $f \circ g$ by examining the composite function itself, $(f \circ g) (x) = \sqrt{x - 3}$ . Since any real number can be an input for g, the only restriction on $f \circ g$ is that the radicand must be nonnegative. We have

$\begin{array}{rcl} x - 3 & \geq & 0 \\ x & \geq & 3. \end{array}$

Again, we see that the domain of $f \circ g$ is ${x | x \geq 3}$ , or $[3, \infty)$ . The graph in Fig. 1 confirms this.

Figure 1.

The inputs of $g \circ f$ are outputs of f, so the domain of $g \circ f$ consists of the values of x in the domain of f, $[0, \infty)$ , for which $g (x)$ is defined. Since g can accept any real number as an input, any output from f is acceptable, so the entire domain of f is the domain of $g \circ f$ . That is, the domain of $g \circ f$ is ${x | x \geq 0}$ , or $[0, \infty)$ .

We can also examine the composite function itself to find its domain. First, recall that the domain of f is ${x | x \geq 0}$ , or $[0, \infty)$ . Then consider $(g \circ f) (x) = \sqrt{x} - 3$ . The radicand cannot be negative, so we have $x \geq 0$ . As above, we see that the domain of $g \circ f$ is the domain of f, ${x | x \geq 0}$ , or $[0, \infty)$ . The graph in Fig. 2 confirms this.

Figure 2.

Now Try Exercise 27.

Example 3

Given that $f (x) = \frac{1}{x - 2}$ and $g (x) = \frac{5}{x}$ , find $f \circ g$ and $g \circ f$ and the domain of each.

Solution

We have

$\begin{array}{rcl} (f \circ g) (x) = f (g (x)) & = & f (\frac{5}{x}) = \frac{1}{\frac{5}{x} - 2} = \frac{1}{\frac{5 - 2 x}{x}} = \frac{x}{5 - 2 x}; \\ (g \circ f) (x) = g (f (x)) & = & g (\frac{1}{x - 2}) = \frac{5}{\frac{1}{x - 2}} = 5 (x - 2) . \end{array}$

Values of x that make the denominator 0 are not in the domains of these functions. Since $x - 2 = 0$ when $x = 2$ , the domain of f is ${x | x \neq 2}$ . The denominator of g is x, so the domain of g is ${x | x \neq 0}$ .

The inputs of $f \circ g$ are outputs of g, so the domain of $f \circ g$ consists of the values of x in the domain of g for which $g (x) \neq 2$ . (Recall that 2 cannot be an input of f.) Since the domain of g is ${x | x \neq 0}$ , 0 is not in the domain of $f \circ g$ . In addition, we must find the value(s) of x for which $g (x) = 2$ . We have

$\begin{array}{rcll} g (x) & = & 2 \\ \frac{5}{x} & = & 2 & Substituting \frac{5}{x} for g (x) \\ 5 & = & 2 x \\ \frac{5}{2} & = & x . \end{array}$

This tells us that $\frac{5}{2}$ is also not in the domain of $f \circ g$ . Then the domain of $f \circ g$ is ${x | x \neq 0$ and $x \neq \frac{5}{2}}$ , or $(- \infty, 0) \cup (0, \frac{5}{2}) \cup (\frac{5}{2}, \infty)$ .

We can also examine the composite function $f \circ g$ to find its domain. First, recall that 0 is not in the domain of g, so it cannot be in the domain of $(f \circ g) (x) = x / (5 - 2 x)$ . We must also exclude the value(s) of x for which the denominator of $f \circ g$ is 0. We have

$\begin{array}{rcl} 5 - 2 x & = & 0 \\ 5 & = & 2 x \\ \frac{5}{2} & = & x . \end{array}$

Again, we see that $\frac{5}{2}$ is also not in the domain, so the domain of $f \circ g$ is ${x | x \neq 0 and x \neq \frac{5}{2}}$ , or $(- \infty, 0) \cup (0, \frac{5}{2}) \cup (\frac{5}{2}, \infty)$ .

Since the inputs of $g \circ f$ are outputs of f, the domain of $g \circ f$ consists of the values of x in the domain of f for which $f (x) \neq 0$ . (Recall that 0 cannot be an input of g.) The domain of f is ${x | x \neq 2}$ , so 2 is not in the domain of $g \circ f$ . Next, we determine whether there are values of x for which $f (x) = 0 :$

$\begin{array}{rcll} f (x) & = & 0 \\ \frac{1}{x - 2} & = & 0 & Substituting \frac{1}{x - 2} for f (x) \\ (x - 2) \cdot \frac{1}{x - 2} & = & (x - 2) \cdot 0 & Multiplying by x - 2 \\ 1 & = & 0. & False equation \end{array}$

We see that there are no values of x for which $f (x) = 0$ , so there are no additional restrictions on the domain of $g \circ f$ . Thus the domain of $g \circ f$ is

${x | x \neq 2}, or (- \infty, 2) \cup (2, \infty) .$

We can also examine $g \circ f$ to find its domain. First, recall that 2 is not in the domain of f, so it cannot be in the domain of $(g \circ f) (x) = 5 (x - 2)$ . Since $5 (x - 2)$ is defined for all real numbers, there are no additional restrictions on the domain of $g \circ f$ . The domain is

${x | x \neq 2}, or (- \infty, 2) \cup (2, \infty) .$

Now Try Exercise 23.

Decomposing a Function as a Composition

In calculus, one often needs to recognize how a function can be expressed as the composition of two functions. In this way, we are “decomposing” the function.

Example 4

If $h (x) = {(2 x - 3)}^{5}$ , find $f (x)$ and $g (x)$ such that $h (x) =$ $(f \circ g) (x)$ .

Solution

The function $h (x)$ raises $(2 x - 3)$ to the 5th power. Two functions that can be used for the composition are

$f (x) = x^{5} and g (x) = 2 x - 3.$

We can check by forming the composition:

$h (x) = (f \circ g) (x) = f (g (x)) = f (2 x - 3) = {(2 x - 3)}^{5} .$

This is the most “obvious” solution. There can be other less obvious solutions. For example, if

$f (x) = {(x + 7)}^{5} and g (x) = 2 x - 10,$

then

$\begin{array}{rcl} h (x) & = & (f \circ g) (x) = f (g (x)) \\ = & f (2 x - 10) \\ = & {[2 x - 10 + 7]}^{5} = {(2 x - 3)}^{5} . \end{array}$

Now Try Exercise 39.

Example 5

If $h (x) = \frac{1}{{(x + 3)}^{3}}$ , find $f (x)$ and $g (x)$ such that $h (x) = (f \circ g) (x)$ .

Solution

Two functions that can be used are

$f (x) = \frac{1}{x} and g (x) = {(x + 3)}^{3} .$

We check by forming the composition:

$h (x) = (f \circ g) (x) = f (g (x)) = f ({(x + 3)}^{3}) = \frac{1}{{(x + 3)}^{3}} .$

There are other functions that can be used as well. For example, if

$f (x) = \frac{1}{x^{3}} and g (x) = x + 3,$

then

$h (x) = (f \circ g) (x) = f (g (x)) = f (x + 3) = \frac{1}{{(x + 3)}^{3}} .$

Now Try Exercise 41.

2.3 Exercise Set

Given that $f (x) = 3 x + 1$ , $g (x) = x^{2} - 2 x - 6$ , and $h (x) = x^{3}$ , find each of the following.

1. $(f \circ g) (- 1)$
2. $(g \circ f) (- 2)$
3. $(h \circ f) (1)$
4. $(g \circ h) (\frac{1}{2})$
5. $(g \circ f) (5)$
6. $(f \circ g) (\frac{1}{3})$
7. $(f \circ h) (- 3)$
8. $(h \circ g) (3)$
9. $(g \circ g) (- 2)$
10. $(g \circ g) (3)$
11. $(h \circ h) (2)$
12. $(h \circ h) (- 1)$
13. $(f \circ f) (- 4)$
14. $(f \circ f) (1)$
15. $(h \circ h) (x)$
16. $(f \circ f) (x)$

Find $(f \circ g) (x)$ and $(g \circ f) (x)$ and the domain of each.

17. $f (x) = x + 3, g (x) = x - 3$
18. $f (x) = \frac{4}{5} x, g (x) = \frac{5}{4} x$
19. $f (x) = x + 1, g (x) = 3 x^{2} - 2 x - 1$
20. $f (x) = 3 x - 2, g (x) = x^{2} + 5$
21. $f (x) = x^{2} - 3, g (x) = 4 x - 3$
22. $f (x) = 4 x^{2} - x + 10, g (x) = 2 x - 7$
23. $f (x) = \frac{4}{1 - 5 x}, g (x) = \frac{1}{x}$
24. $f (x) = \frac{6}{x}, g (x) = \frac{1}{2 x + 1}$
25. $f (x) = 3 x - 7, g (x) = \frac{x + 7}{3}$
26. $f (x) = \frac{2}{3} x - \frac{4}{5}, g (x) = 1.5 x + 1.2$
27. $f (x) = 2 x + 1, g (x) = \sqrt{x}$
28. $f (x) = \sqrt{x}, g (x) = 2 - 3 x$
29. $f (x) = 20, g (x) = 0.05$
30. $f (x) = x^{4}, g (x) = \sqrt[4]{x}$
31. $f (x) = \sqrt{x + 5}, g (x) = x^{2} - 5$
32. $f (x) = x^{5} - 2, g (x) = \sqrt[5]{x + 2}$
33. $f (x) = x^{2} + 2, g (x) = \sqrt{3 - x}$
34. $f (x) = 1 - x^{2}, g (x) = \sqrt{x^{2} - 25}$
35. $f (x) = \frac{1 - x}{x}, g (x) = \frac{1}{1 + x}$
36. $f (x) = \frac{1}{x - 2}, g (x) = \frac{x + 2}{x}$
37. $f (x) = x^{3} - 5 x^{2} + 3 x + 7, g (x) = x + 1$
38. $f (x) = x - 1, g (x) = x^{3} + 2 x^{2} - 3 x - 9$

Find $f (x)$ and $g (x)$ such that $h (x) = (f \circ g) (x)$ . Answers may vary.

39. $h (x) = {(4 + 3 x)}^{5}$
40. $h (x) = \sqrt[3]{x^{2} - 8}$
41. $h (x) = \frac{1}{{(x - 2)}^{4}}$
42. $h (x) = \frac{1}{\sqrt{3 x + 7}}$
43. $h (x) = \frac{x^{3} - 1}{x^{3} + 1}$
44. $h (x) = | 9 x^{2} - 4 |$
45. $h (x) = {(\frac{2 + x^{3}}{2 - x^{3}})}^{6}$
46. $h (x) = {(\sqrt{x} - 3)}^{4}$
47. $h (x) = \sqrt{\frac{x - 5}{x + 2}}$
48. $h (x) = \sqrt{1 + \sqrt{1 + x}}$
49. $h (x) = {(x + 2)}^{3} - 5 {(x + 2)}^{2} + 3 (x + 2) - 1$
50. $h (x) = 2 {(x - 1)}^{5 / 3} + 5 {(x - 1)}^{2 / 3}$
51. Ripple Spread. A stone is thrown into a pond, creating a circular ripple that spreads over the pond in such a way that the radius is increasing at a rate of $3 f t / s e c$ .
1. Find a function $r (t)$ for the radius in terms of t.
2. Find a function $A (r)$ for the area of the ripple in terms of the radius r.
3. Find $(A \circ r) (t)$ . Explain the meaning of this function.
52. The surface area S of a right circular cylinder is given by the formula $S = 2 π r h + 2 π r^{2}$ . If the height is twice the radius, find each of the following.
1. A function $S (r)$ for the surface area as a function of r
2. A function $S (h)$ for the surface area as a function of h
53. Blouse Sizes. A blouse that is size x in Japan is size $s (x)$ in the United States, where $s (x) = x - 3$ . A blouse that is size x in the United States is size $t (x)$ in Australia, where $t (x) = x + 4$ . (Source: www.onlineconversion.com) Find a function that will convert blouse sizes in Japan to blouse sizes in Australia.
54. A manufacturer of tools, selling rechargeable drills to a chain of home improvement stores, charges $6 more per drill than its manufacturing cost m. The stores then sell each drill for 150% of the price that it paid the manufacturer. Find a function $P (m)$ for the price at the home improvement stores.

Skill Maintenance

Consider the following linear equations. Without graphing them, answer the questions in Exercises 55–62. [1.3], [1.4]

$y = x$
$y = - 5 x + 4$
$y = \frac{2}{3} x + 1$
$y = - 0.1 x + 6$
$y = 3 x - 5$
$y = - x - 1$
$2 x - 3 y = 6$
$6 x + 3 y = 9$

55. Which, if any, have y-intercept (0, 1)?
56. Which, if any, have the same y-intercept?
57. Which slope down from left to right?
58. Which has the steepest slope?
59. Which pass(es) through the origin?
60. Which, if any, have the same slope?
61. Which, if any, are parallel?
62. Which, if any, are perpendicular?

Synthesis

63. Let $p (a)$ represent the number of pounds of grass seed required to seed a lawn with area a. Let $c (s)$ represent the cost of s pounds of grass seed. Which composition makes sense: $(c \circ p) (a) or (p \circ c) (s)$ ? What does it represent?
64. Write equations of two functions f and g such that $f \circ g = g \circ f = x$ . (In Section 5.1, we will study inverse functions. If $f \circ g = g \circ f = x$ , functions f and g are inverses of each other.)

Mid-Chapter Mixed Review

Determine whether the statement is true or false.

$f (c)$ is a relative maximum if $(c, f (c))$ is the highest point in some open interval containing c. [2.1]
If f and g are functions, then the domain of the functions $f + g, f - g$ , fg, and $f / g$ is the intersection of the domain of f and the domain of g. [2.2]
In general, $(f \circ g) (x) \neq (g \circ f) (x)$ . [2.3]
Determine the intervals on which the function is (a) increasing; (b) decreasing; (c) constant. [2.1]
Using the graph shown below, determine any relative maxima or minima of the function and the intervals on which the function is increasing or decreasing. [2.1]
Determine the domain and the range of the function graphed in Exercise 4. [2.1]
Window Design. Lucas is designing a window for the peak of an A-frame house. The base is 4 ft more than the height h. Express the area of the window as a function of the height. [2.1]
For the function defined as

$f (x) = {\begin{array}{l} x - 5, & for x \leq - 3, \\ 2 x + 3, & for - 3 < x \leq 0, \\ \frac{1}{2} x, & for x > 0, \end{array}$

$f (- 5), f (- 3), f (- 1)$ , and $f (6)$ . [2.1]
Graph the function defined as

$g (x) = {\begin{array}{l} x + 2, & for x < - 4, \\ - x, & for x \geq - 4. \end{array}$

[2.1]

Given that $f (x) = 3 x - 1$ and $g (x) = x^{2} + 4$ , find each of the following, if it exists. [2.2]

$(f + g) (- 1)$
$(f g) (0)$
$(g - f) (3)$
$(g / f) (\frac{1}{3})$

For each pair of functions in Exercises 14. and 15:

Find the domains of $f, g, f + g, f - g, f g, f f, f / g$ , and $g / f$ .
Find $(f + g) (x), (f - g) (x), (f g) (x), (f f) (x), (f / g) (x)$ , and $(g / f) (x)$ . [2.2]

$f (x) = 2 x + 5, g (x) = - x - 4$
$f (x) = x - 1, g (x) = \sqrt{x + 2}$

For each function f, construct and simplify the difference quotient

$\frac{f (x + h) - f (x)}{h} .$

[2.2]

$f (x) = 4 x - 3$
$f (x) = 6 - x^{2}$

Given that $f (x) = 5 x - 4, g (x) = x^{3} + 1$ , and $h (x) = x^{2} - 2 x + 3$ , find each of the following. [2.3]

$(f \circ g) (1)$
$(g \circ h) (2)$
$(f \circ f) (0)$
$(h \circ f) (- 1)$

Find $(f \circ g) (x)$ and $(g \circ f) (x)$ and the domain of each. [2.3]

$f (x) = \frac{1}{2} x, g (x) = 6 x + 4$
$f (x) = 3 x + 2, g (x) = \sqrt{x}$

Collaborative Discussion and Writing

If $g (x) = b$ , where b is a positive constant, describe how the graphs of $y = h (x)$ and $y = (h - g) (x)$ will differ. [2.2]
If the domain of a function f is the set of real numbers and the domain of a function g is also the set of real numbers, under what circumstances do $(f + g) (x)$ and $(f / g) (x)$ have different domains? [2.2]
If f and g are linear functions, what can you say about the domain of $f \circ g$ and the domain of $g \circ f$ ? [2.3]
Nora determines the domain of $f \circ g$ by examining only the formula for $(f \circ g) (x)$ . Is her approach valid? Why or why not? [2.3]

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Table of Contents for 2.3 The Composition of Functions

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Table of Contents for
2.3 The Composition of Functions